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.
Contents
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
\[\begin{align} \frac{d}{dt}\big(z^{2}\big)&=\frac{d}{dt}\big(x^{2}+y^{2}\big)\\ 2z\frac{dz}{dt}&=2x\frac{dx}{dt}+2y\frac{dy}{dt}. \end{align}\]
Since we are given
\[\begin{array} &\frac{dx}{dt} =40 \text{ km/hr}, &\frac{dy}{dt}=-30 \text{ km/hr}, &x =5 \text{ km}, &y=12 \text{ km}, &z=\sqrt{5^{2}+12^{2}}=13 \text{ km}, \end{array}\]
we have
\[13\frac{dz}{dt} = \big(5(40)+12(-30)\big) \implies \frac{dz}{dt} \approx -12.31 \text{ km/hr}. \ _\square\]
When ships come in to dock where the berths are extremely narrow, the ships need to be towed in to prevent damage to both the dock and the ship. A metal cable is attached to the front of the ship, looped through a pulley that is hinged at the edge of the dock, and attached to an automated winch.
If the automated winch winds the cable at a constant rate, which of the following describes the speed at which the boat will be pulled in (prior to reaching the dock)?
In triangle \( \text{ABC},\) \(\overline{\text{AB}}=2\) and \(\overline{\text{BC}}+\overline{\text{CA}}=2\overline{\text{AB}}.\) Let \(\angle \text{B}=\theta\) and suppose that the rate of change of the angle \(\theta\) with respect to time is \(\frac{d\theta}{dt}=2\ (\text{rad/sec}).\)
Then, if \(S\) is the area of the triangle, what is the rate of change of \(S\) with respect to time when \( \angle \text{A}=\frac{\pi}{2}?\)
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
\[\begin{align} \frac{d}{dt}(V) &= \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\\ \frac{dV}{dt} &= 4\pi r^2\frac{dr}{dt}. \end{align}\]
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
\[100 = 4\pi (2)^2\frac{dr}{dt} \hspace{.1cm} \implies \hspace{.1cm} \frac{dr}{dt} = \frac{100}{16\pi} \hspace{.1cm} \approx \hspace{.1cm} \color{blue}{1.989} \text{ (centimeters per second)}. \]
If \(r=20,\) then
\[100 = 4\pi (20)^2\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{100}{1600\pi} \approx \color{red}{0.01989} \text{ (centimeters per second)}. \]
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
\[\begin{align} \frac{dV}{dt}&=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right) = \color{blue}{\frac{2}{3}\pi hr\frac{dr}{dt}}.\\ \\ \end{align}\]
\(\text{b)}\) Similarly, we have
\[\begin{align} \frac{dV}{dt}&=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right)=\color{red}{\frac{1}{3}\pi r^{2}\frac{dh}{dt}}.\\ \\ \end{align}\]
\(\text{c)}\) This time, we have
\[\begin{align} \frac{dV}{dt}&=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right) = \underbrace{\color{blue}{\frac{2}{3}\pi hr\frac{dr}{dt}} + \color{red}{\frac{1}{3}\pi r^{2}\frac{dh}{dt}}}_{\text{ Product Rule }}. \ _\square \end{align}\]
Popeye loves his spinach. He squeezes a can of his favorite spinach in such a way that it retains the shape of a cylinder and its volume remains constant, but the radius of the can decreases at \(1 \text{ cm}\) per second.
How fast is the height of the can changing at the moment the can has a radius of \(4 \text{ cm}\) and a height of \(10 \text{ cm}?\)
A conical reservoir with its vertex pointing downward has a radius of 10 feet and a depth of 20 feet. Suddenly, a leak springs and water begins to empty the cone at its vertex and into an empty cylindrical basin with radius 6 feet and height 40 feet.
At the moment the depth of the reservoir reaches 16 feet and is decreasing by 2 feet per minute, how fast is the height of water in the basin changing?
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
\[\begin{align} \frac{x_{p}^{2} + a^{2} }{a^{2} } \frac{d\theta}{dt} &= \frac{v_{p}} {a}\\ \Rightarrow v_{r} &=R \cdot \dfrac{d\theta}{dt}\\ &= R \cdot \dfrac{av_{p}}{x_{p}^{2} + a^{2}}. \ _\square \end{align} \]
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
\[\begin{align} V &= \pi\int_{-1}^{h}\big(\sqrt{y + 1}\big)^2dy\\ \frac{dV}{dt} &= \pi(h + 1)\frac{dh}{dt} \\ 2 &= 5\pi\frac{dh}{dt} \qquad (\text{since }h = 4 \text{ to make total depth }5)\\ \frac{dh}{dt} &= \frac{2}{5\pi}. \ _\square \end{align}\]
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:
\[\begin{align} \frac{ d }{ dt } \left( \tan \theta _{ o } \right) &= \frac{ d }{ dt } \left(\frac{ h }{ d } \right) \\ \sec^2 \theta _{ o } \frac{ d \theta_{ o } }{ dt } &= \frac{ dh } { dt } \cdot \frac{ 1 } { d }. \end{align} \]
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.
Case \(\text{ i)}\) This case is relatively easier to solve. The answer is simply \(v = \omega {r}.\)Case \(\text{ii)}\) Let's try to simplify the problem. We can rotate the coordinate axes so that the point \((h, k)\) comes on one of the axis. In the new coordinate system, the point will be \(\big(\sqrt{h^{2}+ y^{2}}, 0\big)\). We can let \(d = \sqrt{h^{2}+ y^{2}}\), and hence in the new coordinate system the light source will be at \((d, 0)\).
To solve this problem, we can make use of polar coordinates. It will make the calculations much simpler:
\[R \sin \theta = \tan \alpha (R \cos \theta - d).\]
[better image to be added]
Differentiating both sides yields
\[ R \cos \theta \frac{ d \theta }{ dt } = \frac{ d \alpha } { dt } \sec^{2} \alpha(R \cos \theta - d) + \tan \theta \left(- R \sin \theta \frac{ d \theta }{ dt } \right).\]
Using the relations \(v = R \frac{ d \theta }{ dt }\) and \( \omega = \frac{ d \alpha } { dt } \), we can simplify this to
\[v \cos \theta = \omega \sec^{2} \alpha ( R \cos \theta - d ) - v \tan \alpha \sin \theta.\]
Using sine rule in the triangle, we get a relation between \(\theta \) and \( \alpha: \)
\[\frac{ \sin \alpha }{ R } = \frac{ \sin( \alpha - \theta ) }{ d } .\]
With this we can eliminate \( \theta \) and be left with a equation in \(R, v, \omega, d\) and \( \alpha\). \(_\square\)