Finding the Rate of Change of Area with Decreasing Length and Increasing Width

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SUMMARY

The discussion focuses on calculating the rate of change of the area of a rectangle given the rates of change of its length and width. The length (L) is decreasing at 3 cm/sec, while the width (W) is increasing at 2 cm/sec. Using the formulas for differentiation, the area (A) is expressed as A = W * L, and the derivative A' is calculated as A'(t) = -3(W0 + 2t) + 2(L0 - 3t). The specific values for L and W are 15 cm and 8 cm, respectively, leading to a definitive calculation of the area’s rate of change.

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Hi,

I was practising differentiation and came across a some what rather easy question which I could not do. Can anyone direct me as usual. (this place is becoming my e-home). Mind you I am enjoying using LaTex converters haha.

Q. The length L of a rectangle is decreasing at the rate of 3 cm/sec, and the width W is increasing at the rate of 2 cm/sec. Find the rate of change of the area when L = 15 cm and W = 8 cm.

Thanks.
 
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dL/dt = -3 and dW/dt = +2. The area A is equal to W * L. Differentiate wrt t and plug in the values you were given.
 
W(t) = W0 + 2t
L(t) = L0 - 3t

A(t) = W(t)L(t)
A'(t) = W(t)L'(t) + W'(t)L(t)
A'(t) = -3(W0 + 2t) + 2(L0 - 3t)
 

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