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Related Rates

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Introduction

In this section we are going to look at an application of implicit differentiation. Most of the applications of derivatives are in the next chapter, however there are a couple of reasons for placing these problems in this chapter as opposed to grouping them with the other applications.

The first reason is that a related rates problem is an application of implicit differentiation, which we just learned in the last section. Practicing implicit differentiation now in the context of some applied problems will help us cement the technique in our memories. The other reason is simply that after doing all these derivatives for their own sake, we should remind ourselves that there really are actual applications to derivatives. Sometimes it's easy to forget that there is a reason for what we're learning.

Let's jump right into some problems and see how they work.

Examples

Air is being pumped into a spherical balloon at a rate of 5 cm3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

The first thing we need to do is to identify what information we've been given and what we want to find. Before we do that, notice that both the volume of the balloon and the radius of the balloon will vary with time and so are really functions of time, $V(t)$ and $r(t).$

We know that air is being pumped into the balloon at a rate of 5 cm3/min. This is the rate at which the volume is increasing. Recall that rates of change are nothing more than derivatives and so we know that $$V'(t) = 5$$

We want to determine the rate at which the radius is changing. Again, rates are derivatives and so it looks like we want to determine $r'(t)$ when $r(t) = \frac{d}{2} = 10$ cm. Note that since we are asked about the rate of change of radius, we want to convert the given diameter value to a radius value.

Now that we've identified what we have been given and what we want to find, we need to relate these two quantities to each other. In this case we can relate the volume and the radius with the formula for the volume of a sphere. $$V(t) = \frac43 \pi[r(t)]^3$$

As in the previous section when we looked at implicit differentiation, we will typically not use the $(t)$ part of functions in formulas, but since this is the first time through one of these problems we will do that to remind ourselves that they are really function of $t.$

Now we don't just want a relationship between the volume and the radius. What we really want is a relationship between their derivatives. We can do this by differentiating both sides of the formula with respect to $t.$ This requires us to use implicit differentiation. Doing this gives $$V'(t) = 4\pi [r(t)]^2 r'(t)$$ Now all we need to do is substitute what we know into the derivative and solve for what we want to find.

$$5=4\pi(10^2)r' \hspace{.15in} \Rightarrow \hspace{.15in} r'=\frac{1}{80\pi}\text{cm/min}$$

We can find the units of the derivative by recalling that $$r'(t) = \frac{dr}{dt}$$ The units of the derivative will be the units of the numerator divided by the units of the denominator.


In this last example, we used a familiar formula (volume of a sphere) to relate the two changing quantities. In the next example, we will use a familiar theorem, the Pythagorean Theorem, to express the relationship we need. This theorem is used in many examples that involve right triangles.

A 15 foot ladder is resting against a wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of $\frac14$ ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

The first thing to do in this problem is to create a sketch that shows us what is happening.

This image shows a right triangle with the right angle at the bottom left of the image. The hypotenuse of the triangle is labeled 15. The horizontal side of the triangle is labeled x, and beneath it is an arrow pointing left with the label dx/dt = -1/4. The vertical side of the triangle is labeled y, and to the left of it is an arrow pointing up with the label dy/dt = ?

An important step in creating a sketch for a related rates problem is to decide which quantities are fixed and which are changing. In this situation, we drew a right triangle and labeled its sides. Two of the sides represent the distance of the bottom of the ladder from the wall and the distance of the top of the ladder from the floor. These quantities are changing, so we labeled them with variables, $x$ and $y.$ Since they change over time, they are really functions of $t$, we have just chosen not to include the $(t)$ for simplicity. The hypotenuse is the ladder itself, and since the length of the ladder is fixed, we labeled this side with a constant.

Next, we need to identify what we know and what we want to find. We know that the rate at which the bottom of the ladder is moving towards the wall. This is $$\frac{dx}{dt} = -\frac14$$ Note as well that the rate is negative since the ladder is moving towards the wall so the distance from the wall, $x$, is decreasing. We always need to be careful with signs in these problems.

We want to find the rate at which the top of the ladder is moving away from the floor. This is $\frac{dy}{dt}.$ Note as well that this quantity should be positive since $y$ is increasing.

As with the first example, we need a relationship between $x$ and $y.$ We can use the Pythagorean Theorem to write this relationship. $$x^2+y^2 = (15)^2 = 225$$ All we need to do at this point is to differentiate both sides with respect to $t$, remembering that $x$ and $y$ are really functions of $t$, and so we will need to do implicit differentiation. Doing this gives an equation that shows the relationship between the derivatives. $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ Next, we want to substitute any known quantities and rates into this equation, then solve for the rate we were asked to find. We were given $\frac{dx}{dt}$ and were asked to find $\frac{dy}{dt}.$ That leaves $x$ and $y$ as quantities that we need to find values for.

Determining $x$ and $y$ is actually fairly simple. We know that initially $x=10$ and the bottom of the ladder is being pushed in towards the wall at a rate of $\frac14$ ft/sec. We are interested in what has happened after 12 seconds. We know that

distance = rate $\times$ time
$\hspace{1in}= \left(\frac14\right)(12) = 3$

So the end of the ladder has been pushed in 3 fee t and so after 12 seconds we must have $x=7.$ Note that we could have computed this in one step as follows: $$x = 10 - \frac14(12) = 7$$ To find $y$ (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the value of $x$ that we just found above: $$y=\sqrt{225 - x^2} = \sqrt{225 - 49} = \sqrt{176}$$ Now all we need to do is substitute $\frac{dx}{dt}$, $x$ and $y$ into the derivative above and solve for $\frac{dy}{dt}.$

$$2(7)\left(-\frac14\right)+2\sqrt{176}\hspace{1pt} \frac{dy}{dt} = 0\Rightarrow \frac{dy}{dt}=\frac{7/4}{\sqrt{176}} = \frac{7}{4\sqrt{176}}\approx 0.1319 \text{ ft/sec}$$

Note that the rate is positive here because the top of the ladder is moving up the wall as the bottom of the ladder is being pushed towards the wall.



Before working another example, we need to make a quick comment about the set up of the previous problem. As we observed already, we labeled the hypotenuse of the triangle as a constant rather than with a letter, like $z.$ We did this because the length of the ladder is fixed.

If we had labeled the hypotenuse $z$, we could still have worked the problem correctly, but it would have taken a little more work. In this case, the equation given by the Pythagorean Theorem and its derivative would have been $$x^2+y^2 = z^2\hspace{1in} 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$$ There is nothing wrong with this, but it does require that we find the values of $z$ and $\frac{dz}{dt}.$ Because $z$ is just the length of the ladder, we have $z=15.$ The problem some students run into then is with finding $\frac{dz}{dt}.$ We have to remember that because the ladder has a fixed length, the rate of change of the length is zero. So $\frac{dz}{dt} = 0.$

What if we had used $z$ to label the hypotenuse instead of 15 ft in example 2?

When we put these values into the derivative, we end up with the same equation that we found in the example. It just took more work to get there. In general, if we label a fixed quantity with a letter, it may be easy to forget that it is a constant and that its derivative is therefore zero. If you don't remember this, then there is an "extra" rate left in the problem. So, in any problem with a quantity that is fixed and that will never change over the course of the problem, it is best to just label it with its value rather than with a letter.

Of course, if we'd had a sliding ladder that was allowed to change its length, then we would have had to label it with a letter. In a problem like that, where all three sides of the right triangle are changing, we would need some additional information in order to actually do the problem. The practice problems in this section have several problems like this. Be sure you try them.

The rate of change for a fixed quantity is 0 (because it's not changing)!

Two people are 50 feet apart. One of them starts walking north at a rate that makes the angle shown in the diagram below change at the rate of $0.01$ radians per minute. At what rate is the distance between the two people changing when $\theta = 0.5$ radians?

This image shows a right triangle with the right angle at the bottom left. The top left vertex is labeled, 'moving person' and there is an arrow to the left, pointing up. The bottom right vertex is labeled, 'stationary person.' The hypotenuse is labeled x, and the angle at the bottom right is labeled theta. The length of the horizontal side of the triangle is labeled 50.

