MCV4U: Related Rates

How It Works

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

1. Easy

   The radius of a circle grows at $3$ cm/s. How fast is the area growing when the radius is $5$ cm?

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   1. Given $\frac{dr}{dt} = 3$. Want $\frac{dA}{dt}$ when $r = 5$.
   2. Link: $A = \pi r^2$.
   3. Differentiate with respect to $t$: $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$.
   4. Substitute: $\frac{dA}{dt} = 2\pi(5)(3) = 30\pi$.

   Answer: $30\pi \approx 94.2$ cm²/s

2. Easy

   A snowball melts so that its volume drops by $12$ cm³/min. How fast is its radius shrinking when the radius is $3$ cm?

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   1. Given $\frac{dV}{dt} = -12$ (negative: it is shrinking). Want $\frac{dr}{dt}$ when $r = 3$.
   2. Link: $V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.
   3. Substitute: $-12 = 4\pi(3)^2\frac{dr}{dt} = 36\pi\frac{dr}{dt}$, so $\frac{dr}{dt} = -\frac{1}{3\pi}$.

   Answer: The radius shrinks at $\frac{1}{3\pi} \approx 0.106$ cm/min.

3. Medium

   Car A is $4$ km north of an intersection, driving south toward it at $60$ km/h. Car B is $3$ km east of the intersection, driving east at $50$ km/h. How fast is the distance between them changing?

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   1. Let $y$ be A's distance north, $x$ be B's distance east and $r$ the distance between them. Then $\frac{dy}{dt} = -60$ and $\frac{dx}{dt} = 50$.
   2. Link: $r^2 = x^2 + y^2$. Right now $r = \sqrt{3^2 + 4^2} = 5$.
   3. Differentiate: $2r\frac{dr}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$, so $r\frac{dr}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$.
   4. Substitute: $5\frac{dr}{dt} = 3(50) + 4(-60) = -90$, so $\frac{dr}{dt} = -18$.

   Answer: The distance is decreasing at $18$ km/h.

4. Medium

   A tank is an upside-down cone with a top radius of $2$ m and a depth of $6$ m. Water pours in at $3$ m³/min. How fast is the water level rising when the water is $4$ m deep?

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   1. Given $\frac{dV}{dt} = 3$. Want $\frac{dh}{dt}$ when $h = 4$.
   2. Similar triangles: the water's radius and height keep the tank's ratio, $\frac{r}{h} = \frac{2}{6}$, so $r = \frac{h}{3}$.
   3. Volume in $h$ alone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}$.
   4. Differentiate: $\frac{dV}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt}$.
   5. Substitute: $3 = \frac{16\pi}{9}\frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{27}{16\pi}$.

   Answer: $\frac{27}{16\pi} \approx 0.537$ m/min

5. Test level

   A spotlight on the ground shines on a wall $12$ m away. A $2$ m tall student walks from the spotlight straight toward the wall at $1.6$ m/s. How fast is the student's shadow on the wall shrinking when the student is $4$ m from the wall?

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   1. Let $x$ be the student's distance from the spotlight and $H$ the shadow's height on the wall. Given $\frac{dx}{dt} = 1.6$. Want $\frac{dH}{dt}$ when $x = 12 - 4 = 8$.
   2. Similar triangles, light to student and light to wall: $\frac{H}{12} = \frac{2}{x}$, so $H = \frac{24}{x}$.
   3. Differentiate: $\frac{dH}{dt} = -\frac{24}{x^2}\frac{dx}{dt}$.
   4. Substitute: $\frac{dH}{dt} = -\frac{24}{64}(1.6) = -0.6$.

   Answer: The shadow shrinks at $0.6$ m/s.

Learn It on Martian Lab

Martian Lab teaches this in the Related Rates lesson of its MCV4U Optimization unit. Watch it, then take the Optimization practice tests to see if it stuck.

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