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Calculus Level 3

An 8.5m long ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at 0.5 m/s 0.5\text{ m/s} . How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 4 m from the wall?

− 3 15 \frac{-3}{15} m s \frac{m}{s} − 4 7.5 \frac{-4}{7.5} m s \frac{m}{s} − 4 15 \frac{-4}{15} m s \frac{m}{s}

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1 solution

We wish to find d y d t \frac{dy}{dt} when x = 4 m x=4m

By Pythagorean Theorem,

y 2 = 8. 5 2 − x 2 y^2=8.5^2-x^2

Differentiate both sides with respect to t,

d y d t \frac{dy}{dt} = = − x y \frac{-x}{y} d x d t \frac{dx}{dt}

when x = 4 x=4

y = y= 8. 5 2 − 4 2 \sqrt{8.5^2-4^2} = = 7.5 7.5

substitute,

d y d t \frac{dy}{dt} = = − 4 7.5 \frac{-4}{7.5} ( 0.5 ) (0.5) = = − 4 15 \frac{-4}{15} m / s m/s

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