Question

A ladder 25 feet long is leaning againstthe wall of a house. The base of the ladder ispulled away from the wall at a rate of 2 feet per second
(a) How fast is the top of the ladder moving down the wallwhen its base is 7 feet, 15 feet, and 24 feet from the wall?(b) Consider the triangle formed by the side of the house, theladder, and the ground. Find the rate at which the areaof the triangle is changing when the base of the ladder is7 feet from the wall.(c) Find the rate at which the angle between the ladder and thewall of the house is changing when the base of the ladderis 7 feet from the wall

Answer 1

Answer: a) -7/12 ft/s, -20 ft/s, -7 ft/s

              b) 527/24 Ft²/s

              c) 2/25 rad/sec

Step-by-step explanation:

a) b=7 ; a^2+b^2 = 25^2

             a^2+b^2 = 625

Differentiate w.r.t = 2a (da/dt) + 2b ( db/dt) = 0

                                    $b(db/dt) = -a(da/dt)$

As found from above put a=24 b=7,15 and 24. You will get the reuqired answer.

b) Area of the triangle = 1/2 * b*h

differentiate wrt to time leads to the following relation:

$2dA/dt= h(db/dt)+b(dh/dt) \\dA/dt=24-49/24 = 527/24 Ft^2/sec$

c) sin (Θ) = b/25

differentiate it again to give

$cos(0.284)(d\theta/dt) = (1/25) (db/dt) \\=2/25 rad/sec$