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Calculus/rate of change in rectangle

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Expert: Scotto - 7/10/2008


Question
I'm having difficulty setting these equations up.  Any help would be greatly appreciated!

Robert

The length, L, of a rectangle is decreasing at the rate of 2cm/second while the width, W, is increasing at the rate of 2cm/second.  When L = 12cm and W = 5cm, find the rates of change of the area, the perimeter, and the lengths of the diagnol of the rectangle.

Answer
dL/dt = -2, dW/dt = 2, L=12, W=5

Area is A=WL, perimeter is P=2W+2L, lengths of the diagonal of the rectangle is C=Sqrt(W²+L²).

Note that A, P, and d are all functions of W and L.  Using the chain rule, find dA/dt, dP/dt, and dC/dt.

dA/dt = (dW/dt)L +W(dL/dt), dP/dt=2(dW/dt) + 2(dL/dt).
dC/dt = (W+L)/Sqrt(W²+L²).

Put in the value given at the top into the equations just given to get the answer.

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Scotto

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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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