Related Rates of Change

Watch the animation closely. Water is being added to the conical cup at a constant rate. What do you notice about the rate at which the water level increases? If you said that it's slowing down, then you are right! The conical cup seems to fill more slowly as the water level reaches the top. We know that the same volume of water is being added to the cup every second but that volume of water is dispersing over larger cross-sectional areas at higher heights so the rate of increase of water level slows down.

This wiki will focus on measuring such phenomena using the language of derivatives. It will help us discover unknown rates of change as they relate to other known rates of change. We will first learn the basics; then look at a wide range of examples using 2D/3D geometry and also a few applications of the same in physics and chemistry.

It's time to begin solving examples! In this section, we shall mainly look at examples which involve 2D geometry. For example, we shall see how changing the radius of a circle affects its area, or how changing the angles in a right triangle changes the ratio of sides. We will be using concepts such as the Pythagorean theorem and basic trigonometry, so be sure that you understand these concepts well before moving on.

Space shuttle launches can be very exciting, but camera crews need to know precisely where to move their equipment to accurately track a moving rocket. In the animation below, we see that our view slowly tilts upward as the shuttle shoots off into space. But just how fast is the camera rotating?

Suppose that a tracking camera is positioned 5 kilometers away from the launch pad and the shuttle is ascending vertically at a constant rate of 500 meters per second. How fast should the camera rotate to stay focused on the rocket at the instant the rocket is 2 kilometer from the launchpad?

While the Pythagorean theorem could be used to relate the sides of the right triangle depicted in this scenario, we find that it will not be useful here because the formula does not contain any angle measurements. Instead, we will use trigonometric ratios to produce a relationship between the angle whose rate of change we wish to find and the sides whose rates of change we already have:

\[\begin{align*} \tan \theta &= \frac{y}{x} = \frac{y}{5} \qquad (\text{substitute in } x = 5 \text{ because it's constant}) \\
\frac{d}{dt}(\tan \theta) &= \frac{d}{dt}\left(\frac{y}{5}\right) \\ \sec^2 \theta \frac{d\theta}{dt} &= \frac{1}{5}\frac{dy}{dt}. \end{align*} \]

Though we were not explicitly given $\theta$ , we know that $\sec \theta$ can be found by using the sides of the right triangle depicted by this scenario, which is why a diagram can be extremely useful:

$\frac{29}{25}\frac{d\theta}{dt} = \frac{1}{5}(0.5) \implies \frac{d\theta}{dt} = \frac{25}{290} = 0.0862 \text{ (radians per second)}. \ _\square$

A car traveling south at $30 \text{ km/hr}$ is approaching a fixed point $A$ , and another car traveling west at $40 \text{ km/hr}$ is moving away from $A$ . How is the distance between the cars changing, expressed in $\text{km/hr},$ when the car traveling south is $12 \text{ km}$ from $A$ and the car traveling west is $5 \text{ km}$ from $A?$

Let the distance between the two cars be $z$ , then we have

$z^{2}=x^{2}+y^{2},$

where $x$ is the distance between $A$ and the car traveling west and $y$ is the distance between $A$ and the car traveling south. Differentiating both sides with respect to $t$ , we have

$\begin{aligned} \frac{d}{dt}\big(z^{2}\big)&=\frac{d}{dt}\big(x^{2}+y^{2}\big)\\ 2z\frac{dz}{dt}&=2x\frac{dx}{dt}+2y\frac{dy}{dt}. \end{aligned}$

Since we are given

$\begin{array}{c}&\frac{dx}{dt} =40 \text{ km/hr}, &\frac{dy}{dt}=-30 \text{ km/hr}, &x =5 \text{ km}, &y=12 \text{ km}, &z=\sqrt{5^{2}+12^{2}}=13 \text{ km}, \end{array}$

we have

$13\frac{dz}{dt} = \big(5(40)+12(-30)\big) \implies \frac{dz}{dt} \approx -12.31 \text{ km/hr}. \ _\square$

Let's move on to examples using 3D Geometry. Now we can analyze various 3D shapes such as a cone, sphere, cylinder… By the end of this section, you will be able to visualize clearly how the rate of change of one variable—for example, the radius of a cone—is related to the rate of change of another variable like the cone's volume.

Just by looking at the animation, you might already have an intuition about when the radius of the balloon is increasing fastest. Notice that the balloon seems to inflate rapidly at first but then slows down as more gas is pumped into the balloon. Let's verify this with related rates!

A spherical balloon is being filled with gas at a constant rate of 100 cubic centimeters per second.

Will the radius of the balloon be changing faster when its radius is 2 centimeters than when its radius is 20 centimeters?

