Mathematics, 21.07.2019 10:30 liddopiink1
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
Answers: 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\frac{dh}{dt}} \right)-\cfrac{\pi }{3}(\stackrel{chain~rule}{3h^2\frac{dh}{dt}}) \\\\\\ \begin{cases} \frac{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{\frac{cm}{sec}}{900\pi }](/tpl/images/0115/2527/2201c.png)
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