If the radius of the circle changes at the rate of <span class="math-tex">\(\rm-\frac{2}{\pi}\ m/sec\)</span>, at what rate does the circle's area change when the radius is 10 m?
Step-by-step Solution:
To find the rate at which the circle's area is changing when the radius is 10 m, we proceed as follows: The area of a circle is given by: \[ A = \pi r^2 \] Differentiating both sides with respect to time \( t \): \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \] We are given: \[ \frac{dr}{dt} = -\frac{2}{\pi} \, \text{m/sec}, \quad r = 10 \, \text{m} \] Substitute the values into the equation: \[ \frac{dA}{dt} = 2\pi (10) \left(-\frac{2}{\pi}\right) \] Simplify: \[ \frac{dA}{dt} = 20\pi \cdot \left(-\frac{2}{\pi}\right) \] \[ \frac{dA}{dt} = -40 \, \text{m}^2/\text{sec} \]