## anonymous one year ago Sand is falling on a pile , having shape pf a cone, at the rate 1.5 m3/min. assime diameter of base is 3 times the altitiude, at what rate is the altitude increasing when the altitude is 2m?

1. anonymous

@ganeshie8 @ParthKohli @Loser66

2. anonymous

whats is "r" equal too ?

3. anonymous

the question says 3d=h

4. anonymous

5. anonymous

other way around. d = 3h

6. anonymous

so r = 3h/2

7. anonymous

and V = πr²h/3, so make the substitution for r then take the derivative

8. anonymous

okey i did eberyhing i am getting 1/9pi and u ?

9. anonymous

i did somthing wrong

10. anonymous

yeah I got something different? What was your derivative?

11. anonymous

did u get 1/6pi?

12. anonymous

yes

13. jim_thompson5910

If you mean $\Large \frac{1}{6\pi}$ then you are correct

14. anonymous

how ?

15. anonymous

oh, that's what I thought you meant

16. jim_thompson5910

what did you get for dV/dt ?

17. anonymous

my der was 9/12pih3 dh/dt?

18. anonymous

after putiing in 3/2h

19. jim_thompson5910

the h^3 part is incorrect

20. anonymous

how

21. jim_thompson5910

you will have h^3 in the V volume formula but after you derive, it turns into 3h^2

22. anonymous

the formula for cone is x=1/3r2h

23. anonymous

when 3/2h goes in it become h2 then h3 by timising the outside h

24. anonymous

|dw:1433903048615:dw|

25. anonymous

|dw:1433903137752:dw|

26. anonymous

ok what are the steps take the derv fist then sub?

27. anonymous

Sub for r first because we don't have a value for dr/dt, then take the derivative

28. anonymous

wait so (3h/2) doesnt equal 9h2/4? it eiqls 9h/4?

29. anonymous

oh wait i think i get it now

30. anonymous

you good?

31. anonymous

h is 2 right

32. anonymous

yes

33. anonymous

then 9(2)^2 = 36

34. anonymous

hold on the 1/3 is gone beacuse of derv right

35. anonymous

well the 1/3 was gone before the derivative. It was 3/4 after the substitution, and 9/4 after the derivative

36. anonymous

damn i am going to take 5 min brake

37. anonymous

$V = \frac{1}{3} \pi r^2 h$ $r = \frac{3}{2} h$ $\rightarrow V = \frac{1}{3} \pi (\frac{3}{2}h)^2 h = \frac{3}{4} \pi h^3$ Take derivative with respect to time $\frac{dV}{dt}= \frac{9}{4} \pi h^2 \frac{dh}{dt}$ plug in dV = 1.5, h =2 , then solve for dh/dt $\frac{dh}{dt} = \frac{1.5}{9 \pi}$