AdrianCavinder
Jan 31, 2012, 07:49 AM
HI again. Sorry, I have another problem I can’t seem to work out:
A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 12cm, the liquid is dripping from the funnel at a rate of 0.2cm^3s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?
Now, from the chain rule, I understand that dV/dt = dV/dh x dh/dt, where V = Volume h = depth and t = time.
From the formula, V = (1/3)Ï€r^2h, dV/dh = (1/3)Ï€r^2, so 0.2 = (1/3)Ï€r^2 x dh/dt. As the depth is twelve, I believe the radius to be 8 because of the proportion rule 30:12 = 20:8
So dh/dt = 0.6/64Ï€ = 0.00298cms^-1, but the answer is 0.004cms^-1.
Can you show me where I’m zagging when I should be zigging? Cheers.
A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 12cm, the liquid is dripping from the funnel at a rate of 0.2cm^3s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?
Now, from the chain rule, I understand that dV/dt = dV/dh x dh/dt, where V = Volume h = depth and t = time.
From the formula, V = (1/3)Ï€r^2h, dV/dh = (1/3)Ï€r^2, so 0.2 = (1/3)Ï€r^2 x dh/dt. As the depth is twelve, I believe the radius to be 8 because of the proportion rule 30:12 = 20:8
So dh/dt = 0.6/64Ï€ = 0.00298cms^-1, but the answer is 0.004cms^-1.
Can you show me where I’m zagging when I should be zigging? Cheers.