Related Rates: Water is leaking out of an inverted conical t

[chucknorrisfish]

Water is leaking out of an inverted conical tank at a rate of 500 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 26 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

I know the volume of the cone (with radius "r" and height "h") is given by the formula V = (1/3)pi r^2 h .

I get an answer of 413.5121, and it's not right. Neither is 4135.121 or 41351.2. I don't know what I'm doing wrong.

 

[stapel]

I don't know what I'm doing wrong.

Since you haven't shown your work, I'm afraid we can't tell what you're doing wrong, either. Sorry.

Please reply showing all of your work and reasoning. Thank you.

Eliz.

 

[chucknorrisfish]

r/h = 2.25/10 gives you r=2.25h

V=pi(2.25h/10)^2 h gives you pi5.0625h^3/300

using the power rule you get dV/dt=3(pi)(5.0625)(h^2)/300
simplified you get dV/dt=(pi)(5.0625)(h^2)/100

since heighth of water = 1 m = 100 cm so:

dV/dt=(pi)(5.0625)(100^2)/100
then (dV/dt)(26)=41351.2133
and it's not right

 

[stapel]

r/h = 2.25/10 gives you r=2.25h

No; it gives r = 0.225h.

V=pi(2.25h/10)^2 h gives you pi5.0625h^3/300

Where is the "300" coming from?

using the power rule you get dV/dt = 3(pi)(5.0625)(h^2)/300

Where is your "dh/dt"?

Thank you.

Eliz.

 

[chucknorrisfish]

No; it gives r = 0.225h.
r=2.25h/10
300 comes from the 1/3 in the formula...

V=pi(2.25h/10)^2 h gives you (1/3)pi5.0625h^3/100
goes to pi5.0625h^3/300

dV/dt = (3(pi)(5.0625)(h^2)/300)(dh/dt)[/quote]

 

[wjm11]

at the same time that water is being pumped into the tank at a constant rate. The tank has height 10 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 26 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

i know V=1/3pi r^2 h
i get: 413.5121 and it's not right, neither is 4135.121 or 41351.2 i dont know what im doing wrong.

Stapel has steered you quite close. Here's a bit more:

Start by drawing a sketch and labeling it with the given info. From the geometry of the cone, one can establish the ratio between height and radius. Let’s use

r = .225h

Thus, our volume eqn becomes

V = (1/3)(pi)(.225h)^2*h = .016875(pi)h^3

Differentiate with respect to time:

dV/dt = 3(.016875(pi))h^2(dh/dt)

And we know that dh/dt = 26 cm/min when h = 1 m = 100 cm
(It’s important to keep all our units consistent. Therefore, work in cm and cm^3.)

Let’s label our inflow as dV2/dt and our outflow as dV1/dt. We know that

dV1/dt = -500cc/min

dV/dt = dV2/dt + dV1/dt

You should be able to finish up from here.

 

[galactus]

\(\displaystyle \L\\\frac{r}{h}=\frac{\frac{9}{4}}{10}\)

Solve for r: \(\displaystyle r=\frac{9h}{40}\)

\(\displaystyle \L\\V=\frac{1}{3}{\pi}r^{2}h\)

Sub: \(\displaystyle \L\\V=\frac{1}{3}{\pi}(\frac{9h}{40})^{2}h=\frac{1}{3}{\pi}(\frac{81h^{3}}{1600})=\frac{27{\pi}h^{3}}{1600}\)

\(\displaystyle \L\\\frac{dV}{dt}=\frac{81{\pi}h^{2}}{1600}\)

You know \(\displaystyle \frac{dV}{dt}; \;\ \frac{dV_{out}}{dt}=-500; \;\ \frac{dh}{dt}=26; \;\ h=1\)

Find \(\displaystyle \frac{dV_{in}}{dt}\)