A ladder slipping in oil

MediumCalculus~6m

A $50$ -foot ladder leans against a vertical wall. Its base slides away from the wall at a constant $4$ ft/min. At the instant the base is $30$ ft from the wall, at what rate (in radians per minute) is the angle $\theta$ between the ladder and the ground decreasing?

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Use $x=50\cos\theta$ , so $\tfrac{dx}{dt}=-50\sin\theta\,\tfrac{d\theta}{dt}$ .
At $x=30$ (a $30$ – $40$ – $50$ triangle) $\sin\theta=\tfrac45$ ; solve $4=-40\,\tfrac{d\theta}{dt}$ .

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A ladder slipping in oil

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