3.3 Related Rates
When two quantities are tied together by an equation, their rates of change are tied together too — differentiate the equation with respect to time.
For quantities that change over time, the rates of change are derivatives. If two quantities are related, their rates of change are related too. When a balloon is filled with air, both its radius and its volume increase, and $\dfrac{dV}{dt}$ and $\dfrac{dr}{dt}$ are related rates because $V$ is related to $r$. In this section we compute one rate of change from another.
Setting up Related-Rates Problems
✏️ Example 1 — Inflating a balloon
A spherical balloon is being filled with air at the constant rate of $2\ \text{cm}^3/\text{sec}$. How fast is the radius increasing when the radius is $3$ cm?
💡 Show solution▶
Step 1 The volume of a sphere of radius $r$ is $V=\tfrac43\pi r^3$. Both $V$ and $r$ are functions of time: $V(t)=\tfrac43\pi[r(t)]^3$.
Step 2 Differentiate with respect to $t$ using the chain rule: $$V^\prime(t)=4\pi[r(t)]^2\cdot r^\prime(t).$$
Step 3 We know $V^\prime(t)=2$, so $2=4\pi[r(t)]^2\,r^\prime(t)$, which gives $$r^\prime(t)=\frac{1}{2\pi[r(t)]^2}\ \text{cm/sec}.$$
Step 4 When $r=3$: $$r^\prime(t)=\frac{1}{18\pi}\approx0.0177\ \text{cm/sec}.$$ Notice the radius grows more and more slowly — the same air spreads over an ever larger surface.
🧭 Problem-Solving Strategy — Solving a related-rates problem
- Assign symbols to all variables involved in the problem. Draw a figure if applicable.
- State, in terms of the variables, the information that is given and the rate to be determined.
- Find an equation relating the variables introduced in step 1.
- Using the chain rule, differentiate both sides of that equation with respect to the independent variable (usually $t$).
- Substitute all known values into the equation from step 4, then solve for the unknown rate of change.
⚠️ Do not substitute too soon
If the value of a changing quantity is substituted before differentiating, that quantity behaves as a constant and its derivative disappears from the equation. Differentiate first (step 4), substitute second (step 5). Only genuine constants — like the fixed height of the plane in Example 2 — may be put in before differentiating.
Examples of the Process
✏️ Example 2 — An airplane flying at a constant elevation
An airplane is flying overhead at a constant elevation of $4000$ ft. A man is viewing the plane from a position $3000$ ft from the base of a radio tower. The airplane is flying horizontally away from the man at $600$ ft/sec. At what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?
💡 Show solution▶
Step 1 Let $x$ be the horizontal distance from the man to the point on the ground directly below the plane, and $s$ the distance from the man to the plane. The $4000$ ft height is constant, so it gets no variable. The three lengths form a right triangle.
Step 2 Given: $\dfrac{dx}{dt}=600$ ft/sec. Want: $\dfrac{ds}{dt}$ when $x=3000$ ft (the plane is over the tower).
Step 3 Pythagoras: $[x(t)]^2+4000^2=[s(t)]^2$.
Step 4 Differentiate (the constant $4000^2$ has derivative $0$): $$2x\frac{dx}{dt}=2s\frac{ds}{dt}\ \Longrightarrow\ x\frac{dx}{dt}=s\frac{ds}{dt}.$$
Step 5 At $x=3000$: $s=\sqrt{3000^2+4000^2}=5000$ ft. So $(3000)(600)=(5000)\dfrac{ds}{dt}$, and $$\frac{ds}{dt}=\frac{3000\cdot600}{5000}=360\ \text{ft/sec}.$$
✏️ Example 3 — Chapter opener — a rocket launch
A rocket is launched so that it rises vertically. A camera is positioned $5000$ ft from the launch pad. When the rocket is $1000$ ft above the launch pad, its velocity is $600$ ft/sec. Find the necessary rate of change of the camera\'s angle so that it stays focused on the rocket.
💡 Show solution▶
Step 1 Let $h$ be the height of the rocket and $\theta$ the angle between the camera lens and the ground.
Step 2 Want $\dfrac{d\theta}{dt}$ when $h=1000$ ft, given $\dfrac{dh}{dt}=600$ ft/sec.
Step 3 The right triangle gives $\tan\theta=\dfrac{h}{5000}$, that is $h=5000\tan\theta$.