Because the problem refers to both the volume and radius of a spherical balloon, it only seems natural to start with the equation of the volume of a sphere, which relates these two quantities:

$V = \frac{4}{3}\pi r^3.$

Differentiate both sides of this equation with respect to time. By chain rule, we'll have

$\begin{aligned} \frac{d}{dt}(V) &= \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\\ \frac{dV}{dt} &= 4\pi r^2\frac{dr}{dt}. \end{aligned}$

Remember that the quantities $\frac{dV}{dt}$ and $\frac{dr}{dt}$ represent the rates of change of the sphere's volume and radius, respectively. So we want to get our hands on $\frac{dr}{dt}$ because that quantity will tell us how fast the balloon's radius is changing. Plugging in what we know about the volume's rate of change and the radii given, we can determine $\frac{dr}{dt}.$

If $r=2,$ then

$100 = 4\pi (2)^2\frac{dr}{dt} \hspace{.1cm} \implies \hspace{.1cm} \frac{dr}{dt} = \frac{100}{16\pi} \hspace{.1cm} \approx \hspace{.1cm} \color{#3D99F6}{1.989} \text{ (centimeters per second)}.$

If $r=20,$ then

$100 = 4\pi (20)^2\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{100}{1600\pi} \approx \color{#D61F06}{0.01989} \text{ (centimeters per second)}.$

Sure enough, the radius increases much more slowly when the balloon is larger. This is because the gas has more volume over which to spread, resulting in a small increase to the balloon's radius. $_\square$

When two changing variables are multiplied, the combined effect of their rates of change can be even greater than if they were separated. For instance, we know that the volume of a cone depends on a product involving its radius $r$ and height $h$ , so we should expect its change in volume to have contributions from both its changes in radius and height! This is where the product rule will come into play.

The equation for the volume of a cone is given by $V=\frac{1}{3}\pi r^{2}h.$

$\qquad \text{a)}$ How is $\frac{dV}{dt}$ related to $\frac{dr}{dt}$ if $h$ is constant?

$\qquad \text{b)}$ How is $\frac{dV}{dt}$ related to $\frac{dh}{dt}$ if $r$ is constant?

$\qquad \text{c)}$ How is $\frac{dV}{dt}$ related to $\frac{dh}{dt}$ and $\frac{dr}{dt}$ if neither $h$ nor $r$ is constant?

$\text{a)}$ From the chain rule, we have

$\begin{aligned} \frac{dV}{dt}&=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right) = \color{#3D99F6}{\frac{2}{3}\pi hr\frac{dr}{dt}}.\\ \\ \end{aligned}$

$\text{b)}$ Similarly, we have

$\begin{aligned} \frac{dV}{dt}&=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right)=\color{#D61F06}{\frac{1}{3}\pi r^{2}\frac{dh}{dt}}.\\ \\ \end{aligned}$

$\text{c)}$ This time, we have

$\begin{aligned} \frac{dV}{dt}&=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right) = \underbrace{\color{#3D99F6}{\frac{2}{3}\pi hr\frac{dr}{dt}} + \color{#D61F06}{\frac{1}{3}\pi r^{2}\frac{dh}{dt}}}_{\text{ Product Rule }}. \ _\square \end{aligned}$

The use of related rates in the physical sciences is imperative because a variety of disciplines require evaluation of rates of change. From speeding cars and falling objects to expanding gas and electrical discharge, related rates are ubiquitous in the realm of science.

If a $1700 \text{ kg}$ car is accelerating at a rate of $6 \text{ m}/\text{s}^2$ , then how fast is its kinetic energy changing when the speed is $30 \text{ m}/\text{s}?$

Recalling the formula for kinetic energy is $E_k = \frac{1}{2}mv^2$ , we'll begin by differentiating both sides with respect to time and obtain

$\begin{aligned} \frac{d}{dt}(E_k) &= \frac{d}{dt}\left(\frac{1}{2}mv^2\right)\\ \Rightarrow \frac{dE_k}{dt} &= mv\frac{dv}{dt} \\ &= mva \\ &= (1700)(30)(6) \\ &= 306000 \text{ J}/\text{s}. \ _\square \end{aligned}$

Two $10$ - $\Omega$ resistors labeled $A$ and $B$ are connected in parallel. $A$ 's resistance is increasing by $2~\Omega / \text{min}$ while $B$ 's resistance is decreasing by $3~\Omega / \text{min}$ . At what rate is the effective resistance of this system changing?

