Solving Related Rates Problems

Distances

Example 1

A delivery drone is

30

meters above its landing pad and descending straight down at

4

meters per second. A person

40

meters from the pad is walking directly away from it at

1.5

meters per second. Find the rate at which the distance between the drone and the person is changing at that moment.

The drone descends toward the pad while the person walks away from it along the ground.

Let

y

be the drone's height,

x

the person's distance from the pad, and

z

the distance between them. Then

\frac{dy}{dt} = -4

, because

y

is shrinking, and

\frac{dx}{dt} = 1.5

. The pad, the drone, and the person form a right triangle:

\begin{aligned} z^2 &= x^2 + y^2, \\[4pt] z\,\frac{dz}{dt} &= x\,\frac{dx}{dt} + y\,\frac{dy}{dt}. \end{aligned}

At this moment

z = \sqrt{40^2 + 30^2} = 50

, so

50\,\frac{dz}{dt} = 40(1.5) + 30(-4) = -60.

\frac{dz}{dt} = -1.2

: the distance between the drone and the person is decreasing at

1.2

meters per second. The person walks away, but the drone's descent outweighs it, so the gap shrinks.

Example 2

At one moment a kite is

60

meters above the ground with

100

meters of string out. It's rising at

0.5

meter per second, and the wind carries it horizontally away from the person holding the string at

2

meters per second. (a) Find how fast the string must be let out at that moment. (b) Find the rate at which the angle

\theta

between the string and the ground is changing at that moment.

The kite at height y, horizontal distance x, and string length s, with s² = x² + y².

(a) Here

x

is the horizontal distance,

y

the height, and

s

the string length. Differentiating

s^2 = x^2 + y^2

as in Example 1 gives

s\,\frac{ds}{dt} = x\,\frac{dx}{dt} + y\,\frac{dy}{dt}.

With

x = \sqrt{100^2 - 60^2} = 80

\frac{ds}{dt} = \frac{80(2) + 60(0.5)}{100} = 1.9.

The string must be let out at

1.9

meters per second.

(b) From the triangle,

\sin\theta = \frac{y}{s}

, and both

y

and

s

change. By the quotient rule,

\cos\theta\,\frac{d\theta}{dt} = \frac{s\,\frac{dy}{dt} - y\,\frac{ds}{dt}}{s^2}.

At this moment

\cos\theta = \frac{80}{100} = 0.8

, so

\frac{d\theta}{dt} = \frac{100(0.5) - 60(1.9)}{100^2 (0.8)} = -0.008.

The angle is decreasing at

0.008

radian per second. The kite is rising, but it drifts out fast enough that the string flattens.

Angles

Example 3

A lighthouse stands

1

kilometer from a straight shore, and its beam turns at

3

revolutions per minute. Let

P

be the point on the shore nearest the lighthouse. Find the speed of the spot of light along the shore when the spot is

2

kilometers from

P

The beam makes angle θ with the perpendicular from the lighthouse to the shore, and meets the shore x kilometers from P.

Let

\theta

be the angle between the beam and the perpendicular to the shore, and

x

the spot's distance from

P

. Then

x = \tan\theta

, so

\frac{dx}{dt} = \sec^2\theta\,\frac{d\theta}{dt}.

Three revolutions per minute is

\frac{d\theta}{dt} = 6\pi

radians per minute. When

x = 2

\sec^2\theta = 1 + \tan^2\theta = 5

, so

\frac{dx}{dt} = 5 \cdot 6\pi = 30\pi \approx 94.248

kilometers per minute.

A point on a curve

Example 4

A particle travels along the curve

y = \sqrt{2x}

so that

\frac{dx}{dt} = 2

centimeters per second. Find the rate at which its distance

D

from the point

(5, 0)

is changing when

x = 2

The particle at (2, 2) on y = √(2x), and its distance D from (5, 0).

Since

y^2 = 2x

on the curve,

\begin{aligned} D^2 &= (x - 5)^2 + y^2 \\[4pt] &= (x - 5)^2 + 2x. \end{aligned}

Differentiating,

2D\,\frac{dD}{dt} = (2x - 8)\,\frac{dx}{dt}

. When

x = 2

, the particle is at

(2, 2)

and

D = \sqrt{9 + 4} = \sqrt{13}

, so

\frac{dD}{dt} = \frac{(-4)(2)}{2\sqrt{13}} = -\frac{4}{\sqrt{13}} \approx -1.109.

The distance is decreasing at about

1.109

centimeters per second, so the particle is getting closer to

(5, 0)

. It travels up and to the right, not straight at

(5, 0)

, but each step still brings it nearer.

Similar triangles

Example 5

A work lamp hangs from a rope and is being lowered at

0.5

meter per second. A

2

-meter post stands

6

meters from the point directly below the lamp. Find the rate at which the post's shadow is lengthening (a) when the lamp is

8

meters high and (b) when it's

4

meters high.

The ray from the lamp past the top of the post ends at the tip of the shadow. The two right triangles are similar.

(a) Let

L

be the lamp's height and

s

the length of the shadow, so

\frac{dL}{dt} = -0.5

. The big triangle (the lamp, height

L

, base

6 + s

) and the small one (the post, height

2

, base

s

) are similar:

\frac{L}{6 + s} = \frac{2}{s}, \quad \text{so} \quad Ls = 12 + 2s.

Differentiating both sides, with the product rule on the left,

\begin{aligned} \frac{dL}{dt}\,s + L\,\frac{ds}{dt} &= 2\,\frac{ds}{dt}, \\[4pt] \frac{ds}{dt} &= -\frac{s}{L - 2}\,\frac{dL}{dt}. \end{aligned}

When

L = 8

, the equation gives

8s = 12 + 2s

, so

s = 2

. Then

\frac{ds}{dt} = -\frac{2}{6}(-0.5) = \frac{1}{6}

. The shadow is lengthening at

\frac{1}{6}

meter per second.

(b) When

L = 4

4s = 12 + 2s

gives

s = 6

, and

\frac{ds}{dt} = -\frac{6}{2}(-0.5) = 1.5

. The shadow is lengthening at

1.5

meters per second. The same descent stretches it nine times as fast. Since

s = \frac{12}{L - 2}

, the rate scales with

\frac{1}{(L - 2)^2}

, and

L - 2

has shrunk from

6

2