Topic 12

Related Rates

t

5 kinds of question. Open the one you are stuck on.

Solve a related rates problem with distances (Pythagoras)

e.g. “ A ladder slides down a wall \dots how fast is the top moving? ”

▾

1
$Draw the right triangle; label the changing sides with letters$ $Measure each moving point from a fixed point on its path (the wall's foot, the intersection), so its speed is that length's rate. Lengths that never change (the ladder, a fixed height) go on the picture as numbers.$
2
$List the given and wanted rates, with signs$ $\Rightarrow$
3
$x^2 + y^2 = z^2$ $0$
4
$t$ $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 2z\dfrac{dz}{dt}$
5
$Find the missing side at the instant, plug in, solve$ $If the instant is a time, first find each side at that time (distance = speed \times time). Use Pythagoras for the missing side. Plug numbers in only now, after differentiating.$

Remember: Picture, rates, Pythagoras, d/dt, plug in.

See it in a problem3 worked examples▾

10

Picturing the problem is part of the solution. Try to sketch it yourself first; on the exam there's no button.

1
$Draw the right triangle; label the changing sides with letters$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dx}{dt} = 0.4$
3
$x^2 + y^2 = z^2$ $x^2 + y^2 = 10^2$
4
$t$ $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$
5
$Find the missing side at the instant, plug in, solve$ $x = 6$

Try it yourself first.

\dfrac{dy}{dt} = -0.3

30

1
$Draw the right triangle; label the changing sides with letters$ $y$
2
$List the given and wanted rates, with signs$ $\dfrac{dy}{dt} = +50$
3
$x^2 + y^2 = z^2$ $x^2 + y^2 = z^2$
4
$t$ $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 2z\dfrac{dz}{dt}$
5
$Find the missing side at the instant, plug in, solve$ $z = \sqrt{40^2 + 30^2} = 50$

Try it yourself first.

\dfrac{dz}{dt} = -18

30

1
$Draw the right triangle; label the changing sides with letters$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dx}{dt} = 2$
3
$x^2 + y^2 = z^2$ $x^2 + 30^2 = z^2$
4
$t$ $2x\dfrac{dx}{dt} = 2z\dfrac{dz}{dt}$
5
$Find the missing side at the instant, plug in, solve$ $z = 50$

Try it yourself first.

\dfrac{dz}{dt} = 1.6

Common mistakes3▾

Plugging the instant's numbers in before differentiating: $6^2 + y^2 = 100$ has no $\dfrac{dx}{dt}$ left in it.
Dropping the sign: a side that shrinks has a negative rate, and a negative answer means "decreasing".
Forgetting the missing side: you need $x$ , $y$ and $z$ at the instant, so find the third one first.

Solve a related rates problem with an angle

e.g. “ How fast is the angle (of elevation, of the camera) changing? ”

▾

1
$\theta$ $Mark which side is fixed and which one moves. A moving point (the spot, the climber) is measured from a fixed point on its path.$
2
$List the rates with signs; angles in radians$ $1$
3
$\theta$ $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$
4
$t$ $\dfrac{d}{dt}\tan\theta = \sec^2\theta\,\dfrac{d\theta}{dt}$
5
$Find the trig values at the instant, plug in, solve$ $\sec\theta$

Remember: Pick the ratio that uses the fixed side.

See it in a problem3 worked examples▾

0.4

Picturing the problem is part of the solution. Try to sketch it yourself first; on the exam there's no button.

1
$\theta$ $y$
2
$List the rates with signs; angles in radians$ $\dfrac{dy}{dt} = 0.4$
3
$\theta$ $\tan\theta = \dfrac{y}{8}$
4
$t$ $\sec^2\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{8}\dfrac{dy}{dt}$
5
$Find the trig values at the instant, plug in, solve$ $y = 6$

Try it yourself first.

\dfrac{d\theta}{dt} = 0.032

2

1
$\theta$ $x$
2
$List the rates with signs; angles in radians$ $\dfrac{d\theta}{dt} = 3 \cdot 2\pi = 6\pi$
3
$\theta$ $\tan\theta = \dfrac{x}{2}$
4
$t$ $\dfrac{dx}{dt} = 2\sec^2\theta\,\dfrac{d\theta}{dt}$
5
$Find the trig values at the instant, plug in, solve$ $x = 1$

Try it yourself first.

\dfrac{dx}{dt} = 15\pi \approx 47.1

100

1
$\theta$ $h$
2
$List the rates with signs; angles in radians$ $\dfrac{d\theta}{dt} = -0.05$
3
$\theta$ $\sin\theta = \dfrac{h}{100}$
4
$t$ $\dfrac{dh}{dt} = 100\cos\theta\,\dfrac{d\theta}{dt}$
5
$Find the trig values at the instant, plug in, solve$ $\theta = \dfrac{\pi}{6}$

Try it yourself first.