This example is not as tricky as it might at first appear. Let's call the distance between the people at any point in time $x$, as noted on the diagram. We have also labeled the angle between the people $\theta.$ Note that as the moving person walks north, both $x$ and $\theta$ change. Since the given rate is the rate of change of one of the angles in the triangle, we will use a trigonometric identity rather than the Pythagorean Theorem to relate the quantities in the problem. Also, since the two rates we wish to relate are $\frac{dx}{dt}$ and $\frac{d\theta}{dt}$, we want an identity that relates $x$ and $\theta.$ There are two that will work: $$\cos\theta = \frac{50}{x}\hspace{.25in}\text{OR}\hspace{.25in}\sec\theta = \frac{x}{50}$$ Either identity will work, but it turns out the if we use secant, things work out a little more simply. Taking the derivative using implicit differentiation gives $$\sec\theta\tan\theta\hspace{2pt}\frac{d\theta}{dt} = \frac{1}{50}\frac{dx}{dt}$$ Notice that there are no $x$'s in this equation. Since we have been asked to find $\frac{dx}{dt}$ and we were given $\theta$ and $\frac{d\theta}{dt}$, all we have to do is substitute and solve. Before we do that, though, we have one decision to make. We were told that $\theta$ is changing at the rate of $0.01$ rad/min, but we were not told whether this quantity should be positive (i.e. $\theta$ is increasing) or negative (i.e. $\theta$ is decreasing). In fact, based on the diagram, we know that $\theta$ is increasing and so we should enter its rate into the equation as a positive quantity. We have $$(50)(0.01)\sec(0.5)\tan(0.5) = \frac{dx}{dt}\hspace{.15in}\Rightarrow\hspace{.15in} \frac{dx}{dt} \approx 0.311254 \text{ ft/min}$$


So far we have seen three related rates problems. While each one looked different, the process was essentially the same in each. In each problem, we identified what we were given and what we wanted to find. We next wrote down a relationship among all the various quantities and used implicit differentiation to arrive at a relationship among the various rates in the problem. Finally, we substituted the known quantities into the equation to find the value we needed.

So, in a general sense, each problem was worked in pretty much the same manner. The only real difference was in coming up with the relationship among the quantities. This is often the hardest part of the problem. In many problems, the best way to come up with the relationship is to sketch a diagram that shows the situation. This often seems like a silly step, but it can make all the difference in whether we can find the relationship or not.

Let's work another problem that uses some different ideas and shows some of the different kinds of things that can show up in related rates problems.

Related Rates:

1. Identify givens and what we're solving for.

2. Write a formula that describes the relationship. Implicitly differentiate (usually with respect to $t$, time).

3. Substitute in known quantities to solve for unknown.

A tank of water in the shape of a cone is leaking water at a constant rate of 2 cubic feet per hour. The base radius of the tank is 5 feet and the height of the tank is 14 feet.

  1. At what rate is the depth of the water in the tank changing when the depth of the water is 6 feet?

  2. At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 feet?

This image shows a cone with its vertex pointing down. About halfway between the vertex and the top of the cone there is a circle that marks the top of another, smaller cone inside the first one. This smaller cone is shaded blue to represent the water in the tank. The top radius of the cone is labeled,

Even though we have a picture of the tank, we should probably start with a sketch that shows all the various quantities involved here. We will draw the sketch as if we were looking at the tank from directly in front of it (i.e. we'll sketch a cross-section of the tank).

This sketch shows an isosceles triangle with the vertex between the two equal sides at the bottom of the sketch. There is a line about 1/3 of the way down from the top edge of the triangle, that creates a second isosceles triangle inside of the original one. This smaller triangle is shaded blue, indicating the water in the tank. There is a dashed vertical line dividing the triangles in half. The distance from the dashed line to the upper right vertex of the large triangle is labeled 5. The distance from the dashed line to the upper right vertex of the smaller triangle is labeled r. The height of the large triangle is labeled 14 and the height of the smaller triangle is labeled h.

As we can see, the water in the tank actually forms a smaller cone/triangle (depending on which image we are looking at) with the same central angle as the tank itself. The radius of the "water cone" at any time is given by $r$ and the height of the "water cone" at any time is given by $h.$ The volume of water in the tank at any time $t$ is then given by $$V = \frac13\pi r^2 h$$ and we've been given that $V'(t) = -2.$

  1. For this part we need to determine $h'$ when $h = 6.$ Now since $V'$ is the given rate and $h'$ is the desired rate, we need a relationship between $V$ and $h.$ Taking the derivative of the volume formula, we have $$V' = \frac23\pi r r' h + \frac13 \pi r^2 h'$$ This equation certainly relates $h'$ to $V'$, but it also includes both $r$ and $r'$, neither of which is known. Since we really have no way of finding $r'$ (without knowing $h'$), we will find a way to eliminate $r$ from the volume formula, so that it relates only $h$ to $V$ and then take the derivative.

    This is easier than it might look at first. If we go back to our sketch above and look at just the right half of the tank, we see that we have two similar triangles. Recall that two triangles are similar if their angles are identical, which is the case here. When we have two similar triangles, the ratios of any pair of sides are equal. So we have $$\frac{r}{h} = \frac{5}{14}\hspace{.15in}\Rightarrow\hspace{.15in} r=\frac{5}{14}h$$ Substituting this back into the volume formula yields $$V = \frac13\pi r^2 h = \frac13\pi\left(\frac{5}{14} h\right)^2 h = \frac{25}{588}\pi h^3$$ So now we have a volume formula that only involves the volume and the height of the water. Note, though, that this volume formula is only valid for the cone in this problem, so don't be tempted to use it for other cones! If we differentiate this volume formula we have $$V' = \frac{25}{196}\pi h^2 h'$$ At this point all we need to do is substitute what we know and solve for $h'.$

    $$-2=\frac{25}{196}\pi(6^2)h'\hspace{.15in}\Rightarrow\hspace{.15in} h'=-\frac{98}{225\pi}\approx -0.1386$$

    So it looks like the depth of the water is decreasing at a rate of approximately $0.1386$ feet per hour.

  2. Now we are being asked for $r'.$ There are two ways we can approach this, an easier way and a (slightly) more difficult way. The "difficult" way is to redo the work in part (a) only this time use the similar triangles to find $h$ in terms of $r$, $$\frac{h}{r} = \frac{14}{5} \hspace{.15in}\Rightarrow\hspace{.15in} h=\frac{14}{5}r$$ and then write the volume formula in terms of $V$ and $r$ and proceed as before.

    That's not too difficult, but it is more work than we need to do. Recall from the first part that we have $$r=\frac{5}{14}h$$ and so we can find that $$r' = \frac{5}{14}h'$$ So, if we use the relationship between $h$ and $r$ that we found in part (a), we can differentiate it to find a relationship between $h'$ and $r'.$ At this point all we need to do is use the result from part (a) to find $$r' = \frac{5}{14}\left(-\frac{98}{225\pi}\right) = -\frac{7}{45\pi}\approx -0.04951$$ So the radius is decreasing at the rate of approximately $0.04951$ feet per hour.

    This was much quicker than redoing all the work from the first part. Note, though, that we were only able to do this because we were being asked for $r'$ at exactly the same time that we were asked for $h'$ in the first part. If we hadn't been using the same time, then we would have had no choice but to do this the "difficult" way.


In the second part of the previous problem we saw an important idea in dealing with related rates. In order to find the desired rate, all we need is an equation that relates the rate we're looking for to a rate that we already know. Sometimes there are multiple equations that we can use and sometimes one will be easier than another.

If there are multiple equations that describe the situation, try to select the easier one!

Also, this problem showed us that we will often have an equation that contains more variables than the two we need to relate. In these cases, we will need to eliminate one (or more) of the variables. In this problem we eliminated the extra variable using a property of similar triangles. This will not always be how we do this, but many of these problems do use similar triangles, so make sure you are familiar with them.

If we have extra variables, try to eliminate the extras, maybe with a handy substitution.

A trough of water is 8 meters in length, and its ends are in the shape of isosceles triangles whose width is 5 meters and height is 2 meters. If water is being pumped in at a constant rate of 6 m3/sec, at what rate is the height of the water changing when the water has a height of 120 cm? At what rate is the width of the water changing when the water has a height of 120 cm?

This image shows two equal sized isosceles triangles, one above and somewhat to the right of the other. The corresponding vertices of the triangles are connected by lines. Inside this trough, there are two smaller isosceles triangles, each one inside one of the larger triangles. The corresponding sides of the smaller triangles are also connected by lines. This smaller trough is shaded blue to indicate the water in the larger trough. The width of the large trough is labeled 5, the height is labeled 2 and the length is labeled 8.

Note that an isosceles triangle is just a triangle in which two of the sides are the same length. In our case, the sides of the tank have the same length.

Let's add some dimensions for the water to the sketch from above.

This image shows two equal sized isosceles triangles, one above and somewhat to the right of the other. The corresponding vertices of the triangles are connected by lines. Inside this trough, there are two smaller isosceles triangles, each one inside one of the larger triangles. The corresponding sides of the smaller triangles are also connected by lines. This smaller trough is shaded blue to indicate the water in the larger trough. The width of the large trough is labeled 5, the height is labeled 2 and the length is labeled 8. In addition, the width of the water is labeled w and the height of the water is labeled h.

