If volume of a spere is changing at a rate of 20cm^3 / sec, then at rate radius will changing at the instant when radius is 5 cm.
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Answered by
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HEY.......
HERE IS U R ANSWER
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VOLUME OF SPHERE = 4/3 π r ^3
differentiate with time
dV / dt = 4/3 π * 3r^2 dr /dt
dV/dt = 4πr^2 dr/dt
dr/dt = dV/dt / 4πr^2
= 20 / (4π * 25)
= 1/5π
===================================
HOPE IT WILL HELP YOU A LOT....
THANKU.....
HERE IS U R ANSWER
===================================
VOLUME OF SPHERE = 4/3 π r ^3
differentiate with time
dV / dt = 4/3 π * 3r^2 dr /dt
dV/dt = 4πr^2 dr/dt
dr/dt = dV/dt / 4πr^2
= 20 / (4π * 25)
= 1/5π
===================================
HOPE IT WILL HELP YOU A LOT....
THANKU.....
Answered by
0
the answer is 1/5π
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