- #1
FiveAlive
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A water glass (10 cm diameter at the top, 6 cm diameter at the bottom, 20 cm in height) is being filled at a rate of 50 cm^3/min. Find the rate of change of the height of the water after 5 seconds.
V=(∏* (h/3)) * (R^2 + r^2 +(R*r))
I've been messing around with this for a bit and made a bit of leeway but I'm unsure of how to eliminate R and r from my equation. If it was a cone I would use the r/h=5/20 and rearrange to 5h/20 and replace my 'r' then differentiate. However that ratio doesn't seem to work on a truncated cylinder and I have two radius-es. Any suggestions?
V=(∏* (h/3)) * (R^2 + r^2 +(R*r))
I've been messing around with this for a bit and made a bit of leeway but I'm unsure of how to eliminate R and r from my equation. If it was a cone I would use the r/h=5/20 and rearrange to 5h/20 and replace my 'r' then differentiate. However that ratio doesn't seem to work on a truncated cylinder and I have two radius-es. Any suggestions?