Math / Calculus

QuestionThe area of a circle is increasing at a constant rate of 178 square feet per second. At the instant when the radius of the circle is 4 feet, what is the rate of change of the radius? Round your answer to three decimal places.

Studdy Solution

STEP 1

1. The area $A$ of a circle is given by the formula $A = \pi r^2$ , where $r$ is the radius.
2. The rate of change of the area with respect to time is constant at 178 square feet per second.
3. We need to find the rate of change of the radius $\frac{dr}{dt}$ when the radius $r$ is 4 feet.

STEP 2

1. Relate the area of the circle to its radius.
2. Differentiate the area with respect to time to find a relationship between $\frac{dA}{dt}$ and $\frac{dr}{dt}$ .
3. Solve for $\frac{dr}{dt}$ using the given information.

STEP 3

The area $A$ of a circle is given by the formula:
$A = \pi r^2$

STEP 4

Differentiate both sides of the equation $A = \pi r^2$ with respect to time $t$ :
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2)$
Using the chain rule, we have:
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

STEP 5

We know $\frac{dA}{dt} = 178$ square feet per second and $r = 4$ feet. Substitute these values into the differentiated equation:
$178 = 2\pi (4) \frac{dr}{dt}$
Simplify and solve for $\frac{dr}{dt}$ :
$178 = 8\pi \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{178}{8\pi}$
Calculate $\frac{dr}{dt}$ and round to three decimal places:
$\frac{dr}{dt} \approx \frac{178}{25.1327} \approx 7.082$
The rate of change of the radius is approximately:
$\boxed{7.082} \text{ feet per second}$

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