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#1 2007-01-06 06:50:09

erndoglai
Guest

related rates help...

basically the problem is as follows

coffee is poured at a uniformrate of 20cmcubed/sec into a cup shaped like a truncated cone if the upper and lower radii of the cup are 4cm and 2 cm and the height of the cup is 6 cm, how fast will the cofee level be rising when the cofee is halfway up

i know this is asking what the related rate will be at height 3 cm nd radius 3 but i'm not sure what to take of it.... i know the cone untruncated is 12 cm high (if thats of any use) i've tried multiple ways of getting it and i get like soe random answers ranging from .7 - 1.4 so i'm really kind of lost for now

the formula of the volume of a truncated cone is

V = (1/3)(pi)(h)(r1^2  + r2^2 + r1r2)

where r = radius and h is the height

i'm not sure what relation i should put for the radius to make it all in terms of h ... b/c w/ the untruncated cone, the relation is 3:1 nd the truncated relation is 3:2

#2 2007-01-08 15:04:12

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: related rates help...

I think you should work out the UNITS that will make this thing divide out right.
For example, say you could make an equation of the volume given the height of the liquid.
The UNITS for this equation would be output of VOLUME or CC's, which is a milliliter.
The input into this equation is height of liquid.
Now you know the volume per second that is filling the container.
Now if you could get the inverse of the first equation, then
you could have the height of liquid expressed in terms of volume for input.
(I haven't looked to see how hard this is to do.)
Let's simplify the example to see if we get anywhere, and to see if we need calculus or not...
For pretend, Let's say that Height=total Volume times a constant.
This would be true for a cylinder being filled with water or anything with vertical sides.
Now the question is, given the variable "Volume", can we simply
divide both sides of the equation by some form of time??
What would be the rules of the algebra??
Would it be a constant multiplier??
Let's step back a bit.
We have a graph with Height on the y-axis and total Volume on the X-axis.
Next we could create a second graph, a simple graph in this case, of
total Volume on the Y-axis and Elapsed Filling Time of the X-axis.
(to be continue)


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#3 2007-01-08 15:10:12

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: related rates help...

If you have A in terms of B, and B in terms of C.
The question is, now you have A in terms of C, right?
And if you want to combine the equations, this is up to you.
Since B in terms of C is simple, the volume in terms of time,
then you can just substitute sides of the equation, forming
A in terms of C.
Then you can do as Newton has taught us, and get the 1st
derivative with respect to time.  And you need to know the
exact time volume at the point of interest, so you can get the
time from the volume, perhaps.
Anyhow, the really interesting thing I find here is what can
be done if we keep the two graphs separated.
A good reason to do this would be if both graphs were complicated,
and combining them would result in a messy equation.
I'll work on this in my spare time.  Good Luck now.

Last edited by John E. Franklin (2007-01-08 15:26:28)


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#4 2007-01-08 15:28:54

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: related rates help...

Maybe the chain rule deals with this??
Like y = f(g(x)), find dy/dx.
I'm rusty on calculus, but I think the chain rule might be this.


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