# 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.

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## 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.

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:

$R \sin \theta = \tan \alpha (R \cos \theta - d).$

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 $\alpha$. $_\square$

**Cite as:**Related Rates of Change.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/related-rates-of-change/