**Application Problems of Rate of Change in Calculus :**

In this section, we will see some practice questions on application problems of rate of change.

**Example 1 :**

Newton's law of cooling is given by θ = θ₀° e^{⁻kt}, where the excess of
temperature at zero time is θ₀° C and at time t seconds is θ° C.
Determine the rate of change of temperature after 40 s given that θ₀ =
16° C and k = -0.03.(e^{1.2} = 3.3201)

**Solution :**

Newton's law of cooling θ = θ₀° e^{⁻kt}

θ₀° = 16° C

k = -0.03

rate of change of temperature with respect to time

dθ/dt = - k θ₀° e^(⁻kt)

t = 40

dθ/dt = - (-0.03) (16)e^(⁻0.03) (40)

= 0.48 e^{⁻1.2}

= 0.48 (3.3201)

= 1.5936° C/s

**Example 2 :**

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm².

**Solution :**

Let "b" and "h" be the base and height of the triangle ABC

Area of triangle ABC

A = (1/2) b h

h = 10 area = 100

100 = (1/2) x b x 10

b = (100 x 2)/10

b = 20

dA/dt = (1/2) [b (dh/dt) + h (db/dt)]

h (db/dt) = 2 (dA/dt) - b (dh/dt)

db/dt = (2/h) (dA/dt) - (b/h) (dh/dt)

db/dt = (2/10) (2) - (20/10) (1)

db/dt = (4/10) - (20/10)

db/dt = -16/10

db/dt = -1.6 cm/min

**Example 3 :**

At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35 km/hr and ship B us sailing north at 25 km/hr. How fast is the distance between the ship changing at 4.00 p.m

**Solution :**

Let P and Q are the starting position of the ships A and B.

Let x, y and z be the distance between QA and QB and AB respectively.

z^{2} = x^{2} + y^{2}

Here x, y and z are changing with respect to time.

2z (dz/dt) = 2x (dx/dt) + 2y (dy/dt)

Dividing each side by 2.

z (dz/dt) = x (dx/dt) + y (dy/dt)

dx/dt = speed of ship A = 35 km/hr

dy/dt = speed of B = 25 km/hr

x = 40, y = 100

z = √(40)^{2} + (100)^{2}

z = √1600 + 10000

z = √2600

z = 20√29

20√29 (dz/dt) = 40 (35) + (100) (25)

20√29 (dz/dt) = 1400 + 2500

20√29 (dz/dt) = 3900

dz/dt = 3900 /(20√29)

dz/dt = 195/√29

There the rate of change of distance between two ships is 195/√29 km/hr.

We hope that the students would have understood how to solve word problems in rate of change.

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