The formula governing the relationship between resistors in parallel is given by

$\frac{1}{R_{e}} = \frac{1}{R_A} + \frac{1}{R_B}.$

Because we will need $R_e$ down the road, we go ahead and find that $R_e = 5~\Omega$ using this formula and the fact that $R_A = R_B = 10~\Omega$ . Then, differentiating this with respect to time gives us

$\begin{aligned} \frac{d}{dt}\left(\frac{1}{R_{e}}\right) &= \frac{d}{dt}\left(\frac{1}{R_A} + \frac{1}{R_B}\right)\\ -\frac{1}{R_{e}^2} \frac{dR_{e}}{dt} &= -\frac{1}{R_{A}^2} \frac{dR_{A}}{dt} - \frac{1}{R_{B}^2} \frac{dR_{B}}{dt}\\ -\frac{1}{5^2} \frac{dR_{e}}{dt} &= -\frac{1}{10^2}(2) - \frac{1}{10^2}(-3)\\ \Rightarrow \frac{dR_{e}}{dt} &= -0.25 ~\Omega /\text{min}. \ _\square \end{aligned}$

An internal combustion engine uses a piston to compress combustible air in a cylinder before igniting it to generate energy in the form of mechanical work. Suppose the volume of such a cylinder is $600 \text{ cm}^3$ at $1 \text{ atm}$ of pressure. Suddenly, the piston begins to compress the volume of the air-fuel mixture at a rate of $8000 \text{ cm}^3/\text{s}.$ What is the corresponding rate of change of pressure in the cylinder? Because we'll assume the temperature doesn't change in this scenario, use Boyle's law, which says that $PV = k$ , where $k$ is a constant.

Differentiating both sides of the given equation yields

\[\begin{align} \frac{d}{dt}(PV) &= \frac{d}{dt}(k)\\ \underbrace{P\frac{dV}{dt} + V\frac{dP}{dt}}_{\text{Product Rule}} &= 0\\
(600)\left(\frac{dP}{dt}\right) + (1)(-8000) &= 0\\ \Rightarrow \frac{dP}{dt} &\approx 13.333 \text{ atm} / \text{s}. \ _\square \end{align} \]

Advanced Examples

Related rates can become very involved and may borrow techniques and formulas from a wide variety of disciplines, so check out these advanced examples to see just how complicated (and powerful) related rates can be. These examples are advanced because it is not very easy to see how to go about solving the problem. Often you may have to introduce an intermediate variable to proceed and use other advanced techniques in math such as integration and polar coordinates, to name a few.

Suppose the ground is the $xy$ -plane. A policeman on a motorcycle drives along a straight horizontal road $y = a$ with constant velocity $v_{p}$ . There happens to be a circular pillar with radius $R$ centered at the origin. A robber is standing on the circumference. He moves along the circumference such that the line joining him and the policeman always passes through the origin.

With what velocity $v_{r}$ should the robber move when the policeman is at the point $(x_{p}, a)?$

Assume $R < a$ .

In this problem, we have to relate the speed of the robber to the $x$ -coordinate of the policeman. The direct relation between these two variables is not very clear; so it would be super useful to introduce a new variable that directly relates to $x_{p}$ and to $v_{r}$ . We see that the line joining the robber and the policeman is rotating. It connects the robber and the policeman. So it should not be surprising that a property of the line—the angle that it makes with the $y$ -axis—connects $x_{p}$ and $v_{r}$ .

Let's make the following construction: we draw a perpendicular line from the center of the circle to the line $y = a$ .

Now, let the angle that it makes with the $y$ -axis be called $\theta$ . Then we get the following relation: $v_{r} = R \cdot \frac{d\theta}{dt}$ . From the triangle, we can see that $\tan \theta = \frac{x_{p}}{a}$ . Using implicit differentiation, we have

$\sec ^{2} \theta \dfrac{d\theta }{dt} = \dfrac{dx_{p}}{dt} \cdot \dfrac{1}{a} .$

Therefore, using $\sec \theta = \frac{\sqrt{x_{p}^{2} + a^{2}}}{a}$ from the triangle, we have

$\begin{aligned} \frac{x_{p}^{2} + a^{2} }{a^{2} } \frac{d\theta}{dt} &= \frac{v_{p}} {a}\\ \Rightarrow v_{r} &=R \cdot \dfrac{d\theta}{dt}\\ &= R \cdot \dfrac{av_{p}}{x_{p}^{2} + a^{2}}. \ _\square \end{aligned}$

This example uses the disc method.

The curve $y = x^2 - 1$ is revolved around the $y$ -axis to form a container. If liquid flows into the container at a rate of 2 cubic units per minute, then how fast is the depth of the liquid changing at the moment that the liquid's depth is 5 units?