Now, in this problem, we know that $V' = 6$ m3/sec and we want to determine $h'$ when $h = 1.2$ m. Note that because $V'$ is in terms of meters we converted $h$ into meters as well. We need an equation that will relate these two quantities, and the volume of the tank will do that.

Recall that for this kind of shape (called a "prism"), the volume is the area of the end times the depth. The ends of the trough (and therefore of the water in the trough) are isosceles triangles, so their area is $\frac12$ (base)$\times$ (height). For our problem, the volume of the water in the trough is then

\begin{align} V&= \frac12 hw (8)\\ &= 4hw\\ \end{align}

As with the previous example, we have an extra quantity, $w$, which is also changing with time. To eliminate it from the volume equation, we will again make use of two similar triangles. If we look at one end of the trough, we see the triangles that are the end of the trough itself, and the water in the trough. These triangles have the same angles and are therefore similar. This means that the ratios of any pair of sides must be equal. In this case we have $$\frac{w}{5} = \frac{h}{2}\hspace{.15in}\Rightarrow\hspace{.15in} w=\frac52 h$$ Putting this into the volume equation gives us a formula for the volume of the water in this trough (and only for this trough!) that depends only on the height of the water. $$V=4hw = 4h\left(\frac52 h\right) = 10h^2$$ We can now differentiate this to find $$V'=20hh'$$ Finally, all we need to do is substitute what we know and solve for $h'.$ $$6=20(1.2)h'\hspace{.15in}\Rightarrow\hspace{.15in} h'=0.25$$ So the height of the water is rising at a rate of 0.25 m/sec.

To answer the second part of this question, we can use everything we have already set up. Since we need $w'$, we can use the relationship between $h$ and $w$ from the similar triangles to give us a relationship between $w'$ and $h'.$ $$w = \frac52 h\hspace{.15in}\Rightarrow\hspace{.15in} w'=\frac52 h'$$ From our work above, we know the value of $h'$ when $h$ is 120 cm, so all we need to do is substitute this into the equation and we'll have the answer. $$w'=\frac52(0.25) = 0.625$$ Therefore, the width is increasing at the rate of 0.625 m/sec.


A light is on the top of a 12 foot tall pole, and a 5 foot 6 inch tall person is walking away from the pole at a rate of 2 ft/sec.

  1. At what rate is the tip of the shadow moving away from the pole when the person is 25 feet from the pole?

  2. At what rate is the tip of the shadow moving away from the person when the person is 25 feet from the pole?

In this sketch, there is a right triangle with the right angle at the bottom left of the sketch. There is a vertical line inside the triangle, making a smaller right triangle inside of the first triangle. The upper left vertex of the large triangle is labeled, 'lamp.' The upper left vertex of the smaller triangle is labeled, 'person,' and there is an arrow above it, pointing right. The horizontal side of the smaller triangle is labeled, 'shadow.' The height of the large triangle is labeled 12, and the height of the small triangle is labeled 5.5.

Let's start by adding all the relevant quantities to the sketch above.

In this sketch, there is a right triangle with the right angle at the bottom left of the sketch. There is a vertical line inside the triangle, making a smaller right triangle inside of the first triangle. The upper left vertex of the large triangle is labeled, 'lamp.' The upper left vertex of the smaller triangle is labeled, 'person,' and there is an arrow above it, pointing right. Above that is the label 'x-sub-p-prime = 2 ft/sec.' The horizontal side of the smaller triangle is labeled, 'shadow.' The height of the large triangle is labeled 12, and the height of the small triangle is labeled 5.5. The horizontal side of the large triangle is labeled x. The horizontal side of the smaller triangle is labeled x-sub-s. The horizontal distance between the two vertical sides of the triangles is labeled x-sub-p.

Here $x$ is the distance from the tip of the shadow to the pole, $x_p$ is the distance from the person to the pole, and $x_s$ is the length of the shadow. Also note that we converted the person's height to 5.5 feet, since all the other measurements are in feet.

The tip of the shadow is defined by the rays of light that just pass the person's head. As we see in the diagram, this creates a pair of similar triangles, which will be useful to us in a little bit.

  1. In this question, we want to determine $x'$ when $x_p = 25$, given that $x'_p = 2.$ From the diagram, we see that $x = x_p + x_s$, which relates $x$ to $x_p.$ We will need to eliminate the extra variable, $x_s$, though. To do this, we can make use of the similar triangles. They give us $$\frac{5.5}{12} = \frac{x_s}{x}$$ Note that $$\frac{5.5}{12} = \frac{\frac{11}{2}}{12} = \frac{11}{24},$$ so we have $$x_s = \frac{11}{24}x$$ We can then substitute this into the equation above and solve for $x$ as follows: $$x = x_p+x_s = x_p +\frac{11}{24}x \hspace{.15in}\Rightarrow\hspace{.15in} x = \frac{24}{13}x_p$$ Now all we need to do is differentiate this, substitute in values and solve for $x'.$ $$x' = \frac{24}{13}x'_p \hspace{.15in}\Rightarrow\hspace{.15in} x' = \frac{24}{13}(2)\approx 3.6923$$ The tip of the shadow is then moving away from the pole at an approximate rate of 3.6923 ft/sec. Notice as well that we never had to use the fact that $x_p = 25$ for this problem. This will happen on rare occasions.

  2. This part is actually pretty straightforward if we have the answer from (a) in hand, which we do, of course. In this case, we know that $x_s$ represents the length of the shadow, or the distance from the tip of the shadow to the person. It looks like we want to determine $x'_s$ when $x_p = 25.$

    Again, we can use $x = x_p + x_s.$ Unlike in the first part, though, we now know that $x'_p = 2$ and $x'\approx 3.6923$ ft/sec. So all we need to do is differentiate the equation and substitute in the known values.

    \begin{align} x' &= x'_p + x'_s\\ 3.6923 &= 2 + x'_s \hspace{.15in}\Rightarrow\hspace{.15in} x'_s \approx 1.6923\\ \end{align}

    The tip of the shadow is moving away from the person at an approximate rate of 1.6923 ft/sec.


A spot light is on the ground 20 feet away from a wall, and a 6 foot tall person is walking towards the wall at a rate of 2.5 ft/sec. How fast is the height of the person's shadow changing when the person is 8 feet from the wall? Is the shadow increasing or decreasing in height at this time?

In this sketch, there is a right triangle with the right angle at the bottom left of the sketch. There is a vertical line inside the triangle, making a smaller right triangle inside of the first triangle. The upper left vertex of the large triangle is labeled, 'wall.' The upper left vertex of the smaller triangle is labeled, 'person,' and there is an arrow above it, pointing left. The vertical side of the larger triangle is labeled, 'shadow.' The shared vertex of both triangles, at the bottom right, is labeled, 'spotlight.' The height of the small triangle is labeled 6 and the horizontal side of the large triangle is labeled 20.

We first add all the relevant quantities to the sketch, as shown below. As with the previous problem, since the rays of light passing just over the head of the person define the person's shadow, we again have a pair of similar triangles.

In this sketch, there is a right triangle with the right angle at the bottom left of the sketch. There is a vertical line inside the triangle, making a smaller right triangle inside of the first triangle. The upper left vertex of the large triangle is labeled, 'wall.' The upper left vertex of the smaller triangle is labeled, 'person,' and there is an arrow above it, pointing left. Above that is the label 'x-prime = 2.5 ft/sec.' The vertical side of the larger triangle is labeled, 'shadow.' The height of the large triangle is labeled y, and the height of the small triangle is labeled 6. The horizontal side of the large triangle is labeled 20. The horizontal side of the smaller triangle is labeled x. The horizontal distance between the two vertical sides of the triangles is labeled 20 minus x.

In this problem, we want to determine $y'$ when the person is 8 feet from the wall. Using the distances labeled on the sketch, this is equivalent to $x = 12$ feet. Also, if the person is moving towards the wall at 2.5 ft/sec, then the person must be moving away from the spotlight at 2.5 ft/sec. So we also know that $x'=2.5.$

In all the previous problems in which we made use of similar triangles, we did so to eliminate a variable from the main equation we were working with. In this case, however, we can create the equation that relates $x$ and $y$ directly from the similar triangles. $$\frac{y}{6} = \frac{20}{x} \hspace{.15in}\Rightarrow\hspace{.15in} y = \frac{120}{x}$$ Now all we need to do is differentiate and substitute values to solve for $y'.$ $$y'=-\frac{120}{x^2} x' \hspace{.15in}\Rightarrow\hspace{.15in}y' = -\frac{120}{12^2}(2.5) \approx -2.0833$$

The height of the shadow is then decreasing at a rate of approximately 2.0833 ft/sec.


