A rectangle has a length of 5 inches and a width of 8 inches whose sides are changing. The length is decreasing by 7 in/sec and the width is shrinking at 3 in/sec.

Question

What is the rate of change of the area?

Answer 1

This is a related rate problem. Let's look at this from the point of view of the rectangle...

A_rectangle = Length * Width = L*W

Calculus allows us to look at the rates of change of various components, so in this case we want to know the rate of change of area of this rectangle...

dA/dt = d/dt ( L*W )

Well, this is product rule for the right-hand side.

dA/dt = dL/dt * W + L * dW/dt --> which has units of in²/sec

We know the length and width of the rectangle (5 and 8 inches respectively), the rate of change of the length and width are (-7 in/sec and -3 in/sec respectively). Note shrinking/decreasing means a negative sign...

Therefore: dA/dt = -7 in/sec * 8 in + 5 in * -3 in/sec = -56 in²/sec - 15 in²/sec = -71in²/sec

Answer 2

The key to this related rates problem is to keep in mind that all of the variables, Area, length, and width, are functions of time. Therefore, we must take their derivatives with respect to time using the chain rule, prior to plugging in any given values.

We can start with the area of a rectangle:

A = l*w

Then we can take the derivative of both sides with respect to time, t. This will give us dA/dt on the left, which is what we want to find. On the right, we will need to apply the product rule since l and w are both functions of time:

dA/dt = l*dw/dt + w*dl/dt

Since we know all four of the values on the right hand side, we may now plug them in and simplify:

dA/dt = (5)*(-3) + (8)*(-7)

dA/dt = -15 - 56

dA/dt = -71

For the final answer, we must consider units. The unit of area will be in² since the length and width are given in inches. Time is given in seconds. Our units for the final answer will be in²/sec:

dA/dt = -71 in²/sec