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Question Number 125322 by john_santu last updated on 10/Dec/20

Commented by john_santu last updated on 10/Dec/20

$$question\left(c\right)$$

Commented by john_santu last updated on 10/Dec/20

Commented by john_santu last updated on 10/Dec/20

$$isitcorrect?$$

Commented by mr W last updated on 10/Dec/20

$$\left(c\right)$$$$V=\pi ({Rh}^2-\frac{h^3}{3})$$$$\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}=\pi h(2R-h)\frac{dh}{dt}...\left(i\right)$$$$A=\pi (2Rh-h^2)$$$$\frac{dA}{dt}=\frac{dA}{dh}\times \frac{dh}{dt}=2\pi (R-h)\frac{dh}{dt}...\left(ii\right)$$$$\frac{\frac{dA}{dt}}{\frac{dV}{dt}}=\frac{2(R-h)}{h(2R-h)}$$$$\Rightarrow \frac{dA}{dt}=\frac{2(R-h)}{h(2R-h)}\times \frac{dV}{dt}$$$$=\frac{2(6-2)}{2(2\times 6-2)}\times 4=1.6{in}^2/s$$

Commented by john_santu last updated on 10/Dec/20

$$okcorrect.thanks$$

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