We've worked quite a few problems that involved similar triangles, and you can see how they might arise in other problems. Be sure you can do these kinds of problems.

The next example is similar to the problems we've done, but is sufficiently different that it can cause problems until you've seen how to do it.

Two people on bikes are separated by 350 meters. Person A starts riding north at a rate of 5 m/sec and 7 minutes later, Person B starts riding south at 3 m/sec. At what rate is the distance separating the two people changing 25 minutes after Person A starts riding?

This sketch shows two dots. The first dot is labeled, 'Person A Initial position' and there is an arrow next to the label and pointing up. The second dot is to the right of the first. It is labeled, 'Person B Initial position' and there is an arrow next to the label and pointing down. The horizontal distance between the dots is labeled 350 meters.

There is a lot to digest in the statement of this problem. Let's start with a sketch of the situation that shows each person's location sometime after both people start riding.

In this sketch there is a diagonal line from the top left to the bottom right. At a point about 2/3 of the way down the diagram there is a horizontal line that extends from a point directly beneath the top of the diagonal line to the right of the sketch, to a point directly above the bottom end of the diagonal line. There is a vertical line at the left of the diagram, connecting the upper endpoint of the diagonal down to the horizontal and forming a right triangle. There is another vertical line at the right of the diagram, connecting the bottom endpoint of the diagonal up to the horizontal line. This forms a second right triangle. The hypotenuses of the right triangles lie on the diagonal line. The upper endpoint of the diagonal line is labeled, 'Person A' and there is an arrow next to the label, pointing up. Next to that arrow there is the text 'x-prime = 5 meters per second.' The lower endpoint of the diagonal is labeled, 'Person B' and there is an arrow next to the label, pointing down. Next to that arrow is the text 'y-prime = 3 meters per second.' The whole diagonal is labeled 'z', the height of the upper left right triangle is labeled 'x,' and the height of the right triangle in the lower right is labeled 'y'. The vertical distance between the endpoints of the diagonal is labeled, 'x + y.' The horizontal distance between the endpoints of the diagonal is labeled, '350 meters.

Now, we know $x' = 5$ and $y'=3$ and we are asked to find $z'$ after Person A has been riding for 25 minutes and Person B has been riding for 25 - 7 = 18 minutes. After converting these times to seconds (because all the given rates are in meters per second), this means that at the time we're interested in, each of the bike riders has ridden $$x = 5(25\times 60) = 7500 \text{ meters} \hspace{.2in} y= 3(18\times 60) = 3240 \text{ meters}$$ Next, the Pythagorean Theorem tells us that $$z^2 = (x+y)^2 + 350^2$$ Therefore, 25 minutes after Person A starts riding, the two bike riders are $$z = \sqrt{(x+y)^2 + 350^2} = \sqrt{(7500 + 3240)^2 + 350^2} \approx 10,745.7015$$ meters apart.

To determine the rate at which the two riders are moving apart all we need to do is differentiate the equation above and substitute all the quantities that we know to find $z'.$ \begin{align} 2zz' &= 2(x+y)(x'+y')\\ 2(10745.7015) z' &= 2(7500 + 3240)(5+3)\\ z' &= 7.9958\\ \end{align}

So, the two riders are moving apart at a rate of approximately 7.9958 m/sec.


Every problem we have worked to this point has come down to needing a geometric formula. Let's do one that is not geometric in nature.

Suppose that we have two resistors connected in parallel with resistances $R_1$ and $R_2$, measured in ohms (denoted $\Omega$). The total resistance, $R$, is then given by $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$$ Suppose that $R_1$ is increasing at a rate of $0.4\hspace{2pt} \Omega$/min and $R_2$ is decreasing at a rate of $0.7 \hspace{2pt}\Omega$/min. At what rate is $R$ changing when $R_1 = 80\hspace{2pt} \Omega$ and $R_2 = 105\hspace{2pt} \Omega$?

We are asked to find $R'$, given that $R'_1 = 0.4$ and $R'_2 = -0.7.$ As always, we need to be careful with the signs when we write rates as values of a derivative.

We will use the given formula as the relation among $R$, $R_1$ and $R_2.$ After we take the derivative, we will need to substitute for $R$, so let's go ahead and find its value at the time we're interested in.

$$\frac{1}{R} = \frac{1}{80}+\frac{1}{105} = \frac{37}{1680}\hspace{.15in}\Rightarrow\hspace{.15in} R = \frac{1680}{37}\approx 45.4054\hspace{2pt}\Omega$$

Now let's differentiate the equation given in the problem statement.

\begin{align} -\frac{1}{R^2}R' &= -\frac{1}{(R_1)^2}R'_1 - \frac{1}{(R_2)^2}R'_2\\ R' &= R^2\left(\frac{1}{(R_1)^2} R'_1 + \frac{1}{(R_2)^2} R'_2\right)\\ \end{align}

Finally, we just need to substitute into this and do some calculations. $$R' \approx (45.4054)^2\left(\frac{1}{80^2}(0.4) + \frac{1}{105^2}(-0.7)\right)\approx -0.002045$$ So $R$ is decreasing at a rate of approximately $0.002045\hspace{2pt}\Omega$/min.


Summary

We've seen quite a few related rates problems that cover a wide variety of possible problems. There are still many more different kinds of related rates problems out there, but they all have the following structure.

  • There are two or more changing quantities that are related in some way.

  • The rates at which some of the quantities are changing are known and at least one of the rates is unknown. The goal is to find one of the unknown rates at a particular point in time.

  • To find the unknown rate, we first find an equation that relates the changing quantities to each other. If there is more than one quantity in the equation whose rate is unknown, we use a secondary equation (like one given by similar triangles) to eliminate it from the equation.

  • We differentiate the equation using implicit differentiation, substitute all the known values and solve for the unknown rate.



Practice Problems

  1. In the following, assume that $x$ and $y$ are both functions of $t.$ Given $x = -2, y = 1$ and $x' = -4,$ determine $y'$ for the following equation: $$6y^2 + x^2 = 2-x^3 \text{e}^{4-4y}$$

    We are being asked to find $y'$ and are given $x'$ and the relationship between $x$ and $y.$ So the first thing we need to do is use implicit differentiation to differentiate the equation with respect to $t.$ $$12yy'+2xx' = -3x^2x'\text{e}^{4-4y}+4x^3\text{e}^{4-4y}y'$$ Then we substitute the given information and solve for $y'$: $$12y'+16=48-32y'\hspace{.15in}\Rightarrow\hspace{.15in} y' = \frac{8}{11}$$

  2. In the following, assume that $x, y$ and $z$ are all functions of $t.$ Given $x=4$, $y=-2,$ $z = 1$, $x' = 9$ and $y'=-3,$ determine $z'$ for the following equation: $$x(1-y) + 5z^3 = y^2z^2 + x^2 - 3$$

    Since we were given the relationship among all the variables, we just need to differentiate with respect to $t$, using implicit differentiation. $$x'(1-y)-xy'+15z^2z' = 2yy'z^2 + 2y^2zz'+2xx'$$ Then we substitute the given information and solve for $z'$: $$27+12+15z'=12+8z'+72 \hspace{.15in}\Rightarrow\hspace{.15in} z'=\frac{45}{7}$$

  3. For a certain rectangle, the length of one side is always three times the length of the other side.

    1. If the shorter side is decreasing at a rate of 2 inches/minute, at what rate is the longer side decreasing?

    2. At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

    1. Let's call the shorter side $x$ and the longer side $y.$ We are given that $x' = -2$, and have been asked to find $y'.$

      We need an equation that relates these two quantities. From the problem statement, we know that the length of the longer side is three times that of the shorter side, so an equation that relates $x$ and $y$ is $$y = 3x$$ Differentiating this with respect to $t$ yields $$y'=3x'$$ Finally, we substitute the known quantity and solve for what we want: $y'=-6.$ So the longer side is decreasing at the rate of 6 in/min.

    2. Again, we will call the shorter side $x$ and the longer side $y.$ Since we are asked to find the rate of change of the area, we will use the area equation $$A = xy$$

      Since we are finding the rate at the same time as we did in part (a), we have two paths we can take here. We can either take the derivative now or we can first substitute $y = 3x$ to eliminate $y$ from the area equation.

      If we do this, we find $$A(x) = 3x^2$$ Then differentiating gives us $$A' = 6xx'$$ If we use the equation in terms of both $x$ and $y$, then we have $$A' = xy'+x'y$$ Now all we need to do is substitute the known quantities and solve for $A'.$ Using the equation in terms of only $x$ is simplest because we already have all the needed quantities from the problem statement itself. Doing this gives $$A' = (6)(6)(-2) = -72$$ If we wish to use the equation in terms of both $x$ and $y$, we need to find $y$ and $y'.$ From part (a), we know that $y' = -6.$ Also, since $x=6$ and $y=3x$, we find that $y=18.$ Substituting these into the equation $A' = xy'+x'y$ gives us the same answer: $$A' = (6)(-6)+(-2)(18) = -72$$ Therefore, the area is decreasing at the rate of 72 in2/min.

