Answer to Question #103901 in Trigonometry for Martin

Question

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.)

Answer 1

We are given:

One side of a rectangle is $9$ cm.

Another side of the rectangle is also given which is $19$ 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 $x$ cm.

Increasing rate of another side is:

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

Now:

We know that, the area of this rectangle is:

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

Derivative of (Equation 1) with respect to $t$ is:

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

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

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

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

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

Hence:

The area of the rectangle is changing at the rate of $45$ square centimeters per minute.