Chapter 6 Applications of Derivatives

Notes

Applications of Derivatives

1. Rate of Change of Quantities

If a quantity y changes with respect to another quantity x, then its rate of change is represented by:

dy/dx = Rate of change of y with respect to x

Some Useful Formulae

(i) Square

• Area: A = a²
• Perimeter: P = 4a

(ii) Rectangle

• Area: A = ab
• Perimeter: P = 2(a + b)

(iii) Circle

• Area: A = πr²
• Circumference: C = 2πr

(iv) Cube

• Surface Area: A = 6a²
• Volume: V = a³

(v) Sphere

• Surface Area: A = 4πr²
• Volume: V = (4/3)πr³

(vi) Cylinder

• Total Surface Area: A = 2πr(r + h)
• Volume: V = πr²h

(vii) Cone

• Total Surface Area: A = πr(l + r)
• Volume: V = (1/3)πr²h

Example 1: Rate of Change of Surface Area of a Cube

The volume of a cube is increasing at the rate of 9 cm³/sec. Find how fast the surface area is increasing when the length of an edge is 10 cm.

Let A be the surface area, V be the volume and a be the side of the cube.

Given:

dV/dt = 9 cm³/sec

To find: dA/dt at a = 10 cm.

V = a³

dV/dt = 3a²(da/dt)

9 = 3a²(da/dt)

Therefore,

da/dt = 9/(3a²)

Now,

A = 6a²

dA/dt = 12a(da/dt)

At a = 10 cm,

dA/dt = 12 × 10 × 9/(3 × 10²)

dA/dt = 18/5 cm²/sec

Example 2: Circular Wave

A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, find how fast the enclosed area is increasing.

Let A be the area and r be the radius of the circular wave.

Given:

dr/dt = 4 cm/sec

To find: dA/dt at r = 10 cm.

A = πr²

dA/dt = 2πr(dr/dt)

= 2π × 10 × 4

dA/dt = 80π cm²/sec

Example 3: Rectangle

The length x of a rectangle is decreasing at the rate of 3 cm/min and width y is increasing at the rate of 2 cm/min. When x = 10 cm and y = 6 cm, find the rate of change of:

1. Perimeter
2. Area

Given:

dx/dt = -3 cm/min

dy/dt = 2 cm/min

(a) Perimeter