  4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec, at what rate is the radius decreasing when the area of the sheet is 12 m2?

    Calling the area of the ice $A$ and the radius $r$, we relate them with the area equation $$A = \pi r^2$$ We are given that $A'=-0.5$ and have been asked to find $r'$ when $A=12.$

    When we differentiate the area equation with respect to $t$, we find $$A'=2\pi r r'$$ Since this equation includes $r$, we need to find the value of the radius at the given time. At this time, the area is 12 m2, so we can solve for $r$ as follows: $$12 = \pi r^2 \hspace{.15in}\Rightarrow\hspace{.15in} r=\sqrt{\frac{12}{\pi}}$$ Now we can substitute and solve for $r'$: $$-0.5 = 2\pi\sqrt{\frac{12}{\pi}} r' \hspace{.15in}\Rightarrow\hspace{.15in} r'=-\frac14\sqrt{\frac{1}{12\pi}}\approx -0.040717$$ Therefore, the radius is decreasing at the rate of approximately 0.040717 m/sec.

  5. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec.

    1. At what rate is the distance between the person and the rocket increasing 20 seconds after liftoff?
    2. At what rate is the distance between the person and the rocket increasing 1 minute after liftoff?

    It is a good idea to create a sketch for this problem, showing the relative positions of the person and the rocket. We'll let $y$ be the height of the rocket, and $z$ the distance between the person and the rocket.

    This sketch is a diagram showing a right triangle. There are dots at the two ends of the hypotenuse. The dot at the top right of the diagram is labeled,

    In both parts of the question, we know that $y'=15$ and we want to determine $z'.$ So we need an equation that relates $y$ and $z.$ From the diagram, we see that the Pythagorean Theorem applies here, giving us $$z^2 = y^2 + 350^2$$ Let's differentiate this with respect to $t$ and solve for $z'.$ $$2zz'=2yy' \hspace{.15in}\Rightarrow\hspace{.15in} z'=\frac{yy'}{z}$$ We have now reached a point where the process will differ for each part.

    1. To answer the first question, we need to determine $y$ (from the speed of the rocket and given time) and $z$ (from the Pythagorean Theorem).

      \begin{align} y &= (15)(20) = 300\\ z &= \sqrt{300^2+350^2} = \sqrt{212500} = 50\sqrt{85}\approx 460.9772\\ \end{align}

      Then we find $$z'\approx\frac{(300)(15)}{460.9772} = 9.76187$$

      The distance is increasing at approximately 9.76187 ft/sec.

    2. This part is nearly identical to the first part, except that the time is now 60 seconds after liftoff. Note that since the given rate is in feet per second, we must convert the time to seconds.

      As we did in the first part, we first need to solve for $y$ and $z$ before we can substitute into the derivative.

      \begin{align} y &= (15)(60) = 900\\ z &= \sqrt{900^2+350^2} = \sqrt{932500} = 50\sqrt{373}\approx 965.6604\\ \end{align}

      We now find $$z' \approx \frac{(900)(15)}{965.6604} = 13.98007$$ Therefore, the distance is increasing at approximately 13.98007 ft/sec.

  6. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s, and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station?

    This diagram shows two dots connected by a diagonal line. One dot is at the top right and is labeled, 'Plane' with an arrow above it and pointing to the left. The other dot is at the bottom left and is labeled, 'Radar Station.' The line between the dots is labeled, 'Find the rate this distance is changing.' The vertical distance between the two dots is labeled 750 m and the horizontal distance between them is labeled 2.5 km.

    1. Let $x$ stand for the horizontal distance between the plane and the radar station and let $z$ stand for the straight-line distance between the plane and the radar station. Notice that, since the plane is approaching the radar station, the horizontal distance is decreasing. So $x'$ must be negative. We are given that $x' = -100$ m/s when $x = 2500$ m. Also notice that we converted $x$ to meters, since all the other quantities are in meters.

      Here is the sketch for this part.

      This diagram shows two dots connected by a diagonal line. One dot is at the top right and is labeled, 'Plane' with an arrow above it and pointing to the left. The other dot is at the bottom left and is labeled, 'Radar Station.' The line between the dots is labeled with the variable z. The vertical distance between the two dots is labeled 750 m and the horizontal distance between them is labeled x.

      We want to determine $z'$ in this part, so we will use the Pythagorean Theorem to relate $x$ and $z.$ This gives $$z^2=x^2+750^2 = x^2 + 562,500$$ We differentiate with respect to $t$ using implicit differentiation then solve for $z'$ to find: $$2zz'=2xx' \Rightarrow z'=\frac{xx'}{z}$$ We know the value of $x'$ and $x$ at the moment we are interested in, but we still need to find the value of $z.$ We use the Pythagorean Theorem to do this: $$z=\sqrt{x^2 + 750^2} = \sqrt{2500^2 +750^2} = \sqrt{6812500}=250\sqrt{109}\approx 2,610.0766$$ Now we can find that $$z' \approx \frac{(2500)(-100)}{2610.0766}\approx -95.7826$$ Therefore, the distance between the plane and the radar station is decreasing at approximately 95.7826 meters per second.

    2. Again, we'll let $x$ be the horizontal distance between the plane and the radar station, and $z$ the straight-line distance between them. The sketch for this part is

      This diagram shows two dots connected by a diagonal line. One dot is at the top left and is labeled, 'Plane' with an arrow above it and pointing to the left. The other dot is at the bottom right and is labeled, 'Radar Station.' The line between the dots is labeled with the variable z. The vertical distance between the two dots is labeled 750 m and the horizontal distance between them is labeled x.

      Notice in this part that since the plane is moving away from the radar station, the horizontal distance between them is increasing and $x'$ is positive.

      We wish to determine $z'.$ We can use the Pythagorean Theorem to do this, just as we did above. In fact, the equations are the same: \begin{align} z^2=x^2+750^2 &= x^2 + 562,500\\ 2zz'=2xx'\hspace{.25in} &\Rightarrow \hspace{.25in}z'=\frac{xx'}{z}\\ \end{align} This time, we are not given $x$, so we must find both $x$ and $z.$ We know that the speed of the plane is 100 m/s and we have the fact that it has flown for 30 seconds after passing over the radar station. So $$x=(100)(30)=3000$$ To find $z$, we will use the Pythagorean Theorem again. $$z=\sqrt{3000^2+750^2} = \sqrt{9562500}=750\sqrt{17}\approx 3092.3292$$ Now we can find that $$z'\approx\frac{(3000)(100)}{3092.3292}\approx 97.0143$$ Therefore, the distance between the plane and the radar station is increasing at approximately 97.0143 meters per second.

  7. Two people are at an elevator. At the same time, one person starts to walk away from the elevator at a rate of 2 ft/sec, and the other person starts going up in the elevator at a rate of 7 ft/sec. At what rate is the distance between the two people changing 15 seconds later?

    This diagram shows two dots connected by a diagonal line. One dot is at the top right and is labeled, 'Elevator' with an arrow to the right and pointing up. The other dot is at the bottom left and is labeled, 'Person' with an arrow below it and pointing left. The line between the dots is labeled, 'Find the rate this distance is changing.' There is a dot below the dot labeled, 'Elevator' and to the right of the dot labeled, 'Person.' It is labeled, 'Initial Position of Person and Elevator.'

    Let $x$ be the horizontal distance between the person walking and the initial position of the elevator, let $y$ be the height of the elevator above ground level, and let $z$ be the distance between the two people. Since both the person and the elevator are moving, all three of these distances are changing.

    Here is the sketch:

    This diagram shows a right triangle with the right angle at the lower right. The top right vertex is labeled, 'Elevator' with an arrow to the right and pointing up. The bottom left vertex is labeled, 'Person on ground' with an arrow below it and pointing left. The hypotenuse is labeled with the variable z, the bottom leg of the triangle is labeled x and the other leg is labeled y.

    We want to determine $z'$ after 15 seconds, given that $x'=2, y'=7$ and assuming that the person and the elevator start at the same time.

