# Related Rates (Applying Implicit Differentiation)

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

In this topic we show how implicit differentiation and the chain rule can be used to calculate the rate of change of one variable in terms of the rate of change of another variable (which may be more easily measured).

## Introduction to Related Rates

The procedure of solving a related rates problem is to find an equation that relates two quantities and then use the chain rule to differentiate both sides with respect to time. Then, from knowing the rate of change of one value at a point in time, we can calculate the rate of change of another quantity at that moment in time.

The equation which describes the application is often derived from a verbal description and is called a mathematical model of the problem.
For these equations we can relate different rates of change by using the chain rule and implicit differentiation.

More precisely, given a function $y=f(x)$ where both $x$ and $y$ are functions of time $t,$ we have $x=x(t)$ and $y=y(t)$; and so we can use the chain rule and implicit differentiation to determine $\frac{d y}{d t}$ and $\frac{d x}{d t}.$
Knowing one of these values we can calculate the other as demonstrated in the following examples.

Example. Find $\frac{d y}{d t}$ when $x=7$ given $y=5\sqrt{x+9}$ and $\frac{d x}{d t}=2.$

Solution. Differentiating with respect to $t$ and using the chain rule, \begin{align*} \frac{d y}{d t}=\frac{5}{2\sqrt{x+9}}\frac{d x}{d t}
\end{align*} and so when $x=7,$ \begin{align*} \left. \frac{d y}{d t} \right|_{x=7} =\frac{5}{2\sqrt{7+9}}(2) =\frac{5}{4}. \end{align*} as desired.

Example. Find $\frac{d y}{d t}$ when $x=1$ given $5 x y=10$ and $\frac{d x}{d t}=-2.$

Solution. Differentiating with respect to $t$ and using the chain rule, \begin{align*} x \frac{d y}{d t}+y \frac{d x}{d t}=0 \end{align*} and so when $x=1,$ then $y=2$ and $(1)\frac{d y}{d t}+(2)(-2) =0.$ Therefore, \begin{align*} \left. \frac{d y}{d t}\right|_{x=1}=4. \end{align*} as desired.

## General vs Specific Situation

Every related rates problem has a general situation (properties that hold true at every instant in time) and a specific situation (properties that hold true at a particular instant in time). Distinguishing between these two situations is often the key to successfully solving a related-rates problem. Here are a few guidelines for solving related rate problems:

• Read the problem carefully and identify all given quantities and unknown quantities to be determined. Introduce notation, make a sketch, and label the quantities.
• Write equations involving the variables whose rates of change either are given or are to be determined.
• Use implicit differentiation, by differentiating both sides with respect to he appropriate variable.
• Substitute known variables and known rates of change into the resulting equation to determine the missing rate of change.

Example. Model a water tank by a cone 40 ft high with a circular base of radius 20 ft at the top. Water is flowing into the tank at a constant rate of $80 \left.\text{ft}^3\right/\min .$ How fast is the water level rising when the water is 12 feet deep?

Solution. Let $x$ be the radius of the top circle of the body of water and $y$ its height. The radius of the top circle is 20, the height of the cone is 40 ft. By similar triangles, $20/40=x/y$ and so $x=\frac{1}{2}y.$ The volume of the body of water is \begin{align*} V=\frac{1}{3}\pi x^2y =\frac{1}{12}\pi y^3. \end{align*} Then \begin{align*} \frac{d V}{d t}=\frac{1}{4}\pi y^2 \frac{d y}{d t}. \end{align*} When $y=12,$ $x=6,$ and $\frac{d V}{d t}=80,$ we have \begin{align*}80=\frac{1}{4}\pi (12)^2 \frac{d y}{d t}\end{align*} and thus \begin{align*} \frac{d y}{d t} & =\frac{80(4)}{\pi (144)} \text{ft.}/\min \\[15px] & = \frac{20}{9 \pi } \text{ft.}/\min \\[15px] & \approx 0.71 \text{ft.}/\min . \end{align*} as desired.

Example. Air is being pumped into a spherical balloon at a rate of 4.5 cubic feet per minute. Find the rate of change of the radius when the radius is 2 feet.