\dfrac{dh}{dt} = -2.5\sqrt{3} \approx -4.33

Common mistakes3▾

Forgetting the chain rule: $\dfrac{d}{dt}\tan\theta$ is $\sec^2\theta\,\dfrac{d\theta}{dt}$ , not $\sec^2\theta$ .
Using rev/min as rad/min. One revolution is $2\pi$ radians.
Calculator in degrees. $\cos(\pi/6)$ means radians.

Solve a related rates problem with a tank or volume

e.g. “ Water is pumped into (drains from) a cylindrical / rectangular tank \dots ”

▾

1
$h$ $r$
2
$\dfrac{dV}{dt}$ $\Rightarrow \dfrac{dV}{dt} < 0$
3
$Write the volume of the water's shape$ $V = \pi r^2 h$
4
$h$ $\dfrac{r}{h} = \dfrac{R}{H}$
5
$t$
6
$Plug in the instant's values and solve$

Remember: Shape, one variable, d/dt, plug in.

See it in a problem3 worked examples▾

2

1
$h$ $h$
2
$\dfrac{dV}{dt}$ $\dfrac{dV}{dt} = 500$
3
$Write the volume of the water's shape$ $V = \pi r^2 h$
4
$h$ $r = 2$
5
$t$ $\dfrac{dV}{dt} = 4\pi\dfrac{dh}{dt}$
6
$Plug in the instant's values and solve$ $h$

Try it yourself first.

\dfrac{dh}{dt} = \dfrac{1}{8\pi} \approx 0.040

10

1
$h$ $h$
2
$\dfrac{dV}{dt}$ $\dfrac{dV}{dt} = -2$
3
$Write the volume of the water's shape$ $V = \dfrac{1}{3}\pi r^2 h$
4
$h$ $\dfrac{r}{h} = \dfrac{4}{10}$
5
$t$ $\dfrac{dV}{dt} = \dfrac{4\pi h^2}{25}\dfrac{dh}{dt}$
6
$Plug in the instant's values and solve$ $h = 5$

Try it yourself first.

\dfrac{dh}{dt} = -\dfrac{1}{2\pi}

2

1
$h$ $h_B$
2
$\dfrac{dV}{dt}$ $\dfrac{dh_B}{dt} = 0.3$
3
$Write the volume of the water's shape$ $V_B = 2 \cdot 5 \cdot h_B$
4
$h$ $V_B = 10h_B$
5
$t$ $\dfrac{dV}{dt} = 10\dfrac{dh_B}{dt}$
6
$Plug in the instant's values and solve$ $\dfrac{dV}{dt} = 10(0.3) = 3$

Try it yourself first.

\dfrac{dh_C}{dt} = \dfrac{3}{4\pi} \approx 0.239

Common mistakes3▾

Differentiating the cone with $r$ still in it: $V = \tfrac{1}{3}\pi r^2 h$ has two changing variables. Replace $r$ first.
Treating a cylinder's fixed radius as a variable: $r = 2$ is a number, so $V = 4\pi h$ .
Liters vs cubic meters, or a missing minus sign for draining.

Solve a related rates problem when the formula is given

P = k\,x^a y^b

▾

1
$Copy the formula; mark which quantities change$ $0$
2
$List the given and wanted rates, with signs$ $a$
3
$Differentiate the formula with respect to the changing variable$ $t$
4
$Plug in the instant's values$ $Find any missing value from the formula itself.$
5
$Solve for the wanted rate; state its units$ $\dfrac{dE}{dL} = \dfrac{dE/da}{dL/da}$

Remember: The formula is the equation — just d/dt it.

See it in a problem3 worked examples▾

x

1
$Copy the formula; mark which quantities change$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dx}{dt} = 3$
3
$Differentiate the formula with respect to the changing variable$ $0 = \dfrac{4}{3}x^{-1/3}y^{1/3}\dfrac{dx}{dt} + \dfrac{2}{3}x^{2/3}y^{-2/3}\dfrac{dy}{dt}$
4
$Plug in the instant's values$ $x^{-1/3} = \tfrac{1}{4}$
5
$Solve for the wanted rate; state its units$ $\dfrac{dy}{dt} = -\dfrac{81}{32} \approx -2.53$

Try it yourself first.

\dfrac{dy}{dt} = -\dfrac{81}{32}

PV = C

1
$Copy the formula; mark which quantities change$ $P$
2
$List the given and wanted rates, with signs$ $\dfrac{dP}{dt} = 20$
3
$Differentiate the formula with respect to the changing variable$ $\dfrac{dP}{dt}V + P\dfrac{dV}{dt} = 0$
4
$Plug in the instant's values$ $20(600) + 150\dfrac{dV}{dt} = 0$
5
$Solve for the wanted rate; state its units$ $\dfrac{dV}{dt} = -\dfrac{12000}{150} = -80$

Try it yourself first.

