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I'm practicing how to differentiate equations.

  1. Jan 5, 2008 #1
    Last edited: Jan 5, 2008
  2. jcsd
  3. Jan 5, 2008 #2

    Gib Z

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    Can you please post the original question? The first link you have is an equation for the Volume, only variables being the height and radii - ie The volume is constant with time. The original question may have said the height or radii were changing with time though, so please tell us.
  4. Jan 5, 2008 #3
    Here's the question that I made up:

    Soda is being filled into a soda cup. The cup is 6 inches tall and has a diameter of 3.5 inches at the opening. The diameter is 2.5 inches at the base of the cup. The flow rate of the soda fountain is 8.47 in^3/s. Find the rate of change in height of the soda at the top of the cup.

    Cup I have looks like this: http://img136.imageshack.us/img136/272/1drs4622gz4.jpg

    Ignoring flow rate, I first need an equation for V. That's what I found. Then I need dV/dt to solve for dh/dt.
    r1 is the radii of base, r2 is radii of top of liquid, which is top of cup. dr1/dt is 0 b/c r1 is constant. dr2/dt is not constant.

    Peter H.
    Last edited: Jan 5, 2008
  5. Jan 6, 2008 #4

    \frac{dV}{dt} = \frac{8.47}{3} = 2.82

    Note however, that this is an 'averaged' rate of flow meaning, we are assuming that if [itex]x[/itex] liters flows in 1 second, [itex]2x[/itex] liters shall flow in 2 seconds i.e. [itex]V = kt[/itex].. i.e. Volume is directly proportional to time.. which i guess holds quite true for most real life situations, particularly a pumped soda fountain.

    At a time 't', some soda will be filled in the cup. That soda will constitute a frustum. Let the radius of this frustum be [itex]x[/itex]. Let the height at that time be [itex]h[/itex] and the total height be [itex]H[/itex]. The base radius will still be [itex]r_1[/itex]. Since we have assumed linear proportionality between Volume and time, the volume at that time 't' would be: [itex]V = 2.82t[/itex].

    Also, volume is equal to:

    V = \left (\frac{1}{2} \left (r_1 + r_2 \right ) \right )^2 \pi h

    which is:

    2.82t = \left (\frac{1}{2} \left (r_1 + r_2 \right ) \right )^2 \pi h

    Let's call this equation (II).

    In this we have two variables [[itex]x[/itex] and [itex]h[/itex]] which change w.r.t time. Hence, it would be quite complicated if we just solving this. The Key Idea here is that, for a particular height [itex]h[/itex], we have a particular radius of the frustum, [itex]x[/itex]. This basically means that we can relate these two quantities.

    The relation is:

    x = \frac{h(r_2 - r_1)}{H} + r_1

    Please see the attached image for how I came to this relation. Now substitute this value of 'x' in the equation (II) and differentiate both sides w.r.t. time to get [itex]\frac{dh}{dt}[/itex].

    I may have made a silly mistake in here.. because it seems like dh/dt will be constant.. which is a bit strange.. but since i haven't really calculated it.. it might be dependent on time.. but nevertheless.. this should give you an idea on how to solve these kinds of questions..

    Attached Files:

  6. Jan 6, 2008 #5
    oops.. i calculated a few stuff.. and found out that my method is sorta... wrong. It's not wrong as in.. it would give you the correct answer... but it's gonna be too much calculation. It is because when you first differentiate w.r.t time, you get dh/dt in terms of 'h', which again will change w.r.t time. So, to get dh/dt purely in terms of 't', we need to make a better substitution in the beginning, or as in the current case.. we'll have to substitute a relation between h and t.. which is quite tedious.

    so.. hang on till someone wiser replies.. i'll try to think of something.. and post if i come up with anything :D
  7. Jan 6, 2008 #6


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    Here's what I would do. The cup is a "truncated" cone so draw a picture of it "seen from the side": draw a triangle having vertex at the origin of a coordinate system, base of length r2 parallel to the x-axis. Essentially, you have two symmetric lines throught the origin. You want the line on the right to pass through (0,0), [itex](r_1/2,\lambda)[/itex] (the base of the cup), and [itex](r_2/2,\lambda+ H)[/itex] (the top of the cup). The equation of any line passing throug (0,0) and [itex](r_1/2, \lambda)[/itex] must be and the equation of any line passing through (0,0) and [itex](r_2/2,\lambda+ H)[/itex] must be [itex]y= ((\lambda+ H)/r_2)x[/itex]. For a line passing through all 3 of those points, we must have [itex]\lambda/r_1= ((\lambda+ H)/r_2[/itex]. From that we get that [itex]\lambda= Hr_1/(r_2- r_1)[/itex]. That is, the equation of the line is y= The water in the cup must conform to the sides of the cup and so must be a truncated cone have lower base with radius [itex]r_1/2[/itex], height h, and upper base radius given by [itex]h= (\lambda/r_1)r_1= (Hr_1^2/(r_2- r_1))[/itex].

    The volume of a cone of height h and base radius r is [itex](1/3)\pi r^2h[/itex] and the volume of the water is the difference of the volumes of the two cones: [itex](1/3)\pi [(Hr_1^2/(r_2- r_1))]^2(h+ Hr_1^2/(r_2-r_1)[/itex] and [itex](1/3)\pi r_1^2 (h+ Hr_1^2/(r_2-r_1)[/itex]. Subtract those and differentiate with respect to h.
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