The radius of a cylinder is decreasing at the rate of 4cm/s, while the height is increasing at the rate of 2cm/s. Find the rate of change of the volume when the radius is 2cm and the height is 6cm? Pls endeavour to show workings. Thank you!

Question

The radius of a cylinder is decreasing at the rate of 4cm/s, while the height is increasing at the rate of 2cm/s. Find the rate of change of the volume when the radius is 2cm and  the height is 6cm? Pls endeavour to show workings. Thank you!

myininaya · Answer

so we have r'=-4
and h'=2
V=h*pi*r^2
now take deriative
we have
V'=pi[h'*r^2+2rr'h]

we have r=2
h=6
just plug in

myininaya · Answer

let me know if further assistance or explanation is needed

amistre64 · Answer

would we use a cylindar volume or a cone volume?

amistre64 · Answer

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myininaya · Answer

its talking about a cylinder not a  cone

amistre64 · Answer

or something more akin to a truncated cone?

amistre64 · Answer

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anonymous · Answer

yeah, it could either be interpreted that it becomes a truncated cone as the height grows and the radius shrinks - or the entire cylinder's radius shrinks as the height grows.

amistre64 · Answer

since h increases at 2 per sec we can drop this to the bottom in 3 secs

amistre64 · Answer

given us a radius at the bottom of 4*3 = 12

myininaya · Answer

oh yeah i understand what amistre is saying now that fiddle speaks lol

amistre64 · Answer

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