Let $h$ be the depth of the liquid in the cup. Then we have

$\begin{aligned} V &= \pi\int_{-1}^{h}\big(\sqrt{y + 1}\big)^2dy\\ \frac{dV}{dt} &= \pi(h + 1)\frac{dh}{dt} \\ 2 &= 5\pi\frac{dh}{dt} \qquad (\text{since }h = 4 \text{ to make total depth }5)\\ \frac{dh}{dt} &= \frac{2}{5\pi}. \ _\square \end{aligned}$

Test yourself by solving the following examples, and then check whether you are correct by clicking on "Reveal Solution". Don't worry if you don't get them at first, they are tough!

A man is standing at a distance of $2d$ from a mirrored wall. A block is attached to a pulley which is midway between the man and the wall. The man pulls the string with constant velocity $v$ . The man sees that the block appears to move with angular velocity $\omega_{o}$ and the image of the block in the mirror moves with velocity $\omega_{i}$ .

Find $\frac{d\omega_{i}}{d\omega_{o}}$ .

Note: You may include $h$ , the vertical distance between the man's eye and the block, in your final answer.

Let's focus on what is important in this problem and try to simplify it. The only things important are the eye of the man, the block, and the image of the block in the mirror. So let's concentrate on these things only and fade everything else.

Keep in mind that the image of the block is always at the same height as the block. Also, the image is at a horizontal distance of $3d$ from the man's eye. These facts can be proven using ray optics. Now we are all set to solve this problem.

Recall that the angular velocity of an object with respect to a point is the rate of change of angular position, i.e. $\omega = \frac{ d\theta }{ dt }$ . The angular positions $\theta_{ o }$ and $\theta_{ i }$ with respect to the eye can be found by drawing triangles.

Here we see that $\tan \theta_{ o } = \frac{ h }{ d }$ . To obtain $\omega_{ o }$ , we can differentiate both sides of the equation with respect to time:

$\begin{aligned} \frac{ d }{ dt } \left( \tan \theta _{ o } \right) &= \frac{ d }{ dt } \left(\frac{ h }{ d } \right) \\ \sec^2 \theta _{ o } \frac{ d \theta_{ o } }{ dt } &= \frac{ dh } { dt } \cdot \frac{ 1 } { d }. \end{aligned}$

Now we know that $\frac{ d \theta_{ o } } { dt }$ is $\omega _{ o }$ and $\frac{ dh }{ dt }$ is nothing but $v$ . Also, we can see from the triangle that $\sec \theta_{ o } = \frac{ \sqrt{ h^{ 2 } + d^{ 2 } } } { d } .$ So let's substitute these values:

$\frac{ h ^ { 2 } + d^{ 2 } }{ d^{ 2 } } \omega_{ o } = \frac{ v }{ d } \implies \omega_{ o } = \dfrac{ d \cdot v }{ h^{ 2 } + d^ { 2 } }.$

On solving similarly for $\omega_{i}$ , we obtain $\omega_{i} = \frac{3d \cdot v}{h^{2} + 9d^{2}}$ . The only step remaining is to find $\frac{ d \omega_{ i } } { d \omega_{ o } }$ . Note that both $\omega_{ o }$ and $\omega_{ i }$ are functions of $h$ . We can use derivative of parametric equations to compute $\frac{ d \omega_{ i } } { d \omega_{ o } }$ .

We have $\frac{ d \omega_{ i } } { dh } = \frac{ -3d \cdot v \cdot 2h } { \left( h^{ 2 } + 9d^{ 2 } \right)^{ 2 } }$ and $\frac{ d \omega_{ o } } { dh } = \frac{ -d \cdot v \cdot 2h } { \left( h^{ 2 } + d^{ 2 } \right)^{ 2 } }$ , and therefore

$\frac{ d \omega_{ i } } {d \omega_{ o } } = \frac{ \frac{d \omega_{ i }}{dh} }{\hspace{3mm} \frac{d \omega_{ o }}{dh}\hspace{3mm} } = \frac{ 3 \left( h^{ 2 } + d^{ 2 } \right)^{ 2 } } { \left( h^{ 2 } + 9d^{ 2 } \right)^{ 2 } },$

and we are done! $_\square$

A light ray emanates from a point source in a circular room. The room has its center at the origin and has radius $r$ . The ray rotates with angular velocity $\omega$ , which is a function of time $t$ . Find the velocity of the point where the light ray strikes the wall in the following cases:

$\qquad \text{ i)}$ the point source is placed at the origin;
$\qquad \text{ii)}$ the point source is placed at a general point $(h, k)$ inside the circle.

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