4. 3vS points 1 Previous Answers LarCalc11 2.6.021. 1/5 Used My Notes O Ask Your

Question

4. 3vS points 1 Previous Answers LarCalc11 2.6.021. 1/5 Used My Notes O Ask Your Te 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? 7 feet away from the wall t/sec 15 feet away from the wall ft/sec 20 feet away from the wall ft/sec (o) Consider the triangle formed by the side the side of the house, ladder, and the ground. Find the rate at which the area of the triangle : 60 is changing when the base of the ladder is 15 feet from the wall. t?/sec (e) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of feet from the wall. the ladder is i5 rad/sec

Explanation / Answer

Solution:

(a) Given dx/dt = 2 ft/sec

find dy/dt , when x = 7ft, 15ft and 20ft

Equation; x^2 + y^2 = 25^2

            => x^2 + y^2 = 625

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

            => dy/dt = -(x/y) dx/dt

Note: When x = 7ft, y = 24 ft and when x = 15 ft, y = 20 ft   and when x = 20 ft, y = 15ft

These values were obtained using the Pythagorean theorem (i.e the “Equation”)

So, when x = 7ft;    dy/dt = -(x/y) dx/dt = -(7/24) * 2 = -7/12 = -0.583 ft/sec

when x = 15ft;        dy/dt = -(x/y) dx/dt = -(15/20) * 2 = -15/10 = -3/2 = -1.5 ft/sec

when x = 20ft;        dy/dt = -(x/y) dx/dt = -(20/15) * 2 = -8/3 = -2.67 ft/sec

(b) Given dx/dt = 2 ft/sec

find dA/dt when x = 15 ft.

Equation A = (1/2) xy

dA/dt = (1/2) (x dy/dt + y dx/dt) = (1/2) (15*(-3/2) + 20 * 2) = (1/2) (-45/2 + 40) = 35/2 ft2/sec

(C) Given dx/dt = 2 ft/sec

find d/dt when x = 15 ft.

Equation ; Tan = x/y => = arctan(x/y)

d/dt = {1/(1 + (x/y)^2)} * {(y dx/dt - x dy/dt) / y^2}

        = {1 / (x^2 + y^2)} * {y dx/dt - x dy/dt}

        = {1 / (15^2 + 20^2)} * {20*2 - 15*(-3/2)}

d/dt = (1/625)(40 + 45/2) = (1/625)(125/2) = 1/10 rad/sec = 0.1 rad/sec

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