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The base diameter and altitude of a right circulAR cone are observed at a certain instant to be 10 and 20 inches respectively. if the lateral area is constant and the base diameter is increasing at a rate of 1 inch per cubic minute, find the rate at ehich the altitude is decreasing

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hmm yes.. he formula is LA = pi*r*L
1 inch per cubic minute .... means what ? i am not understanding ' CUBIC MINUTE'
im sory it's 1 inch per minute

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Other answers:

LA = pi*r*L , and here L is slant height
what happn nxt?
is it 1/ sqrt(2) inch/ min
the answer will be 2.25 in per min @Gowthaman
then somewhere i missed, i check once again
but now i got -.25 inch/min
dh/dr =-1/2 (from condition of constant LS area) and dr/dt = 1/2 inch/min (given in the problem) dh/dt = dh/dr x dr/dt = - 1/2 x 1/2 = - 1/4 = -0.25 inch / min
@bii17 ....?
I hope u can solve it from here...
@sauravshakya hey stil there?
@sauravshakya what is that?? i didnt understand those with circle..
That is radius of the cone
Got it?
@hartnn can u help me in this question?
which part of saurav's explanation did u not understand?
here.. |dw:1349002226151:dw| i used it but i didnt get the correct answer the answer should be 2.25 @hartnn
@bii17 i did get it as 2.25. l^2=425, r=5, h=20, dr/dt=2
|dw:1349002789725:dw| i get is 8 @hartnn
oh sorry, dr/dt = 1/2 and not 2
|dw:1349002976588:dw| my answer is 2.. :( @hartnn can u show me what u did?
did u get that u missed the -*minus* sign in the beginning ?

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