\dfrac{dV}{dt} = -80

L

1
$Copy the formula; mark which quantities change$ $E$
2
$List the given and wanted rates, with signs$ $\dfrac{dE}{dL}$
3
$Differentiate the formula with respect to the changing variable$ $\dfrac{dL}{da} = 2e^{1/a} + 2a e^{1/a}\left(-\dfrac{1}{a^2}\right) = 2e^{1/a}\left(1 - \dfrac{1}{a}\right)$
4
$Plug in the instant's values$ $a = 2$
5
$Solve for the wanted rate; state its units$ $\dfrac{dE}{dL} = \dfrac{2e}{e^{1/2}} = 2\sqrt{e} \approx 3.30$

Try it yourself first.

\dfrac{dE}{dL} = 2\sqrt{e} \approx 3.30

Common mistakes3▾

Forgetting the product rule when both $x$ and $y$ change: $\dfrac{d}{dt}(x^a y^b)$ has two terms.
Treating the constant $P$ as changing: "production is kept constant" gives $\dfrac{dP}{dt} = 0$ .
Dividing $E$ by $L$ instead of their derivatives: $\dfrac{dE}{dL}$ is $\dfrac{dE/da}{dL/da}$ , not $\dfrac{E}{L}$ .

Solve a related rates problem with shadows (similar triangles)

e.g. “ A person walks away from (toward) a lamp post \dots how fast is the shadow lengthening? ”

▾

1
$Draw the light, the person and the shadow; measure from the post$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dy}{dt}$
3
$y$ $\dfrac{\text{lamp}}{y} = \dfrac{\text{person}}{y - x}$
4
$t$
5
$Plug in the instant's values and solve$ $\dfrac{dy}{dt}$

Remember: Measure from the post. Two triangles, one ratio, d/dt.

See it in a problem3 worked examples▾

6

Picturing the problem is part of the solution. Try to sketch it yourself first; on the exam there's no button.

1
$Draw the light, the person and the shadow; measure from the post$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dx}{dt} = 1.4$
3
$y$ $\dfrac{6}{y} = \dfrac{1.8}{y - x} \Rightarrow 6y - 6x = 1.8y \Rightarrow 4.2y = 6x \Rightarrow y = \dfrac{10}{7}x$
4
$t$ $\dfrac{dy}{dt} = \dfrac{10}{7}\dfrac{dx}{dt}$
5
$Plug in the instant's values and solve$ $\dfrac{dy}{dt} = \dfrac{10}{7}(1.4) = 2$

Try it yourself first.

0.6

1

1
$Draw the light, the person and the shadow; measure from the post$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dx}{dt} = -1.2$
3
$y$ $\dfrac{5}{y} = \dfrac{1}{y - x} \Rightarrow 5y - 5x = y \Rightarrow y = \dfrac{5}{4}x$
4
$t$ $\dfrac{dy}{dt} = \dfrac{5}{4}\dfrac{dx}{dt}$
5
$Plug in the instant's values and solve$ $\dfrac{dy}{dt} = \dfrac{5}{4}(-1.2) = -1.5$

Try it yourself first.

\dfrac{d}{dt}(y - x) = -0.3

12

1
$Draw the light, the person and the shadow; measure from the post$ $x$
2
$List the given and wanted rates, with signs$ $\dfrac{dx}{dt} = 1.6$
3
$y$ $\dfrac{2}{x} = \dfrac{H}{12} \Rightarrow H = \dfrac{24}{x}$
4
$t$ $\dfrac{dH}{dt} = -\dfrac{24}{x^2}\dfrac{dx}{dt}$
5
$Plug in the instant's values and solve$ $4$

Try it yourself first.

\dfrac{dH}{dt} = -0.6

Common mistakes3▾

Answering the tip's speed with the shadow's rate (or the reverse): the tip moves at $\dfrac{dy}{dt}$ , the shadow's length changes at $\dfrac{dy}{dt} - \dfrac{dx}{dt}$ .
Forcing in a distance given "at the instant": when $y$ is a constant multiple of $x$ , the rate does not depend on it — it is a distractor.
Mismatched ratios: the big triangle's base is $y$ (post to tip), the small one's is $y - x$ (person to tip).

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Solve a related rates problem with distances (Pythagoras)

Solve a related rates problem with an angle

Solve a related rates problem with a tank or volume

Solve a related rates problem when the formula is given

Solve a related rates problem with shadows (similar triangles)