Question #103901
A rectangle has one side of 9 cm. How fast is the area of the rectangle changing at the instant when the other side is 19 cm and increasing at 5 cm per minute? (Give units.)
1
Expert's answer
2020-03-02T09:46:41-0500

We are given:

One side of a rectangle is 99 cm.

Another side of the rectangle is also given which is 1919 cm. But, we will not take this value to calculate the area of the rectangle because this side is changing.

So, we will assume that another side will be xx cm.

Increasing rate of another side is:

dxdt=5\frac{dx}{dt} = 5 cm./min.

Now:

We know that, the area of this rectangle is:

A=9xA = 9x \hspace{1 cm} (Equation 1)

Derivative of (Equation 1) with respect to tt is:

ddt(A)=ddt(9x)\frac{d}{dt} (A) = \frac{d}{dt} (9x) \\

dAdt=9ddt(x)+xddt(9)\Rightarrow \frac{dA}{dt} = 9 \frac{d}{dt} (x) + x \frac{d}{dt} (9) [ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x))]\left[ \because \frac{d}{dx} (f(x)g(x)) = f(x) \frac{d}{dx} (g(x)) + g(x) \frac{d}{dx} (f(x)) \right]

dAdt=9dxdt+x(0)\Rightarrow \frac{dA}{dt} = 9 \frac{dx}{dt} + x(0) [ddx(c)=0(Where c is a constant)]\left[ \because \frac{d}{dx} (c) = 0 \, \text{(Where c is a constant)} \right]

dAdt=9(5)[dxdt=5]\Rightarrow \frac{dA}{dt} = 9(5) \hspace{1 cm} \left[ \because \frac{dx}{dt} = 5 \right]

dAdt=45\Rightarrow \frac{dA}{dt} = 45

Hence:

The area of the rectangle is changing at the rate of 4545 square centimeters per minute.



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