    Using the Pythagorean Theorem and taking the derivative with respect to $t$ gives us \begin{align} z^2 &= x^2 + y^2\\ 2zz'=2xx'+2yy'\hspace{.25in} &\Rightarrow\hspace{.25 in} z'=\frac{xx'+yy'}{z}\\ \end{align} Before we can find $z'$ we need to find the values of $x, y$ and $z.$ We know the speed of the person and the elevator, and we have the fact that 15 seconds have passed. We also have the Pythagorean Theorem, which will help us find $z.$ $$x=(2)(15)=30 \hspace{.5in} y=(7)(15)=105$$ $$z=\sqrt{30^2+105^2}=\sqrt{11925}=15\sqrt{53}\approx 109.2016$$ Substituting, we find $z'$: $$z'\approx\frac{(30)(2)+(105)(7)}{109.2016}\approx 7.2801$$ Therefore, the distance between the two people is increasing at the approximate rate of 7.2801 feet per second.

  8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a rate of 8 m/sec. Five seconds after the first biker started riding north, the second starts to ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20 seconds after the second person started riding?

    This diagram shows two dots connected by a diagonal line. One dot is at the top left and is labeled, ''North' Biker' with an arrow to the left and pointing up. The other dot is at the bottom right and is labeled, ''East' Biker' with an arrow above it and pointing right. The line between the dots is labeled, 'Find the rate this distance is changing.' There is a dot below the dot labeled, ''North'' and to the left of the dot labeled, ''East'.' It is labeled, 'Initial Position of Both People.'

    Let $x$ be the distance the person riding east has traveled,let $y$ be the distance the person riding north has traveled, and let $z$ be the distance between the two people. Since both people are moving, all three of these distances are changing. Further, since the two people are moving away from each other as well as away from their starting point, all three rates of change are positive.

    Here is the sketch:

    This diagram shows a right triangle with the right angle at the lower left. The top left vertex is labeled, ''North' Biker' with an arrow to the right and pointing up. The bottom right vertex is labeled, ''East' Biker' with an arrow above it and pointing right. The hypotenuse is labeled with the variable z, the bottom leg of the triangle is labeled x and the other leg is labeled y.

    We want to determine $z'$ 20 seconds after the second biker starts riding east, given that $x' = 5$ m/sec, $y'=8$ m/sec, and assuming that they start at the same point.

    It should be clear from the sketch that we need the Pythagorean Theorem. As in the previous problems, we will differentiate with respect to $t$, then solve for $z'.$

    \begin{align} z^2 &= x^2 + y^2\\ 2zz'=2xx'+2yy' \hspace{.25in} &\Rightarrow \hspace{.25in} z'=\frac{xx'+yy'}{z}\\ \end{align}

    To find the value of $z'$, we need to find the values of $x$, $y$, and $z.$ We can use the given rates and the times to find $x$ and $y$, and then use the Pythagorean Theorem to solve for $z.$ Note that the biker riding east has been riding for 20 seconds, while the biker riding north has been riding for 25 seconds because that person started 5 seconds earlier.

    $$x=(5)(20)=100 \hspace{.5in} y = (8)(25) = 200$$ $$z=\sqrt{100^2+200^2} = \sqrt{50000}=100\sqrt{5} \approx 223.5=6068$$

  9. A light is mounted on a wall 5 meters above the ground. A two meter tall person is initially 10 meters from the wall, and is moving towards the wall at a rate of 0.5 m/sec. After 4 seconds, is the tip of the person's shadow moving (a) towards or away from the person, and (b) towards or away from the wall?

    In this image, there is a right triangle whose hypotenuse goes from the upper left to the lower right. At the upper left vertex of the triangle, there is a dot labeled 'Lamp'. Partway down the hypotenuse, there is a dot labeled 'Person' and an arrow pointing to the left. A vertical line connecting this dot to the bottom of the triangle forms a second right triangle inside of the first one. The bottom edge of this smaller triangle is labeled 'Shadow'. The sketch also shows that the distance from the lamp to the ground is 5 meters and the height of the person is 2 meters.

    The quantities that are changing in this situation are the distance between the person and the wall, the length of the person's shadow, and the distance between the tip of the shadow and the wall. We will name these $x_p, x_s$ and $x$, respectively. Note that since the person is walking towards the wall, $x_p$ is decreasing. Hence $x'_p = -0.5$ m/sec for both parts.

    Here is a sketch of the situation that will work for both parts.

    In this image, there is a right triangle whose hypotenuse goes from the upper left to the lower right. At the upper left vertex of the triangle, there is a dot labeled 'Lamp'. Partway down the hypotenuse, there is a dot labeled 'Person' and an arrow pointing to the left. A vertical line connecting this dot to the bottom of the triangle forms a second right triangle inside of the first one. The bottom edge of this smaller triangle is labeled 'Shadow'. The sketch also shows that the distance from the lamp to the ground is 5 meters and the height of the person is 2 meters. The horizontal distance between the vertical sides of the two triangles is labeled x-sub-p, the length of the horizontal side of the small triangle is labeled x-sub-s, and the length of the horizontal side of the large triangle is labeled x.

    1. The tip of the shadow is moving towards the person if $x'_s$ is negative and is moving away from the person if $x'_s$ is positive. So we need to find $x'_s.$

      The two right triangles in the sketch are similar triangles, so we can use proportions of corresponding sides to create an equation.

      $$\frac25 = \frac{x_s}{x} = \frac{x_s}{x_p+x_s}$$

      If we solve this for $x_s$ we arrive at

      \begin{align} \frac25 (x_p+x_s) &= x_s\\ \frac25 x_p +\frac25 x_s &= x_s\\ x_s &= \frac23 x_p\\ \end{align}

      This equation relates $x_s$, whose rate we need, to $x_p$, whose rate we know. Differentiation with respect to $t$ will give us $$x'_s=\frac23 x'_p$$ Thus, $$x'_s=\frac23 (-0.5) = -\frac13$$ Because this rate is negative, we see that the tip of the shadow is moving towards the person (at a rate of $\frac13$ m/s).

    2. In this part we want to determine the sign of $x'.$ From the sketch, we see that $x = x_p+x_s$, which works perfectly for us. Differentiation with respect to $t$ gives $$x' = x'_p+x'_s$$ We were given that $x'_p = -\frac12$ in the problem statement, and we found in part (a) that $x'_s = -\frac13.$ Substituting these into the equation above yields $$x' = -\frac12 + \left( -\frac13\right) = -\frac56$$ Because this rate is negative, we see that the tip of the shadow is moving towards the wall (at a rate of $\frac56$ m/s).

  10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the depth of the water in the tank changing when the radius of the top of the water is 10 meters?

    This image shows a cone with its vertex pointing downwards. Inside the cone is a cone with a smaller height, indicating the water. The radius at the top of the cone is labeled 26 meters, and the height of the cone is shown as 8 meters.

    Let's let $r$ be the radius of the water in the tank and let $h$ be the height of the water in the tank. Here is a sketch (not to scale) of the cross section of the cone showing the related quantities.

    This image shows two isosceles triangles, one inside the other, with a smaller height than the outer triangle. The vertex between the two equal sides of the triangles is at the bottom of the sketch. The inner triangle is shaded blue to represent the water in the tank. There is a dashed vertical line running down the center of the triangles. The distance from this line to the top right vertex of the outer triangle is labeled 26, and the distance from the dashed line to the upper right vertex of the inner triangle is labeled r. The height of the outer triangle is labeled 8, and the height of the inner triangle is labeled h.

    We want to determine $h'$ when $r = 10$ and we know that $V' = 12.$ We will need to use the formula for the volume of a cone: $$V = \frac13\pi r^2 h$$ Since this formula relates $V$ to both $h$ and $r$ but we only want to find $h',$ we would like to eliminate $r$ from it. We can do this if we have another equation that relates only $r$ and $h.$

    We can use the fact that the two triangles in the cross section sketch are similar triangles to do this. By equating ratios of corresponding sides we have $$\frac{r}{h} = \frac{26}{8} \Rightarrow r = \frac{13}{4}h$$ Substituting this into the volume equation then gives $$V=\frac{169}{48}\pi h^3$$ When we differentiate this with respect to $t$ we find $$V'=\frac{169}{16}\pi h^2 h'$$ To finish this problem, we need to find the value of $h$ at the time we are interested in. This can be done using the equation from the similar triangles and the fact that at this time, $r = 10.$

    $$h=\frac{4}{13}r = \frac{4}{13}(10) = \frac{40}{13}$$

    The rate of change of the height of the water is then \begin{align} V' &= \frac{169}{16}\pi h^2 h'\\ 12 &= \frac{169}{16}\pi \left(\frac{40}{13}\right)^2 h'\\ &= 100\pi h'\\ \end{align} $$\Rightarrow h'=\frac{3}{25\pi}$$ The depth of the water is increasing at the rate of $\frac{3}{25\pi}$ meters per second.

  11. The angle of elevation is the angle formed by a horizontal line and a line joining the observer's eye to an object above the horizontal line.

