## anonymous 5 years ago water is running out a conical funnel at the rate of 1cu in per sec. if the radius of the base of the funnel is 4an and the altitude is 8in. find the rate at which the water level is dropping when it is 2 in from the top.

1. amistre64

change of rates eh....

2. amistre64

v' = 1 ft^3 per sec; or simply 1 the rate of h is what we wanna determine

3. amistre64

volume of a cone = (1/3)(base area)(height)

4. amistre64

the relation of height to raduis is given by the cross section of the cone

5. amistre64

h = -2r+8 right?

6. anonymous

can you show the solution?

7. amistre64

v = (1/3)(pi 2^2)(-2r +8)

8. amistre64

im working on it; havent got to the end yet lol

9. amistre64

pi 2^2 was meant to be pi r^2

10. amistre64

lets re work that with h as the variabe so we can see the rate of change with respect to h and not r

11. amistre64

r = (h-8)/-2 or (8-h)/2

12. amistre64

v = (1/3)(pi ((8-h)/2)^2)(h) perhaps?

13. amistre64

$v = \frac{h}{3}*\frac{\pi.(8-h)^2}{4}$

14. amistre64

64hpi +h^3pi -16h^2pi --------------------- = v derive now 12

15. amistre64

64pi +3h^2pi -32hpi --------------------- = v' derive now 12

16. amistre64

pi(64 +3h^2 -32h) = 12 64 +3h^2 -32h = 12/pi right?

17. anonymous

ang gulo wah

18. amistre64

maybe i mess that up alittle dv/dh = that up there dv/dt = 1 and dv/dh dh/dt = dv/dt

19. amistre64

we wanna find dh/dt sooo dh/dt = dh/dv dv/dt ; gotta take the inverse if thats gonna work

20. amistre64

dh/dt is the inverse of dh/dv right :)

21. amistre64

on paper i get dh/dt = -1/(9pi)