A 15 foot ladder is leaning against a wall of a house. The base of the ladder is pulled away from the wall at a rate of 3 ft/sec. Find the rate at which the area of the triangle is changing when the base of the ladder is 9 ft from the wall.

7.875 ft/s

Explanation:

L = 15 ft

dx/dt = 3 ft/s

x = 9 ft

Let the top of ladder is coming down with the rate of dy/dt.

use Pythagorean theorem

L^2 = x^2 + y^2 ... (1)

Differentiate both sides with respect to t

0 = 2 x dx/dt + 2y dy/dt

x dx/dt = - y dy/dt

When, x = 9 ft then y = ?. Put this in equation (1)

15^2 = 9^2 + y^2

225 - 81 = y^2

y = 12 ft

So

dy/dt = - x (dx/dt) / y = - 9 (3) / 12 = - 9/4 ft/s

Let A be the area of the triangle

A = 1/2 (base) (height)

A = 1/2 (x y)

Differentiate both sides with respect to t

dA/dt = 0.5 (y dx/dt + x dy/dt)

dA/dt = 0.5 [ 12 (3) - 9 (9/4) ]

dA/dt = 0.5 (36 - 81 / 4) = 31.5 / 4 = 7.875 ft/s