Question

A 13​-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?

Answer 1

Answer:

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,

$l^2 = x^2 + y^2------(1)$

Differentiating with respect to t ( time ),

$2l\frac{dl}{dt} = 2x \frac{dx}{dt}+2y\frac{dy}{dt}$

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

$\implies \frac{dl}{dt}=0$

$\implies 0= 2x \frac{dx}{dt}+2y\frac{dy}{dt}-----(2)$

We have,

y = 5, $\frac{dy}{dt}=12\text{ ft per sec}$,

Again by equation (1),

$13^2 = x^2 + 5^2\implies 169 = x^2 + 25\implies 169 - 25 = x^2\implies 144 = x^2\implies x = 12\text{ ft}$

From equation (2),

$0 = 2(12) \frac{dx}{dt} + 2(12)(5)$

$0= 24\frac{dx}{dt}+120$

$-120 =24\frac{dx}{dt}$

$\implies \frac{dx}{dt}=-\frac{120}{24}=-5\text{ ft per sec}$

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

$\implies A = \frac{1}{2}\times y\times x$

Differentiating with r. t. t,

$\frac{dA}{dt}=\frac{1}{2}(y\frac{dx}{dt}+x\frac{dy}{dt})$

$= \frac{1}{2}(5(-5) + 12(12))$

$=\frac{1}{2}(-25 + 144)$

$=\frac{119}{2}$

= 59.5 ft² per sec

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