A13-ft ladder is leaning against a house when its base starts to slide away. by the time the base is 5 ft from the house, the base is moving away at the rate of 12 ft/sec. a. what is the rate of change of the height of the top of the ladder? b. at what rate is the area of the triangle formed by the ladder, wall, and ground changing then?

a. -5 ft per sec

b. 59.5 ft² per sec

Step-by-step explanation:

a. Suppose l represents the ladder, x represents the height of the top of ladder from the base of house and y represents the distance of base of ladder from the base of house.

By the Pythagoras theorem,

Differentiating with respect to t ( time ),

But, height of ladder, l = 13 ft = constant,

We have,

y = 5, ,

Again by equation (1),

From equation (2),

Hence, the rate of change of the height of the top of the ladder is -5 ft per sec.

b. Now area of the triangle = 1/2 × base × height

Differentiating with r. t. t,

= 59.5 ft² per sec

Hence, area of the triangle is changing with the rate of 59.5 ft per sec.