# RATE OF CHANGE WORD PROBLEMS IN CALCULUS

The derivative can also be used to determine the rate of change of one variable with respect to another. A few  examples are population growth rates, production rates, water flow rates, velocity, and acceleration.

A common use of rate of change is to describe the motion of an object moving in a straight line. In such problems, it is customary to use either a horizontal or a vertical line with a designated origin to represent the line of motion.

On such lines, movements in the forward direction considered to be in the positive direction and movements in the backward direction is considered to be in the negative direction.

Problem 1 :

The radius of a circular plate is increasing in length at 0.01 cm per second. What is the rate at which the area is increasing when the radius is 13 cm? Solution :

Let A and r be the area and the radius.

Rate of change = 0.01 cm

Area of circular plate A = πr2

Differentiate with respect to r

dA/dt = π2r(dr/dt)

here r = 13 cm and dr/dt = 0.01 cm

dA/dt = π2r(dr/dt)

dA/dt = π2(13) (0.01)

dA/dt = π(26)(0.01)

dA/dt = π(0.26)

dA/dt = 0.26π cm2/sec

Problem 2 :

A square plate is expanding uniformly each side is increasing at the constant rate of 1.5 cm/min. Find the rate at which  the area is increasing when the side is 9 cm.

Solution :

Let a be the side of the square and A be the area of the square.

Here the side length is increasing with respect to time.

da/dt = 1.5 cm/min

Now we need to find the rate at which the area is increasing when the side is 9 cm.

That is,

We need to determine dA/dt when a = 9 cm.

Area of square = a2

A = a2

differentiate with respect to t

dA/dt = 2a(da/dt)

dA/dt = 2(9)(1.5)

dA/dt = 18(1.5)

dA/dt = 27 cm2/min

Problem 3 :

A stone thrown into still water causes a series of concentric ripples. If the radius of outer ripple is increasing at the rate of 5 cm/sec,how fast  is the area of the distributed water increasing when the outer most ripple has the radius of 12 cm/sec.

Solution :

Let A be the area and r be the radius.

Here the radius is increasing with respect to time.

Now we need to find dA/dt when radius  =  12 cm/sec

dr/dt = 5 cm/sec

Area of circle = πr2

A = πr2

dA/dt = 2πr(dr/dt)

dA/dt = 2π(12)(5)

dA/dt = 24π(5)

dA/dt = 120π cm2/sec

Problem 4 :

The radius of a spherical balloon is increasing at the rate of 4 cm/sec. Find  the rate of increases of the volume and surface area when the radius is 10 cm.

Solution :

Let V be the volume of spherical balloon and S be the surface area.

Here, we need to find dV/dt and dS/dt.

dr/dt = 4 cm/sec and r = 10 cm.

Volume of the spherical balloon (V) = (4/3)πr3.

Differentiate with respect to t.

dV/dt = (4/3)π3r2(dr/dt)

dV/dt = (4/3)π3(10)2(4)

dV/dt = 16π(100)

dV/dt = 1600π cm3/sec

Surface area of the spherical balloon S = 4πr2.

differentiate with respect to t.

dS/dt = 4π2r(dr/dt)

dS/dt = 8πr(dr/dt)

dS/dt = 8π(10)(4)

dS/dt = 80π(4)

dS/dt = 320πcm2/sec

Problem 5 :

A balloon which remains spherical is being inflated by pumping in 90 cm3/sec. Find the rate at which the surface area of the balloon is increasing when the radius is 20 cm.

Solution :

Let V be the area and r be the radius of the balloon.

dV/dt = 90 cm3/sec.

V = (4/3)πr3

dV/dt = (4/3)π3r2(dr/dt)

90 = 4π(20)2(dr/dt)

90 = 4π(400)(dr/dt)

90 = 1600π(dr/dt)

90/(1600π) = (dr/dt)

dr/dt = 9/(160π)

Surface area of the spherical balloon (S) = 4πr2

dS/dt = 4π(2r)dr/dt

dS/dt = 4π(2 x 20)(9/160π)

dS/dt = 160π(9/160π)

dS/dt = 9 cm2/sec Kindly mail your feedback to v4formath@gmail.com

## Recent Articles 1. ### Law of Cosines Worksheet

Aug 06, 22 11:56 PM

Law of Cosines Worksheet

2. ### Laws of Cosines

Aug 06, 22 11:31 PM

Laws of Cosines - Proof - Solved Problems