Step 4 Differentiate: $\dfrac{dh}{dt}=5000\sec^2\theta\,\dfrac{d\theta}{dt}$.
Step 5 When $h=1000$ the hypotenuse is $\sqrt{1000^2+5000^2}=1000\sqrt{26}$, so $\sec^2\theta=\left(\dfrac{1000\sqrt{26}}{5000}\right)^2=\dfrac{26}{25}$. Then $$600=5000\cdot\frac{26}{25}\cdot\frac{d\theta}{dt}\ \Longrightarrow\ \frac{d\theta}{dt}=\frac{3}{26}\approx0.1154\ \text{rad/sec}.$$
✏️ Example 4 — Water draining from a funnel
Water is draining from the bottom of a cone-shaped funnel at the rate of $0.03\ \text{ft}^3/\text{sec}$. The height of the funnel is $2$ ft and the radius at the top is $1$ ft. At what rate is the height of the water changing when the height of the water is $\tfrac12$ ft?
💡 Show solution▶
Step 1 Let $h$ be the height of the water, $r$ the radius of its surface and $V$ its volume.
Step 2 Want $\dfrac{dh}{dt}$ when $h=\tfrac12$, given $\dfrac{dV}{dt}=-0.03\ \text{ft}^3/\text{sec}$ (negative, because water leaves).
Step 3 $V=\tfrac13\pi r^2h$. Similar triangles give $\dfrac{r}{h}=\dfrac12$, so $r=\dfrac h2$ and $$V=\frac13\pi\left(\frac h2\right)^2h=\frac{\pi}{12}h^3.$$ Eliminating $r$ first leaves only one changing variable.
Step 4 Differentiate: $\dfrac{dV}{dt}=\dfrac{\pi}{4}h^2\dfrac{dh}{dt}$.
Step 5 At $h=\tfrac12$: $-0.03=\dfrac\pi4\cdot\dfrac14\cdot\dfrac{dh}{dt}=\dfrac{\pi}{16}\dfrac{dh}{dt}$, so $$\frac{dh}{dt}=-\frac{0.48}{\pi}\approx-0.153\ \text{ft/sec}.$$
Exercises
I
Answer the following.
- 1 The side of a cube increases at a rate of $12$ m/sec. Find the rate at which the volume of the cube increases when the side of the cube is $4$ m.
- 2 The volume of a cube decreases at a rate of $10\ \text{m}^3/\text{sec}$. Find the rate at which the side of the cube changes when the side is $2$ m.
- 3 The radius of a circle increases at a rate of $2$ m/sec. Find the rate at which the area of the circle increases when the radius is $5$ m.
- 4 The radius of a sphere decreases at a rate of $3$ m/sec. Find the rate at which the surface area decreases when the radius is $10$ m.
- 5 The radius of a sphere increases at a rate of $1$ m/sec. Find the rate at which the volume increases when the radius is $20$ m.
- 6 The radius of a sphere is increasing at a rate of $9$ cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.
- 7 A vertical cylinder is leaking water at a rate of $1\ \text{ft}^3/\text{sec}$. If the cylinder has a height of $10$ ft and a radius of $1$ ft, at what rate is the height of the water changing when the height is $6$ ft?
- 8 A cylinder is leaking water, but you are unable to determine at what rate. The cylinder has a height of $2$ m and a radius of $2$ m. Find the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is $10$ m/min when the height is $1$ m.
- 9 A $10$-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of $2$ ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is $5$ ft from the wall?
Watch the shape change
💡 Show solutions▶
1 $V=s^3\Rightarrow\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}=3(16)(12)=576\ \text{m}^3/\text{sec}$.
2 $\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}\Rightarrow-10=3(4)\dfrac{ds}{dt}\Rightarrow\dfrac{ds}{dt}=-\dfrac56\ \text{m/sec}$ — the side shrinks at $\tfrac56$ m/sec.
3 $A=\pi r^2\Rightarrow\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(5)(2)=20\pi\approx62.83\ \text{m}^2/\text{sec}$.
4 $S=4\pi r^2\Rightarrow\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=8\pi(10)(-3)=-240\pi$. The surface area decreases at $240\pi\approx753.98\ \text{m}^2/\text{sec}$.
5 $V=\tfrac43\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}=4\pi(400)(1)=1600\pi\approx5026.5\ \text{m}^3/\text{sec}$.
