## anonymous 3 years ago A large container has the shape of a frustum of a cone with top radius 8m, bottom radius 2m, and height 6m. The container is being filled with water at the constant rate of 2.9m^3/min. At what rate is the level of water rising at the instant the water is 1m deep? Answer: 10.3cm/min

1. anonymous

$\frac{ dV }{ dt } = +2.9, \frac{ dH }{ dt }=?$

2. anonymous

@iop360 Do you have the Volume formula?

3. anonymous

of a regular cone, yes

4. anonymous

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

5. anonymous

No this one, have 2 bases!

6. anonymous

i think what we are supposed to do though is make it into two cones though

7. anonymous

|dw:1353202550485:dw|

8. anonymous

9. anonymous

b and B represent the different volumes?

10. anonymous

how did you derive this formula

11. anonymous

oh ok

12. anonymous

ill try it. thanks!

13. anonymous

hmm

14. anonymous

have you tried it?

15. anonymous

16. anonymous

i got .383...m/min

17. anonymous

answer is supposed to be 10.3 cm/min

18. anonymous

oh wait i think i made an error.. let me recalculate

19. anonymous

my expression for V is: $V = [\frac{ 16 }{ 27 }(\pi)h^3 + \frac{ 4}{ 27 }(\pi)h^3 + \frac{ 1 }{ 9 }(\pi)h^3]$

20. anonymous

what did you get?

21. anonymous

oh wait.. 1/27, not 1/9

22. anonymous

ok yeah thats what i get now

23. anonymous

taking its derivative, you get $\frac{ 7 }{ 3 }(\pi)h^2$

24. anonymous

dh/dt beside it of course

25. anonymous

(2.9)/(7pi/3) = dh/dt after you plug h =1, which does nothing

26. anonymous

0.395...m^3/min

27. anonymous

i think we did it right, not sure why its resulting in the wrong answer

28. anonymous

did you get 0.395...m/min

29. anonymous

doesnt that still result in 39.5cm/min then?

30. anonymous
31. anonymous

it might be a solution, but i dont get it

32. anonymous

hmm is there anything extra in the formula you may have forgotten

33. anonymous

34. anonymous

http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html this link doesnt have a square root in the formula

35. anonymous

bleh, the formula without the sq.root doesnt work either...

36. anonymous

!!! i think i got it

37. anonymous

forgot the pi from the above forumla

38. anonymous
39. anonymous

i got 10.1 though...calculator didnt take exact values

40. anonymous

$V = \frac{ (\pi)h }{ 3 }[R^2 + Rr + r^2]$

41. anonymous

im going to calculate it again..

42. anonymous

im getting 10.05... cm/min that is slightly off

43. anonymous

hm ok

44. anonymous

thanks

45. anonymous

im going to try it with different values

46. anonymous

oh wait...i used base for r in that last formula

47. anonymous

uh oh

48. anonymous

ill try it with r/R

49. anonymous

annnnnnd were back to 21pi/27 h^3

50. anonymous

yeah, i accidently used the B/b for that forumla instead of R/r

51. anonymous

and yet i got close to a right answer

52. anonymous

do we sub in h = 1 or h =6? h = 1 is right isnt it?

53. anonymous

Yes, we calculate the instant rate when h = 1

54. anonymous

the other guy reading this question: are you doing this question?

55. Callisto

*learning* ._.!

56. anonymous

i could switch the values for the question and try it again if you want, haha

57. anonymous

ok i will

58. anonymous

a large container has the shape of a frustum of a cone with top radius 10m, bottom radius 6m, and height 4m. The container is being filled with water at the constant rate of 2.9m^3/min. At what rate is the level of water rising at the instant rate the water is 2m deep?

59. anonymous

60. anonymous

alright

61. anonymous

i get 1.9 ..hmmm

62. anonymous

it could be a glitch with the thing, maybe.

63. anonymous

ok, im pretty sure were doing it right. imma close this down

64. anonymous

thanks for the help

65. anonymous

Check with your teacher about the formula! I'll sure will message you if I find out something interesting :)

66. anonymous

thanks!

67. anonymous

I'm so sorry that I'm underestimate this Related Rate frustum of Cone! Here's the correct logic: V = π/ 3 ( R² + Rr + r² ) H With r = 2 and R is Radius variable and H is the Height variable -> V = π/ 3 ( R² + 2R+ 4 ) H ( 1 ) From the ratio of right triangle: ( R - 2 ) / ( 8 - 2 ) = H/ 6 --> R = H + 2 (2) Plug (2) into (1): => V = π/ 3 [ ( H +2) + 2( H +2) + 4 ] H = π/ 3 [ H³ + 6H² + 12H ] so V' = π/ 3 [ 3H² + 12H + 12 ] H' At H = 1: V' = 27 π/ 3 * H' 2.9 = 9 π * H' Thus H' = 2.9 / 9π = .1025 m/ min = 10.3 cm/ min

68. anonymous

thanks!

69. anonymous

@Chlorophyll one question though.. what do you do exactly during the ratios of the triangle part? eg. say my top radius was 9, bottom was 8, and height was 5 how do you set it up exactly?

70. anonymous

Ratio R and H: ( R - 8 )/ ( 9-8) = H/5 -> R = ( H/5 ) + 8

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