Q1-A-balloon-in-the-shape-of-cone-surrmo-surmounted-by-hemispherical-top-the-daimeter-of-balloon-is-equal-to-hieght-of-cone-find-the-rate-of-change-of-volume-of-the

Question Number 7265 by 2402@gmail.com last updated on 20/Aug/16

Q 1 . A b a l l o o n i n t h e s h a p e o f c o n e s u r r m o

s u r m o u n t e d b y h e m i s p h e r i c a l

t o p . t h e d a i m e t e r o f b a l l o o n i s

e q u a l t o h i e g h t o f c o n e . f i n d t h e

r a t e o f c h a n g e o f v o l u m e o f t h e

b a l l o o n e w i t h r e s p e c t t o i t s t o t a l

h e i g h t h .

Commented by Yozzia last updated on 20/Aug/16

d = h^{'} = h e i g h t o f c o n e

r = r a d i u s o f h e m i s p h e r e s e c t i o n

V = \frac{1}{3} π r^{2} h^{'} + \frac{2 π r^{3}}{3} = \frac{π}{3} (r^{2} h^{'} + 2 r^{3})

h = h^{'} + \frac{d}{2} = \frac{3}{2} h^{'} \Rightarrow h^{'} = \frac{2}{3} h

r = \frac{d}{2} = \frac{h^{'}}{2} = \frac{1}{2} \times \frac{2}{3} h = \frac{1}{3} h

V = \frac{π}{3} (\frac{h^{2}}{9} \times \frac{2}{3} h + 2 \times \frac{1}{27} h^{3})

V = \frac{π}{3} \times \frac{4}{27} h^{3}

\Rightarrow \frac{d V}{d h} = \frac{4 π}{27} h^{2}

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