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#### karush

##### Well-known member

- Jan 31, 2012

- 2,779

If the foot of the ladder is pulled away from the building at a constant rate of 8in per second how fast is the area of triangle formed by the ladder, the building and the ground changing (in feet squared per second) at the instant when the top of the ladder is 12 feet above the ground

View attachment 1520

$$A=\frac{1}{2}BH=\frac{1}{2}xy$$

$$\frac{d}{dt}A=\frac{d}{dt} \left(\frac{1}{2} xy \right)

\Rightarrow \frac{dA}{dt}=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx}{dt}$$

since $$\frac{dy}{dt}\text{ at y }= 12\text { is }\frac{-5 ft}{12 ft}\cdot \frac {2ft}{3 sec} = -\frac {5 ft}{18 sec}$$

then substituting

$$\frac{dA}{dt} = \frac{119}{36}\frac{ft^2}{sec}$$

hopefully