6 $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. Equal rates means $\dfrac{dV}{dt}=\dfrac{dr}{dt}$, so $4\pi r^2=1$ and $r=\dfrac{1}{2\sqrt\pi}\approx0.2821$ cm. (The given $9$ cm/sec cancels — it does not affect the answer.)
7 $V=\pi r^2h=\pi h$ (with $r=1$ fixed). $\dfrac{dV}{dt}=\pi\dfrac{dh}{dt}\Rightarrow-1=\pi\dfrac{dh}{dt}\Rightarrow\dfrac{dh}{dt}=-\dfrac1\pi\approx-0.318$ ft/sec. For a cylinder this rate is the same at every height, so $h=6$ is not needed.
8 $V=\pi(2)^2h=4\pi h\Rightarrow\dfrac{dV}{dt}=4\pi\dfrac{dh}{dt}=4\pi(-10)=-40\pi$. Water leaks out at $40\pi\approx125.66\ \text{m}^3/\text{min}$.
9 Let $x$ be the distance of the bottom from the wall and $y$ the height of the top: $x^2+y^2=100$. Differentiate: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$. At $x=5$, $y=\sqrt{75}=5\sqrt3$ and $\dfrac{dy}{dt}=-2$, so $$\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}=-\frac{5\sqrt3}{5}(-2)=2\sqrt3\approx3.464\ \text{ft/sec}.$$
II
Ladders, shadows and moving vehicles. Draw a picture, name the changing lengths, and relate them before differentiating.
- 1 A $13$-ft ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall at $4$ ft/sec. How fast is the top of the ladder sliding down the wall when the bottom is $5$ ft from the wall?
- 2 A man $6$ ft tall walks away from a $15$-ft street lamp at $5$ ft/sec. How fast is the length of his shadow increasing?
- 3 For the man in Problem 2, how fast is the tip of his shadow moving along the ground?
- 4 Two cars leave an intersection at the same time. One drives east at $80$ km/h and the other drives north at $60$ km/h. How fast is the distance between them increasing $2$ hours later?
- 5 At noon, ship $A$ is $100$ km west of ship $B$. Ship $A$ sails east at $35$ km/h and ship $B$ sails north at $25$ km/h. How fast is the distance between the ships changing at $4{:}00$ p.m.?
Ladders, shadows and vehicles in motion
💡 Show solutions▶
1 Let $x$ = distance of the bottom from the wall and $y$ = height of the top: $x^2+y^2=169$. Differentiate: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$. At $x=5$, $y=\sqrt{169-25}=12$, so $$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{5}{12}(4)=-\frac53\ \text{ft/sec}.$$ The top slides down at $\tfrac53\approx1.667$ ft/sec.
2 Let $x$ = distance from the lamp to the man and $s$ = length of the shadow. Similar triangles (lamp and man): $\dfrac{15}{x+s}=\dfrac{6}{s}\Rightarrow15s=6x+6s\Rightarrow s=\tfrac23x$. Hence $\dfrac{ds}{dt}=\tfrac23\dfrac{dx}{dt}=\tfrac23(5)=\dfrac{10}{3}\approx3.333$ ft/sec — the same at every position.
3 The tip is at distance $x+s$ from the lamp, so its speed is $\dfrac{d}{dt}(x+s)=\dfrac{dx}{dt}+\dfrac{ds}{dt}=5+\dfrac{10}{3}=\dfrac{25}{3}\approx8.333$ ft/sec.
4 After $t$ hours, $x=80t$ and $y=60t$, and $D^2=x^2+y^2$. Differentiate: $D\dfrac{dD}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$. At $t=2$: $x=160$, $y=120$, $D=200$, so $$\frac{dD}{dt}=\frac{160(80)+120(60)}{200}=\frac{20\,000}{200}=100\ \text{km/h}.$$ (In fact $D=100t$, so the rate is $100$ km/h at every time.)
5 Put $B$'s noon position at the origin. After $t$ hours, $A$ is at $x=-100+35t$ on the east–west line and $B$ is at $y=25t$; $D^2=x^2+y^2$ gives $D\dfrac{dD}{dt}=x(35)+y(25)$. At $t=4$: $x=40$, $y=100$, $D=\sqrt{1600+10\,000}=20\sqrt{29}$. $$\frac{dD}{dt}=\frac{40(35)+100(25)}{20\sqrt{29}}=\frac{3900}{20\sqrt{29}}=\frac{195}{\sqrt{29}}\approx36.21\ \text{km/h}.$$ The distance is increasing.
