﻿ Related Rate Worksheet | Problems & Solutions

# Related Rate Worksheet

Related Rate Worksheet
• Page 1
1.
Find the value(s) of $x$ for which the rate of change of $y$ = with respect to $x$ is 9.
 a. - 2 b. - 7, 1 c. - 7, - 1 d. 7, - 1

#### Solution:

Here y = x33 + 4x2 + 16x

The rate of change of y with respect to x = dydx = 9

ddx (x33 + 4x2 + 16x) = 9
[Substitute y.]

x2 + 8x + 7 = 0

(x + 1)(x + 7) = 0
[Factorize.]

x = - 7, - 1
[Solve.]

2.
A small stone is dropped into a quiet pond and circular ripples spread over the surface of water. The radius of each of these ripples increase at the rate of 28 inches per second. Find the rate at which the area inside the circle is increasing at the instant the radius is 7 ft.
 a. $\frac{98}{3}$$\pi$ ft²/s b. 392$\pi$ ft²/s c. $\frac{49}{3}$$\pi$ ft²/s d. $\frac{14}{3}$$\pi$ ft²/s

#### Solution:

Area of the water ring of radius r is A = πr2

The rate of increase of radius r is drdt = 28 inches/sec

= 7 / 3 ft/sec
[One inch = 1 / 12 ft.]

The rate of increase in area A is dAdt.
[Definition.]

= ddt (πr²)
[Substitute A = πr2.]

= 2πrdrdt

At r = 7 ft, the rate of increase in area A is dA / dt.

= 2π(7)(7 / 3)
[Substitute r = 7.]

= 98 / 3π ft²/sec
[Simplify.]

3.
The side length of a square increases at the rate of 4 cm/s. At what rate is the area of the square increasing when its side length is 40 cm?
 a. 8 cm²/s b. 160 cm²/s c. 320 cm²/s d. 4 cm²/s

#### Solution:

The area of square of side x = A = x2

The rate of increase of side x = dxdt = 4 cm/s.

The rate of increase of area A = dAdt
[Definition.]

= ddt(x2)
[Substitute A = x2.]

= 2xdxdt

At x = 40 cm, the rate of increase of area A = dAdt

= 2(40)(4)
[Substitute x = 40.]

= 320 cm²/sec

4.
The volume of a metallic hollow sphere is constant. Its outer radius is increasing at the rate of 2 cm/s. Find the rate at which its inner radius is increasing if the outer and inner radius of the sphere are 16 cm and 8 cm respectively.
 a. 8 cm/s b. 6 cm/s c. 4 cm/s d. 10 cm/s

#### Solution:

The volume of metallic hollow sphere of outer radius R, inner radius r = V = 4 / 3π(R3 -r3)
[Formula.]

dRdt = 2 cm/s.

dVdt = 0
[As V is constant.]

ddt( 4 / 3π(R3 -r3)) = 0
[Substitute V from step1.]

4 / 3(3R2πdRdt-3 r2πdrdt) = 0
[Simplify.]

drdt = R2r2dRdt
[Simplify.]

At R = 16 cm, r = 8 cm, drdt = 16282(2)
[Substitute dRdt = 2 . ]

= 256 / 64 × 2 = 8 cm/s.

5.
A gas balloon contains 1800 cubic ft. of gas at a pressure of 90 lb per ft². The pressure is decreasing at the rate of 0.1 lb per ft². per second. Find the rate at which the volume increases, if the gas obeys Boyle's law, PV = constant.

 a. 20 ft³/s b. 2 ft³/s c. 18 ft³/s d. 180 ft³/s

#### Solution:

Here P is the pressure, V is the volume.

The rate of change of P is dPdt = - 0.1 lb ft²/sec

PV = constant
[BoyleÃ¢â‚¬â„¢s law.]

ddt(PV) = ddt(constant)
[Take derivative.]

P(dVdt) + V(dPdt) = 0
[Product rule of derivative.]

dVdt = - V / PdPdt

At P = 90 lb per ft², V = 1800 ft³

So dVdt = - 1800 / 90 (- 0.1)
[Substitute the values of P, V, dpdt.]

= 2 ft³/s
[Simplify.]

So , the rate of increase of the volume of the balloon = dVdt = 2 ft³/s.

