rate of change, volume of a funnel, height, radius, time

A funnel has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 12 cm, the liquid is dripping from the funnel at a rate of 0.2 (cm^3)(s^(-1)). At what rate is the depth of the liquid in the funnel decreasing at this instant?

Alright so at this instant the volume of water in the funnel is equal to that of a cone with a circular base, having height 12 cm and diameter 8 cm.

Therefore volume= (1/3)(pi)((r)^2)(h) = (1/3)(pi)(16)(h)

Therefore dV/dh = (16/3)(pi)

and we know that dV/dt = -0.2

so dh/dt = (dV/dt)/(dV/dh) = -.0119

But the right answer is (the depth of the liquid in the funnel is decreasing at a rate of) 0.0040 cm (s^(-1))

Where did I go wrong?

Re: rate of change, volume of a funnel, height, radius, time

Quote:

Originally Posted by

**furor celtica** A funnel has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 12 cm, the liquid is dripping from the funnel at a rate of 0.2 (cm^3)(s^(-1)). At what rate is the depth of the liquid in the funnel decreasing at this instant?

Alright so at this instant the volume of water in the funnel is equal to that of a cone with a circular base, having height 12 cm and diameter 8 cm.

Therefore volume= (1/3)(pi)((r)^2)(h) = (1/3)(pi)(16)(h)

Therefore dV/dh = (16/3)(pi)

and we know that dV/dt = -0.2

so dh/dt = (dV/dt)/(dV/dh) = -.0119

But the right answer is (the depth of the liquid in the funnel is decreasing at a rate of) 0.0040 cm (s^(-1))

Where did I go wrong?

you are given the rate of change of volume w/r to **time** ...

$\displaystyle \frac{dV}{dt} = 0.2 \, cm^3/s$

$\displaystyle V = \frac{\pi}{3} r^2 h$

$\displaystyle \frac{r}{h} = \frac{1}{3} \implies r = \frac{h}{3}$

$\displaystyle V = \frac{\pi}{27} h^3$

take the derivative w/r to time, substitute in your given values and calculate $\displaystyle \frac{dh}{dt}$

Re: rate of change, volume of a funnel, height, radius, time

Quote:

Originally Posted by

**furor celtica** A funnel has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 12 cm, the liquid is dripping from the funnel at a rate of 0.2 (cm^3)(s^(-1)). At what rate is the depth of the liquid in the funnel decreasing at this instant?

Alright so at this instant the volume of water in the funnel is equal to that of a cone with a circular base, having height 12 cm and diameter 8 cm.

Therefore volume= (1/3)(pi)((r)^2)(h) = (1/3)(pi)(16)(h)

Therefore dV/dh = (16/3)(pi)

This is incorrect. The radius is not a **constant**, it depends upon the depth of the water. The ratio of radius to height is 10/30= 1/3. Yes, when the height is 12, that is 4 but more generally, when the height is h, r= h/3. So your formula is $\displaystyle \frac{\pi}{3}r^2h= \frac{\pi}{3}\left(\frac{h^2}{9}\right)h= \frac{\pi}{27}h^3$. Differentiate **that** with respect to h.

Quote:

and we know that dV/dt = -0.2

so dh/dt = (dV/dt)/(dV/dh) = -.0119

But the right answer is (the depth of the liquid in the funnel is decreasing at a rate of) 0.0040 cm (s^(-1))

Where did I go wrong?

Re: rate of change, volume of a funnel, height, radius, time

thanks to the both of you