III
Circles, spheres and cones. For a cone, use similar triangles to eliminate the radius before differentiating.
- 1 A stone dropped into a still pond sends out a circular ripple whose radius increases at $60$ cm/sec. How fast is the area inside the ripple increasing when the radius is $120$ cm?
- 2 A spherical balloon is inflated at $100\ \text{cm}^3/\text{sec}$. How fast is its radius increasing when the radius is $25$ cm?
- 3 For the balloon in Problem 2, how fast is the surface area increasing at that moment?
- 4 A water tank is an inverted right circular cone with top radius $2$ m and height $4$ m. Water is pumped in at $2\ \text{m}^3/\text{min}$. How fast is the water level rising when the water is $3$ m deep?
- 5 Gravel falls from a conveyor belt at $30\ \text{ft}^3/\text{min}$ and forms a conical pile whose base diameter always equals its height. How fast is the height of the pile increasing when the pile is $10$ ft high?
Growing circles, spheres and cones
💡 Show solutions▶
1 $A=\pi r^2\Rightarrow\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(120)(60)=14\,400\pi\approx45\,239\ \text{cm}^2/\text{sec}$.
2 $V=\tfrac43\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. So $100=4\pi(625)\dfrac{dr}{dt}$ and $\dfrac{dr}{dt}=\dfrac{100}{2500\pi}=\dfrac{1}{25\pi}\approx0.0127$ cm/sec.
3 $S=4\pi r^2\Rightarrow\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=8\pi(25)\cdot\dfrac{1}{25\pi}=8\ \text{cm}^2/\text{sec}$. (Shortcut: $\dfrac{dS}{dt}=\dfrac{2}{r}\dfrac{dV}{dt}$.)
4 By similar triangles the water surface has radius $r$ with $\dfrac{r}{h}=\dfrac24$, so $r=\dfrac h2$ and $V=\tfrac13\pi r^2h=\dfrac{\pi h^3}{12}$. Then $\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}$, and at $h=3$: $2=\dfrac{9\pi}{4}\dfrac{dh}{dt}\Rightarrow\dfrac{dh}{dt}=\dfrac{8}{9\pi}\approx0.283$ m/min.
5 Diameter $=h$ means $r=\dfrac h2$, so again $V=\dfrac{\pi h^3}{12}$ and $\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}$. At $h=10$: $30=25\pi\dfrac{dh}{dt}\Rightarrow\dfrac{dh}{dt}=\dfrac{6}{5\pi}\approx0.382$ ft/min.
IV
Angles of elevation, distances, and one conceptual check.
- 1 A hot-air balloon rises vertically at $3$ m/sec from a point on level ground $60$ m from an observer. How fast is the angle of elevation of the balloon increasing when the balloon is $80$ m high?
- 2 A plane flying horizontally at an altitude of $5$ km and a speed of $600$ km/h passes directly over a radar station and flies on. How fast is the angle of elevation from the station to the plane changing when the angle is $\dfrac{\pi}{3}$? Give the answer in rad/min.
- 3 A particle moves along the curve $y=\sqrt x$ so that its $x$-coordinate increases at $3$ cm/sec. How fast is its distance from the origin changing when it is at the point $(4,2)$?
- 4 A spotlight on the ground shines on a wall $12$ m away. A man $2$ m tall walks from the spotlight toward the wall at $1.6$ m/sec. How fast is the height of his shadow on the wall changing when he is $4$ m from the wall?
- 5 A student solves Exercise II.1 by first substituting $x=5$ into $x^2+y^2=169$ to get $25+y^2=169$, then differentiates and obtains $2y\dfrac{dy}{dt}=0$, concluding that the top of the ladder is not moving. Explain the error.