    A person is 500 feet away from the launch point of a hot air balloon. The hot air balloon is starting to come back down at a rate of 15 ft/sec. At what rate is the angle of elevation, $\theta$, changing when the hot air balloon is 200 feet above the ground?

    This image shows a right triangle with the right angle at the bottom right of the image. There is a dot labeled 'Balloon' at top right vertex of the triangle, with an arrow pointing down next to it. There is a dot labeled 'Person' at the bottom left vertex of the triangle. The angle at the bottom left vertex is labeled 'theta, angle of elevation'.

    Let $y$ stand for the height of the balloon. We are given that $y$ is decreasing at the rate of 15 ft/sec, so $y'=-15.$ We are also given that the distance between the person and the launch point of the balloon is 500 feet. We summarize all of this on the sketch below.

    This image shows a right triangle with the right angle at the bottom right of the image. There is a dot labeled 'Balloon' at top right vertex of the triangle, with an arrow pointing down next to it. There is a dot labeled 'Person' at the bottom left vertex of the triangle. The angle at the bottom left vertex is labeled 'theta, angle of elevation', the bottom leg of the triangle is labeled 500, and the other side is labeled y.

    We want to determine $\theta'$ when $y=200$, and we are given that $y'=-15.$

    There are a variety of equations that we could use to relate $y$ and $\theta$, but perhaps the easiest one is $$\tan(\theta) = \frac{y}{500}$$ Differentiating with respect to $t$ gives $$\sec^2(\theta) \theta' = \frac{y'}{500}\hspace{.25in} \Rightarrow \hspace{.25in} \theta' = \frac{y'}{500}\cos^2(\theta)$$ Now we need to determine the value of $\theta$ at the time in question. We can either use the original equation to do this, or we could acknowledge that we really need $\cos(\theta).$ We can use right triangle trigonometry to do that.

    Option 1: Use the original equation to find an approximate value for theta.

    $$\tan(\theta) = \frac{200}{500}\hspace{.25in} \Rightarrow \hspace{.25in} \theta = \arctan\left(\frac25\right) \approx 0.38051 \text{ radians}$$ Then $$\theta' \approx \frac{-15}{500}\cos^2(0.38051)\approx -0.02586$$

    Option 2: Use the right triangle in the sketch to find $\cos(\theta).$

    For this option, we first need to use the Pythagorean Theorem to find the hypotenuse of the triangle.

    $$c^2 = 500^2 + y^2\hspace{.25in}\Rightarrow c = \sqrt{250000 + y^2}$$

    We then substitute this into the equation above to write \begin{align} \theta' &= \frac{y'}{500}\cdot\frac{500^2}{250000+y^2}\\ \theta' &= \frac{-15}{500}\cdot\frac{500^2}{250000+40000} = \frac{-3}{116}\approx -0.02586\\ \end{align}

    Whichever option we choose, we find that the angle of elevation is decreasing at a rate of approximately 0.02586 radians per second.


Assignment Problems

  1. In the following, assume that $x$ and $y$ are both functions of $t.$ Given $x=3,$ $y=2$ and $y' = 7,$ determine $x'$ for the following equation: $$x^3 - y^4 = x^2y - 7$$

  2. In the following, assume that $x$ and $y$ are both functions of $t.$ Given $x = \frac{\pi}{6}$, $y = -4$ and $x' = 12,$ determine $y'$ for the following equation: $$x^2(y^2 - 16) -6\cos(2x) = 1 + y$$

  3. In the following, assume that $x$, $y$ and $z$ are all functions of $t.$ Given $x=-1$, $y=8,$ $z=2,$ $x'=-4,$ and $y' = 7,$ determine $z'$ for the following equation: $$x^4 + \frac{y}{z} = 2x^2z^2 - 3$$

  4. In the following, assume that $x, y$ and $z$ are all functions of $t.$ Given $x=-2, y=3, z=4, y'=6,$ and $z' = 0,$ determine $x'$ for the following equation: $$xy^2z^2=x^3-z^4-8y$$

  5. The sides of a square are increasing at a rate of 10 cm/sec. How fast is the area enclosed by the square increasing when the area is 150 cm2?

  6. The sides of an equilateral triangle are decreasing at a rate of 3 in/hr. How fast is the area enclosed by the triangle decreasing when the sides are 2 feet long?

  7. A spherical balloon is being filled in such a way that the surface area is increasing at a rate of 20 cm2/sec when the radius is 2 meters. At what rate is air being pumped into the balloon when the radius is 2 meters?

  8. A cylindrical tank of radius 2.5 feet is being drained of water at a rate of 0.25 ft3/sec. How fast is the height of the water decreasing?

  9. A hot air balloon is attached to a spool of rope that is 125 feet away from the balloon when it is on the ground. The hot air balloon rises straight up in such a way that the length of rope between the balloon and the spool increases at a rate of 15 ft/sec. How fast is the hot air balloon rising 20 seconds after it lifts off?

    In the lower left corner of this image is a dot labeled “Spool of Rope”. In the upper right corner is a dot labeled “Balloon” and it has an arrow next to it pointing directly up. The two dots are connected with an angled line that has the text “This Length Increases at 15 ft/sec” on it. The sketch also shows that the horizontal distance between the two points is 125 feet.

  10. A rock is dropped straight off a bridge that is 50 meters above the ground and falls at a speed of 10 m/sec. Another person is 7 meters away on the same bridge. At what rate is the distance between the rock and the second person increasing just as the rock hits the ground?

    At the top of this image are two dots connected by a horizontal arrow that indicates the distance between them is 7 meters. The left dot is labeled 'Rock dropped here,' and the right dot is labeled, 'Person.' At the bottom left of the image is a dot labeled 'Rock,' with an arrow pointing down. The dots at the top right and bottom left are connected with an angled line that has the text “Find the rate this distance is changing” on it.

  11. A person is 550 meters away from a road, and there is a car that is initially 800 meters away, approaching the person at a speed of 45 m/sec.

    In the lower left corner of this image is a dot labeled “Person”. In the upper right corner is a dot labeled “Car” and it has an arrow next to it pointing to the left. The two dots are connected with a diagonal line that has the text “Find rate this distance is changing” on it. The sketch also shows that the horizontal distance between the two points is initially 800 meters and that the vertical distance between the two point is 550 meters.

    Find the rate at which the distance between the person and the car is changing

    1. 5 seconds after the start,

    2. when the car is directly in front of the person, and

    3. 10 seconds after the car has passed the person.

  12. Two cars are initially 1200 miles apart. At the same time Car A starts driving at 35 mph to the east, Car B starts driving at 55 mph to the north. (See the sketch below for the initial setup.)

    At the bottom of this image are two dots labeled 'Car A initial position' and 'Car B initial position.' There is a horizontal two-headed arrow between the dots that shows the dots are 1200 miles apart. There is a third dot between the first two labeled 'Car A later position' and it has an arrow above it pointing right. Directly above the bottom right dot is a fourth dot labeled 'Car B later position' and there is an arrow next to it pointing up. There is an angled line connecting these two dots and it is labeled, 'Find the rate this distance is changing.'

    At what rate is the distance between the two cars changing

    1. after 5 hours of travel?

    2. after 20 hours of travel?

    3. after 40 hours of travel?

  13. Two cars are initially 1200 miles apart. Four hours after Car B starts driving at 55 mph to the north, Car A starts driving at 35 mph to the east. (See the sketch in the previous problem for the initial setup.)

    At what rate is the distance between the two cars changing

    1. 5 hours after Car B starts?

    2. 20 hours after Car B starts?

    3. 40 hours after Car B starts?

  14. Two people are on a city block. See the sketch below for their relative positions and distances.

    Person A is on the northeast corner and Person B is on the southwest corner. Person A starts walking towards the southeast corner at a rate of 3 ft/sec. Four seconds later, Person B starts walking towards the southeast corner at a rate of 2 ft/sec.

    In the lower left corner of this image is a dot labeled “Person B” and has an arrow next to is pointing to the right. In the upper right corner is a dot labeled “Person A” and it has an arrow next to it pointing down. The two dots are connected with a diagonal line that has the text “Find Rate This Distance Changing” on it. The sketch also shows that the horizontal distance between the two points is initially 500 feet and that the vertical distance between the two point is 300 feet.

    At what rate is the distance between them changing

    1. 10 seconds after Person A starts walking?

    2. after Person A has covered half the distance to the southeast corner?

  15. A person is standing 75 meters away from a kite and has a spool of string attached to the kite. The kite starts to rise straight up in the air at a rate of 2 m/sec and at the same time the person starts to move towards the kite's launch point at a rate of 0.75 m/sec.

    Is the length of the string increasing or decreasing

    1. after 4 seconds?

    2. after 20 seconds?

  16. A person lights the fuse on a model rocket and starts to move away from the rocket at a rate of 3 ft/sec. Five seconds after lighting the fuse, the rocket launches straight up into the air at a rate of 10 ft/sec.

    In the lower left corner of this image is a dot labeled “Person” and it has an arrow under it pointing to the left. In the upper right corner is a dot labeled “Rocket” and it has an arrow next to it pointing up. The two dots are connected with a diagonal line that has the text “Is this distance increasing or decreasing?” on it. In the bottom right corner of the image is a dot labeled 'Initial location of rocket launch.' The dot in the bottom right is directly below the dot labeled 'Rocket,' and directly to the right of the dot labeled 'Person.'

    Is the distance between the person and rocket increasing or decreasing

    1. 6 seconds after launch?

    2. 12 seconds after launch?

  17. A light is on a pole and is being lowered towards the ground at a rate of 9 in/sec. A 6 foot tall person is on the ground and is 8 feet away from the pole. At what rate is the person's shadow increasing when the light is 15 feet above the ground?

    In this image, there is a right triangle whose hypotenuse goes from the upper left to the lower right. At the upper left vertex of the triangle, there is a dot labeled 'Light' and an arrow to its left and pointing down. Partway down the hypotenuse, there is a dot labeled 'Person.' A vertical line connecting this dot to the bottom of the triangle forms a second right triangle inside of the first one. The bottom edge of this smaller triangle is labeled 'Shadow'. The sketch also shows that the distance from the light to the person is 8 feet and the height of the person is 6 feet.

  18. A light is fixed on a wall 10 meters above the floor. Twelve meters away from the wall a pole is being raised straight up at a rate of 45 cm/sec.

    In this image, there is a right triangle whose hypotenuse goes from the upper left to the lower right. At the upper left vertex of the triangle, there is a dot labeled 'Light'. Partway down the hypotenuse, there is vertical line connecting the hypotenuse to the bottom of the triangle, forming a second right triangle inside of the first one. The top of this line is labeled,' Pole' and there is an arrow to the right that points up. The bottom edge of this smaller triangle is labeled 'Shadow'. The sketch also shows that the distance from the light to the pole is 12 meters and the light is 10 meters above the floor.

    When the pole is 6 meters tall,

    1. at what rate is the tip of the shadow moving away from the pole?

    2. at what rate is the tip of the shadow moving away from the wall?

  19. A light is on top of a 15 foot pole. A five-foot tall person starts at the pole and moves away from the pole at a rate of 2.5 ft/sec.

    In this image, there is a right triangle whose hypotenuse goes from the upper left to the lower right. At the upper left vertex of the triangle, there is a dot labeled 'Light'. Partway down the hypotenuse, there is a dot labeled 'Person' and an arrow pointing to the right. A vertical line connecting this dot to the bottom of the triangle forms a second right triangle inside of the first one. The bottom edge of this smaller triangle is labeled 'Shadow'. The sketch also shows that the distance from the light to the ground is 15 feet and the height of the person is 5 feet.

    After moving for 8 seconds,

    1. at what rate is the tip of the shadow moving away from the person?

    2. at what rate is the tip of the shadow moving away from the pole?

  20. A tank of water is in the shape of a cone with the vertex of the cone pointing downwards (see the sketch) and is leaking water at a rate of 35 cm3/sec. The base radius of the tank is 1 meter and the height of the tank is 2.5 meters.

    This image shows a cone with its vertex pointing downwards. Inside the cone is a cone with a smaller height, indicating the water. The radius at the top of the cone is labeled 1 meter, and the height of the cone is shown as 2.5 meters.

    When the depth of the water is 1.25 meters,

    1. at what rate is the depth changing?

    2. at what rate is the radius of the top of the water changing?

  21. A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and whose height is 10 meters. Assume that the two equal length sides of the triangle are the sides of the water trough, and the other side of the triangle is the top of the trough, and is parallel to the ground. Water is being pumped into the trough at a rate of 2 m3/min.

    This image shows two isosceles triangles, one below and to the left of the other. Both triangles are pointing down, with the third, possibly unequal side at the top of each triangle. Each triangle has another line parallel to the top side, cutting off part of the two equal sides, forming a smaller isosceles triangle within the original ones. The vertices of one pair of large and small triangles are connected to the corresponding vertices of the other pair. The whole image indicates a triangular trough with a smaller trough inside it, standing for the water in the tank. The sketch shows that the width of the trough is 7 meters, its height is 10 meters and its length is 20 meters.

    When the water is 6 meters deep,

    1. at what rate is the depth of water changing?

    2. at what rate is the width of the top of the water changing?

  22. A trough of water is 9 feet long and its ends are in the shape of an equilateral triangle whose sides are 1.5 feet long. Assume that the top of the trough is parallel to the ground. If water is being pumped out of the trough at a rate of 2 ft3/sec, at what rate is the depth of the water changing when the depth is 0.75 feet?

    This image shows two equilateral triangles, one below and to the left of the other. Both triangles have one vertex at the bottom, with the third side at the top of each triangle. Each triangle has another line parallel to the top side, cutting off part of the two sides, forming a smaller triangle within the original ones. The vertices of one pair of large and small triangles are connected to the corresponding vertices of the other pair. The whole image indicates a triangular trough with a smaller trough inside it, standing for the water in the tank. The sketch shows that the width of the trough is 1.5 feet, its length is 9 feet, and the sides of the original triangles are all 1.5 feet.

  23. The angle of elevation (depression) is the angle formed by a horizontal line and a line joining the observer's eye to an object above (below) the horizontal line. Two people are on the roofs of buildings separated by a 25 foot wide road. Person A is 100 feet above Person B and drops a rock off the roof of their building. The rock falls at a rate of 3 ft/sec.

    1. At what rate is the angle of elevation changing when the rock is 35 feet above Person B? (Note that the sketch below is not to scale).

      In the upper right of this image there is a dot labeled, 'Person A - dropped rock' and at the bottom left of the image is a dot labeled, 'Person B.' Directly below the dot labeled 'Person A' is another dot labeled 'Rock' and there is an arrow next to it, pointing down. The rock and person B are two vertices of a right triangle and are the endpoints of its hypotenuse. In this image, line connecting the rock to Person B goes from the rock and down and to the left. The angle at the dot labeled 'Person B' is labeled 'theta, angle of elevation.' The sketch also shows that Person A is 100 feet above Person B and that the horizontal distance between Person A and Person B is 25 feet.

    2. At what rate is the angle of depression changing when the rock is 65 feet below Person B? (Note that the sketch below is not to scale).

      In the upper right of this image there is a dot labeled, 'Person A - dropped rock.' Directly below the dot labeled 'Person A' is another dot labeled 'Rock' and there is an arrow next to it, pointing down. Between these two dots there is a horizontal line to the left. At the end of this line is a dot labeled, 'Person B.' The rock and person B are two vertices of a right triangle and are the endpoints of its hypotenuse. In this image, line connecting the rock to Person B goes from Person B, down and to the right. The angle at the dot labeled 'Person B' is labeled 'theta, angle of depression.' The sketch also shows that Person A is 100 feet above Person B and that the horizontal distance between Person A and Person B is 25 feet.

  24. The angle of elevation is the angle formed by a horizontal line and a line joining the observer's eye to an object above the horizontal line. A person is standing 15 meters away from a building, watching an outside elevator move down the face of the building. When the angle of elevation is 1 radian, it is changing at a rate of 0.15 radians/sec. At this point in time, what is the speed of the elevator?

    This image shows a right triangle with the right angle at the bottom right of the image. There is a dot labeled 'Elevator' at top right vertex of the triangle, with an arrow pointing down next to it. There is a dot labeled 'Person' at the bottom left vertex of the triangle. The angle at the bottom left vertex is labeled 'theta, angle of elevation'. The sketch also shows that the person is 15 meters away from the building.

  25. The angle of elevation is the angle formed by a horizontal line and a line joining the observer's eye to an object above the horizontal line. A person is 24 feet away from a building, watching an outside elevator move up the face of the building. The elevator is moving up at a rate of 4 ft/sec and the person is moving towards the building at a rate of 0.75 ft/sec.

    Assuming that the elevator started moving from the ground at the same time that the person started walking, is the angle of elevation increasing or decreasing after 10 seconds?

    This image shows a right triangle with the right angle at the bottom right of the image. There is a dot labeled 'Elevator' at top right vertex of the triangle, with an arrow pointing up next to it. There is a dot labeled 'Person' at the bottom left vertex of the triangle, with an arrow pointing right. The angle at the bottom left vertex is labeled 'theta, angle of elevation'. The sketch also shows that the person is 24 feet away from the building.