Water is leaking from a hemispheric bowl of radius 20cm at the rate of 0.5cm^3 /s. Find the rate at which surface area of the water decreasing when the water level is halfway from the top. Thank you


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Question Number 197640 by Mastermind last updated on 25/Sep/23

$W a t e r i s l e a k i n g f r o m a h e m i s p h e r i c$ $b o w l o f r a d i u s 20 c m a t t h e r a t e o f$ $0.5 {c m}^{3} / s . F i n d t h e r a t e a t w h i c h s u r f a c e$ $a r e a o f t h e w a t e r d e c r e a s i n g w h e n t h e$ $w a t e r l e v e l i s h a l f w a y f r o m t h e t o p .$ $T h a n k y o u$
	Answered by mr W last updated on 25/Sep/23

	Commented by mr W last updated on 25/Sep/23

	$v o l u m e o f w a t e r V = π h^{2} (R - \frac{h}{3})$ $\frac{d V}{d t} = π h (2 R - h) \frac{d h}{d t}$ $\Rightarrow \frac{d h}{d t} = \frac{1}{π h (2 R - h)} \times \frac{d V}{d t}$ $a t h = \frac{R}{2} :$ $\frac{d h}{d t} = \frac{4}{3 π R^{2}} \times \frac{d V}{d t} = \frac{4}{3 π \times 20^{2}} \times (- 0.5)$ $= - 0.00053 c m / s$
	Commented by Mastermind last updated on 25/Sep/23

	$T h a n k y o u s o m u c h$
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