1. ## related rates help

water is poured into a conical funnel at a constant rate of 1 in^3/sec and flows out at the rate of 1/2in^3/sec. the funnel is a right circular cone with height of 4in and radius of 2in at the base. how fast is the water level changing when the water is 2in high? (v=1/3 pi r^2 h)

2. To start you off:

By using similar triangles it is easy to see that r/h = 2/4 where r is the radius at height h.

Use this to write the volume of the cone as a function as h (instead of r and h)

Then differentiate, and substitute in the information given.

3. yea i got that much but would the water be flowing in at 1/2 in^3/sec since its flowing in at 1 and flowing out at 1/2?

4. Yep - that's correct. dV/dt will be 1/2, and you are looking for dh/dt.

5. so far i have

(dv/dt)=2/3pi r (dr/dt) (dh/dt)
1/2=2/3pi 1 (dr/dt) (dh/dt)

i dont kno dr/dt or dh/dt but im trying to find dh/dt

6. Originally Posted by korky0429
so far i have

(dv/dt)=2/3pi r (dr/dt) (dh/dt) ... this is incorrect
1/2=2/3pi 1 (dr/dt) (dh/dt)

i dont kno dr/dt or dh/dt but im trying to find dh/dt
since $\displaystyle \frac{r}{h} = \frac{1}{2}$ , $\displaystyle r = \frac{h}{2}$

as stated earlier by DrSteve, sub this value in for r in the volume formula.

$\displaystyle V = \frac{\pi}{3} \left(\frac{h}{2}\right)^2 \cdot h$

simplify , take the time derivative, then sub in your given values.

7. Korky, you took the derivative incorrectly. If you leave V in terms of both h and r, then you need to use a product rule. Although you can do it this way, skeeter's method is more efficient.

8. is 2pi the right answer? does anyone kno or figure it out?

9. Originally Posted by korky0429
is 2pi the right answer? does anyone kno or figure it out?
2pi is incorrect ... recheck your arithmetic.