iop360
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
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iop360
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\[\frac{ dV }{ dt } = +2.9, \frac{ dH }{ dt }=?\]
Chlorophyll
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@iop360 Do you have the Volume formula?
iop360
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of a regular cone, yes
iop360
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\[V = \frac{ 1 }{ 3 }(\pi)r^2h\]
Chlorophyll
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No this one, have 2 bases!
iop360
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i think what we are supposed to do though is make it into two cones though
iop360
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|dw:1353202550485:dw|
iop360
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R is top radius?
iop360
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b and B represent the different volumes?
iop360
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how did you derive this formula
iop360
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oh ok
iop360
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ill try it. thanks!
iop360
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hmm
iop360
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have you tried it?
iop360
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im getting a wrong answer
iop360
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i got .383...m/min
iop360
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answer is supposed to be 10.3 cm/min
iop360
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oh wait i think i made an error.. let me recalculate
iop360
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my expression for V is:
\[V = [\frac{ 16 }{ 27 }(\pi)h^3 + \frac{ 4}{ 27 }(\pi)h^3 + \frac{ 1 }{ 9 }(\pi)h^3]\]
iop360
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what did you get?
iop360
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oh wait.. 1/27, not 1/9
iop360
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ok yeah thats what i get now
iop360
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taking its derivative, you get
\[\frac{ 7 }{ 3 }(\pi)h^2\]
iop360
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dh/dt beside it of course
iop360
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(2.9)/(7pi/3) = dh/dt after you plug h =1, which does nothing
iop360
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0.395...m^3/min
iop360
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i think we did it right, not sure why its resulting in the wrong answer
iop360
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did you get 0.395...m/min
iop360
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doesnt that still result in 39.5cm/min then?
iop360
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it might be a solution, but i dont get it
iop360
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hmm is there anything extra in the formula you may have forgotten
iop360
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bleh, the formula without the sq.root doesnt work either...
iop360
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!!! i think i got it
iop360
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forgot the pi from the above forumla
iop360
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i got 10.1 though...calculator didnt take exact values
iop360
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\[V = \frac{ (\pi)h }{ 3 }[R^2 + Rr + r^2]\]
iop360
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im going to calculate it again..
iop360
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im getting 10.05... cm/min
that is slightly off
iop360
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hm ok
iop360
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thanks
iop360
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im going to try it with different values
iop360
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oh wait...i used base for r in that last formula
iop360
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uh oh
iop360
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ill try it with r/R
iop360
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annnnnnd were back to 21pi/27 h^3
iop360
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yeah, i accidently used the B/b for that forumla instead of R/r
iop360
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and yet i got close to a right answer
iop360
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do we sub in h = 1 or h =6?
h = 1 is right isnt it?
Chlorophyll
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Yes, we calculate the instant rate when h = 1
iop360
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the other guy reading this question: are you doing this question?
Callisto
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*learning* ._.!
iop360
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i could switch the values for the question and try it again if you want, haha
iop360
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ok i will
iop360
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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?
iop360
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answer: 1.4cm/min
iop360
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alright
iop360
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i get 1.9 ..hmmm
iop360
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it could be a glitch with the thing, maybe.
iop360
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ok, im pretty sure were doing it right. imma close this down
iop360
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thanks for the help
Chlorophyll
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Check with your teacher about the formula!
I'll sure will message you if I find out something interesting :)
iop360
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thanks!
Chlorophyll
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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
iop360
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thanks!
iop360
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@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?
Chlorophyll
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Ratio R and H:
( R - 8 )/ ( 9-8) = H/5
-> R = ( H/5 ) + 8