IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kane

a balloon in the form of a right circular cone surmounted by a hemisphere, having a diametre equal to height of the cone,is being inflated.how fast is its volume changing w.r.t. its total height h,when h=9cm?

answer is 12 $\pi$ cm³/sec

allamraju

Volume of the balloon=cone+hemisphere=pir²h/3+(2/3)pir³.Given,2r=h

So,V=pih³/12+pih³/12=pih³/6.Since the question is asked w.r.t total height,H=h+r=3h/2 $\Rightarrow$ h=2H/3.Substituting this in the above eqn.,we get V=(4/81)piH³

Hence,dV/dH=(4/81)(3pi)(H)²=12picm³/s as H=9cm.

indrajeet_bariar

@kane
HEY DUDE. U ASKED WHTS THE RATE OF CHANGE OF VOLUME W.R.T ITS HEIGHT AND U R GIVING ANSWER IN cm3/sec which means rate of change of volume with respect to time. i think u should reconsider the question or if i m wrong then u can tell me why i m wrong.

kane

no my dear friend the question is absolutely correct.

we have to find dV/dh i.e

dV/dt/dh/dt

since we are differentiating w.r.t to time the unit provided in the answer is correct

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Volume of the balloon=cone+hemisphere=pir2h/3+(2/3)pir3.Given,2r=h

So,V=pih3/12+pih3/12=pih3/6.Since the question is asked w.r.t total height,H=h+r=3h/2h=2H/3.Substituting this in the above eqn.,we get V=(4/81)piH3

Hence,dV/dH=(4/81)(3pi)(H)2=12picm3/s as H=9cm.

Volume of the balloon=cone+hemisphere=pir²h/3+(2/3)pir³.Given,2r=h

So,V=pih³/12+pih³/12=pih³/6.Since the question is asked w.r.t total height,H=h+r=3h/2 $\Rightarrow$ h=2H/3.Substituting this in the above eqn.,we get V=(4/81)piH³

Hence,dV/dH=(4/81)(3pi)(H)²=12picm³/s as H=9cm.