anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ dV }{ dt } = +2.9, \frac{ dH }{ dt }=?\]
anonymous
  • anonymous
@iop360 Do you have the Volume formula?
anonymous
  • anonymous
of a regular cone, yes

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anonymous
  • anonymous
\[V = \frac{ 1 }{ 3 }(\pi)r^2h\]
anonymous
  • anonymous
No this one, have 2 bases!
anonymous
  • anonymous
i think what we are supposed to do though is make it into two cones though
anonymous
  • anonymous
|dw:1353202550485:dw|
anonymous
  • anonymous
R is top radius?
anonymous
  • anonymous
b and B represent the different volumes?
anonymous
  • anonymous
how did you derive this formula
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
ill try it. thanks!
anonymous
  • anonymous
hmm
anonymous
  • anonymous
have you tried it?
anonymous
  • anonymous
im getting a wrong answer
anonymous
  • anonymous
i got .383...m/min
anonymous
  • anonymous
answer is supposed to be 10.3 cm/min
anonymous
  • anonymous
oh wait i think i made an error.. let me recalculate
anonymous
  • anonymous
my expression for V is: \[V = [\frac{ 16 }{ 27 }(\pi)h^3 + \frac{ 4}{ 27 }(\pi)h^3 + \frac{ 1 }{ 9 }(\pi)h^3]\]
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
oh wait.. 1/27, not 1/9
anonymous
  • anonymous
ok yeah thats what i get now
anonymous
  • anonymous
taking its derivative, you get \[\frac{ 7 }{ 3 }(\pi)h^2\]
anonymous
  • anonymous
dh/dt beside it of course
anonymous
  • anonymous
(2.9)/(7pi/3) = dh/dt after you plug h =1, which does nothing
anonymous
  • anonymous
0.395...m^3/min
anonymous
  • anonymous
i think we did it right, not sure why its resulting in the wrong answer
anonymous
  • anonymous
did you get 0.395...m/min
anonymous
  • anonymous
doesnt that still result in 39.5cm/min then?
anonymous
  • anonymous
try reading this http://www.askmehelpdesk.com/mathematics/related-rates-frustum-cone-352086.html
anonymous
  • anonymous
it might be a solution, but i dont get it
anonymous
  • anonymous
hmm is there anything extra in the formula you may have forgotten
anonymous
  • anonymous
http://www.youtube.com/watch?v=1v1Pp-lJSKY this video ?
anonymous
  • anonymous
http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html this link doesnt have a square root in the formula
anonymous
  • anonymous
bleh, the formula without the sq.root doesnt work either...
anonymous
  • anonymous
!!! i think i got it
anonymous
  • anonymous
forgot the pi from the above forumla
anonymous
  • anonymous
http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html
anonymous
  • anonymous
i got 10.1 though...calculator didnt take exact values
anonymous
  • anonymous
\[V = \frac{ (\pi)h }{ 3 }[R^2 + Rr + r^2]\]
anonymous
  • anonymous
im going to calculate it again..
anonymous
  • anonymous
im getting 10.05... cm/min that is slightly off
anonymous
  • anonymous
hm ok
anonymous
  • anonymous
thanks
anonymous
  • anonymous
im going to try it with different values
anonymous
  • anonymous
oh wait...i used base for r in that last formula
anonymous
  • anonymous
uh oh
anonymous
  • anonymous
ill try it with r/R
anonymous
  • anonymous
annnnnnd were back to 21pi/27 h^3
anonymous
  • anonymous
yeah, i accidently used the B/b for that forumla instead of R/r
anonymous
  • anonymous
and yet i got close to a right answer
anonymous
  • anonymous
do we sub in h = 1 or h =6? h = 1 is right isnt it?
anonymous
  • anonymous
Yes, we calculate the instant rate when h = 1
anonymous
  • anonymous
the other guy reading this question: are you doing this question?
Callisto
  • Callisto
*learning* ._.!
anonymous
  • anonymous
i could switch the values for the question and try it again if you want, haha
anonymous
  • anonymous
ok i will
anonymous
  • 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?
anonymous
  • anonymous
answer: 1.4cm/min
anonymous
  • anonymous
alright
anonymous
  • anonymous
i get 1.9 ..hmmm
anonymous
  • anonymous
it could be a glitch with the thing, maybe.
anonymous
  • anonymous
ok, im pretty sure were doing it right. imma close this down
anonymous
  • anonymous
thanks for the help
anonymous
  • anonymous
Check with your teacher about the formula! I'll sure will message you if I find out something interesting :)
anonymous
  • anonymous
thanks!
anonymous
  • 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
anonymous
  • anonymous
thanks!
anonymous
  • 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?
anonymous
  • anonymous
Ratio R and H: ( R - 8 )/ ( 9-8) = H/5 -> R = ( H/5 ) + 8

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