Question

the volume of water in a bowl is given by V=1/3pih^2(60-h), where h is the depth of the water in centimeters. If the depth is increasing at the rate of 3cm/sec when the water is 10 cm deep, how fast is the volume increasing at that instant

Answer 1

`\bf V=\cfrac{1}{3}\pi h^2(60-h)\implies V=\cfrac{1}{3}\pi h^260-\cfrac{1}{3}\pi h^3 \\\\\\ V=20\pi h^2-\cfrac{\pi }{3}h^3\implies \cfrac{dV}{dt}=20\pi \left(\stackrel{chain~rule}{2h(dh)/(dt)} \right)-\cfrac{\pi }{3}(\stackrel{chain~rule}{3h^2(dh)/(dt)}) \\\\\\ \begin{cases} (dh)/(dt)=3\\ h=10 \end{cases}\implies \cfrac{dV}{dt}=20\pi [2(3)(10)]-\cfrac{\pi }{3}[3(10)^2(3)] \\\\\\ \cfrac{dV}{dt}=1200\pi -300\pi \implies \cfrac{dV}{dt}=\stackrel{(cm)/(sec)}{900\pi }`