Implicit differentiation

**deltaxray** · February 19th 2010, 08:12 PM

Hi, I am stuck on this question.

A ladder 8 meters long rests against a wall with its base x metres from the wall and its top y metres high. Explain why $x^2+y^2=64$ and differentiate the equation implicitly with respect to time. Find with three significant figures:

a) the rate at which the top is sliding down when the base is alipping out at a constant rate of 2cm/s and is 2 meters from the wall

b) the rate at which the base is slipping out when the top is slipping down at a constant rate of 2mm/s and is 7 meters high.

One part i really dont get is how to differentiate with respect to time.

Thanks

**drumist** · February 19th 2010, 10:21 PM

Starting with $x^2+y^2=64$

Take the derivative with respect to t:

$\implies~ \frac{d}{dt} (x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt} (64)$

Can you calculate each of these terms?

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