**Application Problems on Rate of Change :**

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

**Example 1 :**

Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at the rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is π/3.

**Solution :**

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

Let θ radian be the angle between the sides AB and AC when b = 5 cm, c = 4 cm

dθ/dt = 0.06 radian/sec.

Let A and t be the area of the triangle and time taken respectively.

Find the value of dA/dt when θ = π/3

Area of triangle ABC = (1/2) b c sin θ

dA/dt = (1/2) b c cos θ (dθ/dt)

= (1/2) (5) (4) (sin π/3) (0.06)

= (1/2) (20) (1/2) (0.06)

= (20/4) (0.06)

= 0.3 m²/sec

Therefore, the area is increasing at the rate of 0.03 m²/sec.

**Example 2 :**

Two sides of a triangle have length 12 m and 15 m. The angle between them is increasing at a rate of 2°/min. How fast is the length of third side increasing when the angle between the sides of fixed length is 60° ?

**Solution :**

Let ABC be the given triangle

Length of the side AB = 12 m

length of the sides BC = 15 m

Let θ be the angle between the sides AB and BC at "t" minutes

dθ/dt = 2°(π/180)'

dθ/dt = π/90

now we have to find db/dt when θ = 60°

b^{2} = c^{2} + a^{2} - 2 a c cos θ

2 b (db/dθ) = 0 + 0 - 2 a c (-sin θ) (dθ/dt)

2 b (db/dθ) = 2 a c (sin θ) (dθ/dt)

b (db/dθ) = a c (sin θ) (dθ/dt) --- (1)

now apply a = 15, c = 12 and θ = 60°

b^{2} = 12^{2} + 15^{2} - 2 (15) (12) cos 60°

b^{2} = 144 + 225 - 360 (1/2)

b^{2} = 369 - 180

b^{2} = 189

b = √189

now let us apply these values in the first equation

b (db/dθ) = a c (sin θ) (dθ/dt) --- (1)

√189 (db/dθ) = 15 ⋅ 12 (sin 60°) (π/90)

db/dθ = 15 ⋅ 12 ⋅ (√3/2) (π/90)/√189

db/dθ = [(180) (√3) (π)]/[2 (√189) (90)]

(db/dθ) = π/√63 m/minutes

Therefore, the rate of change of the third side is π/√63 m/minutes.

**Example 3 :**

Gravel is being dumped from a conveyor belt at a rate of 30 ft³/min and its coarsened such that it forms a pile in the shape of cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?

**Solution :**

Let V = the volume of the cone, t = time, r = radius and h = height of the cone.

Here we have a information that the base diameter and height are always equal.

So, 2r = h

Rate of change of volume (dV/dt) = 30

Now, we have to find how fast is the height of the pile increasing when the pile is 10 ft.

h = 10

Volume of cone V = (1/3) π r² h

V = (1/3) π (h/2)² h

V = (1/3) π (h³/4)

dV/dt = (1/3) π (3 h²/4) (dh/dt)

Apply, h = 10 and dV/dt = 30

30 = (1/3) π [3 (10)²/4] (dh/dt)

(30 x 3 x 4)/300 π = (dh/dt)

6/5π = dh/dt

dh/dt = (6/5 π) ft/min

Therefore, the height is increasing at the rate of (6/5π) ft/min

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

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