Solution. Let $V$ be the volume of the balloon and let $r$ be the radius. Because the volume is increasing at a rate of 4.5 cubic feet per minute, we know that at time $t$ the rate of change of the volume is $$\frac{d V}{d t}=\frac{9}{2}.$$ The equation that relates the radius $r$ to the volume $V$ is $V=\frac{4}{3}\pi r^3.$ So we have $$\frac{d V}{d t}=4\pi r^2\frac{d r}{d t}$$ and by solving for $\frac{d r}{dt}$ we have \begin{align*} \frac{d r}{d t}=\frac{1}{4\pi r^2}\left(\frac{d V}{d t}\right). \end{align*} Finally, when $r=2,$ the rate of change of the radius is \begin{align*} \frac{d r}{d t}=\frac{1}{16 \pi }\left(\frac{9}{2}\right) \approx 0.09 \end{align*} foot per minute.

Example. A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 4 feet, at what rate is the total area $A$ of the disturbed water changing?

Solution. For the area of a circle we use $A=\pi r^2$ and we differentiate implicitly with respect to time $t$ leading to \begin{align*}\frac{d A}{d t}=2\pi r \frac{d r}{d t}\end{align*} by using the chain rule. Since $\frac{d r}{d t}=1$ and when $r=4$ we have \begin{align*} \frac{d A}{d t}=2\pi (4)(1) =8 \pi. \end{align*} as desired.

Example. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Solution. Let $x$ meters be the distance from the bottom of the ladder to the wall and $y$ meters the distance from the top of the ladder to the ground. Since $x$ and $y$ are functions of time and $\frac{d x}{d t}$= 1 ft/s we are asked to find $\frac{d y}{d t}$ when $x=6$ ft. The relationship between $x$ and $y$ is the Pythagorean Theorem, namely $x^2+y^2=100.$ Using implicit differentiation and the chain rule we have \begin{align*} 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0. \end{align*} By solving for $\frac{d y}{d t}$ we have \begin{align*} \frac{d y}{d t}=-\frac{x}{y}\frac{d x}{d t}. \end{align*} When $x=6$ then $y=8$, \begin{align*} \frac{d y}{ d t} & =\frac{-6}{8}(1) \\[15px] & =\frac{-3}{4} \text{ ft/s.} \end{align*} as desired.

Example. Car A is going west at 50 mi/h and car B is going north at 60 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is $0.3$ mi and car B is $0.4$ mi from the intersection?

Solution. At a given time $t,$ let $x$ be the distance from car $A$ to the point of intersection $P$ and let $y$ be the distance from car B to $P$ and let $z$ be the distance between the cars, where $x,$ $y,$ and $z$ are measured in miles. Since $x$ and $y$ are decreasing we take the derivatives to be negative and so we are given \begin{align*} \frac{d x}{d t}=-50 \text{ mi/h}\end{align*} and \begin{align*}\frac{d y}{d t}=-60\text{ mi/h}. \end{align*} To find $\frac{d z}{d t}$ we use the Pythagorean Theorem, namely $x^2+y^2=z^2$ and differentiate with respect to time $t.$ We have \begin{align*} 2 z \frac{d z}{d t}=2 x\frac{d x}{d t}+2 y\frac{d y}{d t} \end{align*} and by solving for $\frac{d z}{d t}$ we have \begin{align*} \frac{d z}{d t}=\frac{1}{z}\left(x\frac{d x}{d t}+ y\frac{d y}{d t}\right). \end{align*} Now when $x=0.3$ mi and $y=0.4$ mi we have $z=0.5$ mi and so \begin{align*} \frac{d z}{d t} & =\frac{1}{0.5}[0.3(-50)+0.4(-60)] \\[15px] & =-78 \text{ mi/h}. \end{align*} Therefore, the cars are approaching each other at a rate of 78 mi/h.

Example. A person 6 ft tall walks away from a streetlight at the rate of 5 ft/s. If the light is 18 ft above ground level, how fast is the person’s shadow lengthening?

Solution. Let $x$ be the length of the shadow and $y$ be the distance of the person from the street light. Using similar triangles, $x/6=(x+y)/18$ and by solving for $y$ we have $y=2x.$ Thus, $d y/d t=2 (d x/ d t)$ and given that $d y/ d t=5 \text{ft}/s$ we find that $d x/ d t=2.5 \text{ft}/s$ is the rate the shadow is lengthening.

Example. At noon, a ship sails due north from a point $P$ at 8 knots. Another ship, sailing at 12 knots, leaves the same point 1 h later on a course $60{}^{\circ}$ east of north. How fast is the distance between the ships increasing at 5 P.M.?

