Question

A 10-foot ladder is placed against a vertical wall. Suppose that the bottom of the ladder slides away from the wall at a constant rate of 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?

Answer 1

Answer: -2.25 ft/s

Explanation:

we have the following parameters

length of the ladder = 10 ft

distance of the base of the ladder (x) = 6 ft

applying Pythagoras theorem

X^2 + Y^2 = 10^2  ............equation 1

X^2 + Y^2 = 100

differentiating the above we have

2X (dx/dt) + 2Y (dy/dt) = 0  ...........equation 2

substituting the value of X into equation 1

6^2 + Y^2 = 10^2

Y^2 = 100 - 36

Y =$\sqrt{64}$

Y = 8

now that we have Y= 8, X = 6 and dx/dt (rate of sliding of the base) = 3 ft/s, we can substitute them into equation 2 to get dy/dt (rate of sliding of the top)

we now have

(2 x 6 x 3 ) + (2 x 8 x dy/dt ) = 0

36 + 16(dy/dt) = 0\

dy/dt = (-36) / 16

= -2.25 ft/s ( - means its sliding down)