the height and radius of a cone are both changing. the radius of the cone is decreasing at a constant rate…

Explanation:

Step1: Differentiate the volume formula with respect to time

Given (V=\frac{1}{3}\pi r^{2}h). Since (V) is constant ((dV/dt = 0)), we use the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u = r^{2}) and (v = h). Differentiating (V) with respect to (t): [ \begin{align*} \frac{dV}{dt}&=\frac{1}{3}\pi\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)\ 0&=2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt} \end{align*} ]

Step2: Find the radius when (h = 3)

We know (V=\frac{1}{3}\pi r^{2}h) and (V = 193), (h = 3). [ 193=\frac{1}{3}\pi r^{2}\times3\Rightarrow r^{2}=\frac{193}{\pi}\Rightarrow r=\sqrt{\frac{193}{\pi}}\approx7.84 ]

Step3: Solve for (\frac{dh}{dt})

We know (\frac{dr}{dt}=- 7) (negative because (r) is decreasing), (h = 3), (r\approx7.84) From (0 = 2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}), we can solve for (\frac{dh}{dt}): [ \begin{align*} r^{2}\frac{dh}{dt}&=-2rh\frac{dr}{dt}\ \frac{dh}{dt}&=-\frac{2h}{r}\frac{dr}{dt}\ \frac{dh}{dt}&=-\frac{2\times3}{7.84}\times(-7)\ \frac{dh}{dt}&=\frac{42}{7.84}\approx5.357 \end{align*} ]

Answer:

(5.357)