6.
An airplane at an altitude of 900 m flying horizontally with a speed of 250 m/sec passes directly over an observer. Find the rate at which the plane approaches the observer, when it is at 936 m away from the observer.
 a. $\frac{1285}{18}$ m/s b. $\frac{\sqrt{66096}}{900}$ m/s c. $\frac{32125}{468}$ m/s d. $\frac{\sqrt{66096}}{90}$ m/s

#### Solution:

Here A is the position of the observer.
[Figure.]

B is the position of the airplane at some instant.
[Figure.]

AC is the horizon.
[Figure.]

BC = a = 900 m

Let AB = s m, AC = x m.

The speed of the airplane = dxdt = 250 m/sec

s2 = x2 + (900)2
[From the right triangle ABC.]

9362 = x2 + 9002
[When s = 936 m.]

x2 = 876096 - 810000 = 66096
[Simplify.]

x = 66096
[Solve for x.]

ddts2 = ddt(x2+(900)2)
[From step 7.]

2sdsdt = 2xdxdt

dsdt = xs (dxdt)

At s = 936 m, x = 66096, the rate at which the airplane approaches the observer = dsdt

= 66096936× 250

= 32125 / 468 m/sec
[Simplify.]

7.
If the semi vertical angle of a right circular cone is 45o and the rate of change of volume of the cone is $\pi$${r}^{k}\frac{dr}{dt}$, then find the value of ($k$ + 2)($k$ + 5)($k$ + 4).

 a. 40 b. 168 c. 160 d. 80

#### Solution:

In a right circular cone with semicircular angle 45o

The base radius r = the height h
[From the right triangle ABC.]

Volume of the cone = V = 1 / 3 πr2h

V = 1 / 3 πr3
[Substitute r = h.]

The rate of change of volume = dVdt
[Definition.]

= ddt(13πr3)
[Substitute from step 4.]

= πr2 drdt= πrkdrdt

k = 2.
[Solve for k.]

Therefore the value of (k + 2)(k + 5)(k + 4) = (2 + 2)(2 + 5)(2 + 4) = 168.

8.
$x$ = and $y$ = , where 0 ≤ $\theta$$\pi$. Find the value of $\theta$ at which the rate of change of $x$ and $y$ with respect to $\theta$ are equal.
 a. $\frac{1}{20}$ b. 3$\pi$ c. $\frac{\pi }{20}$ d. $\frac{3\pi }{20}$

#### Solution:

x = 4 cos 5θ, y = 4 sin 5θ (0 ≤ θπ)

dxdθ =ddθ (4cos 5θ) = - 4(5)sin 5θ
[Substitute x = 4cos 5θ]

dydθ =ddθ (4sin 5θ) = 4(5)cos 5θ
[Substitute y = 4 sin 5θ.]

If dydθ =dxdθ, then

2(5)cos 5θ = -2(5) sin 5θ

tan 5θ = -1
[Solve tanθ = -1, (0 ≤ θπ).]

θ = 3π20
[Simplify.]

9.
The rate of change of the circumference C of a circle with respect to its area A is $k$ $\sqrt{\frac{\pi }{A}}$. Find the value of 9$k$² + 8$k$.
 a. $\frac{41}{16}$ b. 52 c. 17 d. 44

#### Solution:

Area of circle with radius r is A = πr2

Circumference of circle with radius r is C = 2πr

C2 = 4π2r2
[Square on both sides.]

C2 = 4πA
[Substitute πr2 for A.]

ddA (C)2 = ddA (4πA)

2C dCdA = 4π

dCdA = 2πC
[Substitute 4πA for C.]

= 2π4πA
[Substitute 4πA for C.]

=πA = kπA

k = 1.

9k² + 8k = 9(1)² + 8 = 17
[Substitute the value of k.]

10.
The side length of an equilateral triangle is $l$ cm. If the rate of change of area of the incircle with respect to $l$ is $\frac{k\pi l}{6}$, then find the value of 9k + 18.
 a. 27 b. 18 c. 9 d. 28

#### Solution:

Side of the equilateral triangle = l cm

The radius of incircle of the triangle = r = l23 cm.
[Formula.]

The area of the incircle = A = πr2

= πl212
[Substitute the value of r.]

= πl6
[Simplify.]

The rate of change of the area of the incircle with respect to the side l of an equilateral triangle = dAdl =πl 6 =kπl 6

Therefore k = 1.

9k + 18 = 9(1) + 18 = 27