Introduction to Related Rates

When quantities are linked by an equation and each one changes with time, their rates of change are linked too. Differentiating the equation with respect to time

t

gives the link.

Differentiating with respect to time

Treat every quantity that changes as a function of

t

, and differentiate both sides of the equation with respect to

t

. By the chain rule, each changing variable brings its own rate along. The product and quotient rules apply as usual, and a constant differentiates to

0

Term	Its derivative with respect to $t$
$x^2$	$2x\,\frac{dx}{dt}$
$\sqrt{y}$	$\frac{1}{2\sqrt{y}}\,\frac{dy}{dt}$
$xy$	$x\,\frac{dy}{dt} + y\,\frac{dx}{dt}$
$\sin\theta$	$\cos\theta\,\frac{d\theta}{dt}$
$5$ , or any constant	$0$

Equations relating rates

Example 1

Each equation relates quantities that change with time. Write an equation relating their rates of change with respect to

t

\begin{aligned} &\text{(a)}\ \ A = \pi r^2 \\[4pt] &\text{(b)}\ \ K = \tfrac{1}{2}mv^2 \\[4pt] &\text{(c)}\ \ z^2 = x^2 + y^2 \end{aligned}

(a)

\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}

(b) For a rocket burning fuel, the mass

m

and the speed

v

both change. The right side is a product:

\frac{dK}{dt} = \frac{1}{2}v^2\,\frac{dm}{dt} + mv\,\frac{dv}{dt}.

(c)

2z\,\frac{dz}{dt} = 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt}

, or

z\,\frac{dz}{dt} = x\,\frac{dx}{dt} + y\,\frac{dy}{dt}

Example 2

A box has a square base of side

s

centimeters, and its height is always

2

centimeters more than

s

, with time in seconds. (a) Express

\frac{dV}{dt}

in terms of

s

and

\frac{ds}{dt}

. (b) Find

\frac{dV}{dt}

when

s = 3

and

\frac{ds}{dt} = 0.5

(a) Substitute

h = s + 2

before differentiating:

V = s^2(s + 2) = s^3 + 2s^2

, so

\frac{dV}{dt} = \left(3s^2 + 4s\right)\frac{ds}{dt},

in cubic centimeters per second. Differentiating

V = s^2 h

first gives the same result, by the product rule, with

\frac{dh}{dt} = \frac{ds}{dt}

\begin{aligned} \frac{dV}{dt} &= 2sh\,\frac{ds}{dt} + s^2\,\frac{dh}{dt} \\[4pt] &= 2s(s + 2)\frac{ds}{dt} + s^2\,\frac{ds}{dt} \\[4pt] &= \left(3s^2 + 4s\right)\frac{ds}{dt}. \end{aligned}

(b)

\frac{dV}{dt} = (27 + 12)(0.5) = 19.5

cubic centimeters per second.

An angle

Example 3

A drawbridge leaf

13

meters long is hinged at the edge of the road. Let

x

be the horizontal distance from the hinge to the leaf's free end. Let

\theta

be the angle between the leaf and the road, so

\cos\theta = \frac{x}{13}

. As the leaf is lowered,

x

grows at

1.2

meters per second. Find

\frac{d\theta}{dt}

when

x = 5

The leaf (amber) hinged at the road, with its free end x meters out and y meters up, where x² + y² = 13².

Differentiating,

-\sin\theta\,\frac{d\theta}{dt} = \frac{1}{13}\,\frac{dx}{dt}.

When

x = 5

, the free end is

y = \sqrt{169 - 25} = 12

meters above the road, so

\sin\theta = \frac{12}{13}

. With

\frac{dx}{dt} = 1.2

-\frac{12}{13}\,\frac{d\theta}{dt} = \frac{1.2}{13}, \qquad \frac{d\theta}{dt} = -0.1.

The angle is decreasing at

0.1

radian per second.

A point on a curve

Example 4

A rectangle has its base on the

x

-axis and its top corners on the parabola

y = 8 - x^2

. Its corner

(x, y)

in the first quadrant lies on the parabola, so its area is

A = 2xy

. The corner travels along the parabola. When it's at

(2, 4)

\frac{dx}{dt} = -2

units per second. Find

\frac{dA}{dt}

at that moment.

The rectangle under the parabola with corner (2, 4). The corner heads up and to the left along the parabola.

First find

\frac{dy}{dt}

from the parabola:

\frac{dy}{dt} = -2x\,\frac{dx}{dt}

. At

(2, 4)

\frac{dy}{dt} = -2(2)(-2) = 8

. Then the product rule gives

\begin{aligned} \frac{dA}{dt} &= 2\left(\frac{dx}{dt}\,y + x\,\frac{dy}{dt}\right) \\[4pt] &= 2\big((-2)(4) + 2(8)\big) = 16. \end{aligned}

The area is increasing at

16

square units per second.

Equations relating rates

An angle

A point on a curve

Rates with respect to time