1010-foot 0.8 ft divided by s0.8 ft/s. 8 ft8 ft 10 ft10 ft Let x be the distance

Question

1010-foot

0.8 ft divided by s0.8 ft/s.

8 ft8 ft

10 ft10 ft

Let x be the distance from the foot of the ladder to the wall and let y be the distance from the top of the ladder to the ground. Write an equation relating x and y.

A

1010-foot

ladder is leaning against a vertical wall (see figure) when Jack begins pulling the foot of the ladder away from the wall at a rate of

0.8 ft divided by s0.8 ft/s.

How fast is the top of the ladder sliding down the wall when the foot of the ladder is

8 ft8 ft

from the wall?

10 ft10 ft

Explanation / Answer

Let x be the distance from the foot of the ladder to the wall and let y be the distance from the top of the ladder to the ground.
Length of ladder = 10 foot
length of ladder = sqrt (x^2 + y^2)
10 = sqrt (x^2 + y^2)
x^2+y^2 = 10^2
x^2+y^2 = 100

given:
x = 8 ft
y = sqrt(10^2 - 8^2) = 6 foot
dx/dt = 0.8 ft/s

differentiate with respect to t,
x^2+y^2 = 100
2x*dx/dt + 2y*dy/dt = 0
put values,
2*8*0.8 + 2*6*dy/dt = 0
dy/dt = -(2*8*0.8)/(2*6)
= -1.07 ft/s
Answer: top of the ladder is sliding down at 1.07 ft/s

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