Jenny asked in Science & MathematicsMathematics · 5 months ago

# A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 ft/s?

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.

(a) What is the velocity of the top of the ladder when the base is given below?

15 feet away from the wall?

20 feet away from the wall?

24 feet away from the wall?

(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 15 feet from the wall.

(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 15 feet from the wall.

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• 5 months ago

b^2 + h^2 = 25^2

db/dt = 2 ft/sec

2b * db/dt + 2h * dh/dt = 0

b * db/dt + h * dh/dt = 0

h * dh/dt = -b * db/dt

dh/dt = -b * db/dt / h

dh/dt = -b * (db/dt) / sqrt(625 - b^2)

dh/dt = -2 * b / sqrt(625 - b^2)

a1) -2 * 15 / sqrt(625 - 15^2)

a2) -2 * 20 / sqrt(625 - 20^2)

a3) -2 * 24 / sqrt(625 - 24^2)

b)

A = (1/2) * b * h

dA/dt = (1/2) * (b * dh/dt + h * db/dt)

dA/dt = (1/2) * (15 * (-2 * 15 / sqrt(625 - 15^2)) + sqrt(625 - 15^2) * 2)

c)

tan(T) = h/b

sec(T)^2 * dT/dt = (b * dh/dt - h * db/dt) / b^2

dT/dt = (b * dh/dt - h * db/dt) / (b^2 * sec(T)^2)

dT/dt = (b * dh/dt - h * db/dt) / (b^2 * (1 + tan(T)^2))

dT/dt = (b * dh/dt - h * db/dt) / (b^2 * (1 + (h/b)^2))

dT/dt = (b * dh/dt - h * db/dt) / (b^2 + h^2)

dT/dt = (b * dh/dt - h * db/dt) / 25^2

dT/dt = (15 * (-2 * 15 / sqrt(625 - 15^2)) - sqrt(625 - 15^2) * 2) / 625

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