Solution. Let $A$ be the distance travelled by the first ship, and $B$ for the distance travelled by the second ship, $D$ for the distance between them, and $\theta$ the constant angle of $60{}^{\circ}.$ We need to find $\frac{d D}{d t}$ at $t=5.$ The equation that relates all the variables is the law of cosines and is $$D^2=A^2+B^2-2 A B \cos 60^{\circ}.$$ Using the chain rule and implicit differentiation we have \begin{align*} & 2 D \frac{d D}{d t} \\[15px] & = 2 A \frac{d A}{d t}+ 2B \frac{d B}{d t}-2\left(A \frac{d B}{d t}+B \frac{d A}{d t}\right) \left(\frac{1}{2}\right) \end{align*} since $\cos 60{}^{\circ}=\frac{1}{2}.$ Then at $t=5,$ $A=5(8)$ $=\text{40,}$ $B=12(4)$ $=\text{48,}$ $\frac{d A}{d t}=8,$ $\frac{d B}{ d t}=12,$ and \begin{align*} D & =\sqrt{40^2+48^2-2(40)(48)\left(\frac{1}{2}\right)} \\[15px] & =\sqrt{1984}. \end{align*} So we have, \begin{align*} \frac{d D}{d t} & =\frac{2 (40) (8)+ 2(48) (12)-(40) (12)-(48)(8)}{2 \sqrt{1984}} \\[15px] & =\frac{58}{\sqrt{31}} \text{ knots} \\[15px] & \approx 10.4171 \text{ knots}. \end{align*} as desired.

## Exercises on Related Rates

Exercise. Suppose that the radius $r$ and surface area $S=4\pi r^2$ of a sphere are differentiable functions of $t.$ Write an equation that relates $\frac{dS}{dt}$ to $\frac{dr}{dt}.$

Exercise. If $x$, $y,$ and $z$ are lengths of the edges of a rectangular box, the common length of the box’s diagonals is $s=\sqrt{x^2+y^2+z^2}.$ (a) Assuming that $x$, $y,$ and $z$ are differentiable functions of $t,$ how is $ds/dt$ related to $dx/dt,$ $dy/dt,$ and $dz/dt.$

Exercise. The length $l$ of a rectangle is decreasing at the rate of $\text{cm}/\sec$ while the width $w$ is increasing at the rate of $2 \text{cm}/\sec .$ When $l=12 \text{cm}$ and $w=5 \text{cm},$ find the rates of change of (a) the area, (b) the perimeter, (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing and which are increasing?

Exercise. The coordinates of a particle in the metric $x y$-plane are differentiable functions of time $t$ with $dx/dt=-1 m/s$ and $dy/dt=-5 m/s.$ How fast is the particle’s distance from the origin changing as it passes through the point $(5,12)?$

Exercise. A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of $5 \text{ft}/\sec .$ (a) How fast is the top of the ladder sliding down the wall then? (b) At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? (c) At what rate is the angle $\theta$ between the ladder and the ground changing then?

Exercise. Sand falls from a conveyor belt at the rate of $10 \left.m^3\right/\min$ onto the top of a conical pile. The height of the pile is always three-eights of the base diameter. How fast are the (a) height and (b) radius changing when the pile is $4 m$ high?

Exercise. Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to is surface area. Show that under thee circumstances the drop’s radius increase at a constant rate.

Exercise. A balloon is rising vertically above a level straight road at a constant rate of $1 \text{ft}/\sec .$ Just when the balloon is $65 \text{ft}$ above the ground, a bicycle moving at a constant rate of $17 \text{ft}/\sec$ passes under it. How fast is the distance $s(t)$ between the bicycle and balloon increasing 3 sec later?

Exercise. A man 6 ft all walks at a rate of $5 \text{ft}/\sec$ toward a streetlight that is 16 ft above the ground. At what rates the tip of his shadow moving? At what rate is the length of his shadow changing when he is 10 ft from the base of the light?

Exercise. Two commercial airplanes are flying at $40,000$ ft along straight-line courses that intersect at right angles. Plane $A$ is approaching the intersection point at a speed of 442 knots. Plane $B$ is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when $A$ is nautical miles from the intersection point and $B$ is 12 nautical miles from the intersection point?

Exercise. All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each side is (a) 1 centimeter and (b) 10 centimeters?

Exercise. A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep?

Exercise. A balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. What is the rate of change of the angle of elevation of the balloon from the observer when the balloon is 30 meters above the ground?

Exercise. A trough is 10 feet long and its ends have the shape of an isosceles triangles that are 3 feet across at the top and have a height of 1 foot. If the trough is filled with water at a rate of $2 \left.\text{ft}^3\right/\min ,$ how fast does the water level rise when the water is 6 inches
deep?