💡 Show solutions▶
1 Let $h$ be the height of the balloon: $\tan\theta=\dfrac{h}{60}$. Differentiate: $\sec^2\theta\,\dfrac{d\theta}{dt}=\dfrac1{60}\dfrac{dh}{dt}$. At $h=80$ the line of sight has length $100$, so $\sec^2\theta=\left(\tfrac{100}{60}\right)^2=\tfrac{25}{9}$ and $$\frac{d\theta}{dt}=\frac{9}{25}\cdot\frac{3}{60}=\frac{9}{500}=0.018\ \text{rad/sec}.$$
2 Let $x$ be the horizontal distance from the station: $\tan\theta=\dfrac5x$, with $\dfrac{dx}{dt}=600$. Differentiate: $\sec^2\theta\,\dfrac{d\theta}{dt}=-\dfrac{5}{x^2}\dfrac{dx}{dt}$. At $\theta=\tfrac\pi3$: $x=\dfrac{5}{\sqrt3}$, $x^2=\dfrac{25}{3}$, $\sec^2\theta=4$. So $$\frac{d\theta}{dt}=-\frac{5\cdot600}{4\cdot\frac{25}{3}}=-90\ \text{rad/h}=-\frac32\ \text{rad/min}.$$ The angle decreases at $1.5$ rad/min.
3 Distance $D=\sqrt{x^2+y^2}=\sqrt{x^2+x}$ on this curve. Then $\dfrac{dD}{dt}=\dfrac{2x+1}{2\sqrt{x^2+x}}\dfrac{dx}{dt}$. At $x=4$: $\dfrac{dD}{dt}=\dfrac{9}{2\sqrt{20}}(3)=\dfrac{27}{4\sqrt5}=\dfrac{27\sqrt5}{20}\approx3.019$ cm/sec.
4 Let $d$ = distance from the spotlight to the man and $y$ = shadow height on the wall. Similar triangles: $\dfrac{y}{12}=\dfrac{2}{d}\Rightarrow y=\dfrac{24}{d}$. Then $\dfrac{dy}{dt}=-\dfrac{24}{d^2}\dfrac{dd}{dt}$. When he is $4$ m from the wall, $d=8$: $\dfrac{dy}{dt}=-\dfrac{24}{64}(1.6)=-0.6$ m/sec. The shadow shrinks at $0.6$ m/sec.
5 The student substituted a changing quantity before differentiating. Setting $x=5$ turns $x$ into a constant, so the term $2x\dfrac{dx}{dt}$ disappears — but $x$ is changing at $4$ ft/sec. Differentiate $x^2+y^2=169$ first, then substitute $x=5$, $y=12$, $\dfrac{dx}{dt}=4$ to get $\dfrac{dy}{dt}=-\tfrac53$ ft/sec. Only quantities that are constant throughout (here the ladder length $13$) may be put in before differentiating.
🔬 Research–Teaching NexusResearch-based · optional, about 30 minutes
Research question. When water drips steadily into a cone, does the water level rise at the rate the related-rates formula predicts?
Why it matters. Chemical and water engineers use related rates to design tanks, funnels and irrigation channels, including those that feed falaj systems.
What to do.
- Make a paper cone (or use a funnel with the end closed). Measure its top radius $R$ and height $H$.
- Fill it from a bottle with a small hole at a steady rate $Q$ (measure $Q$ in cm$^3$/s with a measuring cup).
- Record the water height $h$ every 10 s.
- Using $V=\frac{\pi}{3}\left(\frac{R}{H}\right)^2h^3$, predict $\frac{dh}{dt}=\frac{Q}{\pi (R/H)^2h^2}$ at two heights. Compare with your measured rates.
Tools: Paper cup, bottle, ruler, phone timer
Hand in: A half-page note: the predicted and measured $\frac{dh}{dt}$ at two heights and one source of error.
Going further. Anywhere: repeat with a cylindrical glass and explain why $\frac{dh}{dt}$ is then constant.
Instructor note: Students should notice the level rises fast at first and slows down, because $\frac{dh}{dt}$ is proportional to $\frac{1}{h^2}$.
🧪 Self-check quiz
0/5 correct
Q1. In a related-rates problem, when should the numerical values of changing quantities be substituted?
Q2. For a balloon filled at $2\ \text{cm}^3/\text{s}$, $\dfrac{dr}{dt}$ when $r=3$ is:
Q3. In the airplane problem, $\dfrac{ds}{dt}$ when $x=3000$ is:
Q4. For a cube, $V=s^3$. If $\dfrac{ds}{dt}=12$ at $s=4$, then $\dfrac{dV}{dt}=$
Q5. A 10-ft ladder\'s top slides down at a constant 2 ft/s. As the bottom moves out toward 10 ft from the wall, the bottom\'s speed: