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Cherry + Cocktail related rate

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A cocktail is being poured into a hemispherical glass, which contains a cherry, at a uniform rate of 1 cm3/s. The cherry has a diameter of 2 cm and the glass has a radius of 3 cm. How fast is the level of the cocktail rising at the moment when half the cherry is submerged?

    [tex]\left( \frac{1}{4 \pi}\right)[/tex] is the answer

    [tex]\frac{dV}{dt} = 1 cm^3 /s[/tex]

    [tex]r_{cherry} = 1[/tex] cm because the diameter of the cherry is 2 cm

    [tex]r_{hemisphereical glass} = 3[/tex] cm

    2. Relevant equations

    I know that the formula for a sphere is [tex]V= \frac{4}{3} \pi r^3[/tex] so for a hemisphere which is half a sphere would be [tex]V= \frac{2}{3} \pi r^3[/tex]

    3. The attempt at a solution
    Here is where the problem arisis.... if i use the original Volume of a sphere equation:

    [tex]V = \frac{4}{3} \pi r^3[/tex]

    [tex]\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}[/tex]

    [tex]1 = 4 \pi (1)^2 \frac{dr}{dt}[/tex]

    [tex]1 = 4 \pi \frac{dr}{dt}[/tex]

    [tex]\frac{1}{4 \pi} = \frac{dr}{dt}[/tex]



    By the way, notice that I didn't use the formula for the hemispherical glass but the formula for the sphere which would mean the cherry but the question asks how fast is the level of the cocktail rising.... which now I become confused...........

    I get the answer alright but if I sub in 1 cm for r then that would mean im calculating for the cherry NOT the cocktail. So what am i doing wrong? My logic is incorrect but my answer is correct and PLUS i didnt even use the equation for a hemisphere .... only a normal sphere

    So what do i do ? is there a better way of solving this problem that actually makes sense and not senselessly putting in values and getting the answer Just like that..... Please help im very confuuuuseeeddddd

    All i know is i got the answer but that doesn't explain my understanding. Is this just a lucky shot or am i actually correct ???? is there a better way of solving the problem that makes much more sense and still arrives at the correct answer? Please help.

    The only thing i understanding is with no cocktail the cherry sits at the bottom by itself. When a little bit of cocktail comes in a little surface at the bottom of the cherry is covered with cocktail and as the cocktail level rises more of the surface is covered starting from the bottom and moving to center of the cherry which is the radius of the cherry. I just dont know how to put it in a quantitative way. I do understand it qualitatively.

    Thanks for your help.
     
  2. jcsd
  3. Apr 16, 2009 #2
    The radius of the hemisphere is FIXED.

    So dr/dt = 0 (i.e. the change of the radius with respect to time is 0.)

    Draw a picture of the hemisphere, the bottom half of a sphere. The level of the cocktail is measured along the vertical axis (either from the bottom, or from the top, but stick with which ever way you decide.) The height of the cocktail level (call it h) is different from the radius.

    To find the volume of a liquid at a certain height, h, you will need to set up an integral. However, you are after dV/dh, so you do not need to integrate. Use the FTC to dV/dh.

    Then dV/dt = dV/dh * dh/dt

    You will know the first two - solve to get dh/dt (change in height of liquid with respect to time.)

    Hope this helps.
     
  4. Apr 16, 2009 #3
  5. Apr 17, 2009 #4
    will somebody please help me out here?
     
  6. Apr 17, 2009 #5
    Try this:

    Draw the glass. Draw a vertical line through the center. We will measure the height of the liquid from the bottom upwards. Use h to represent the height of the liquid.

    Draw a horizontal line somewhere in the glass to represent the height of the liquid at some random moment. Mark that as "h". Since the radius of the glass is 3, the distance from the top of the liquid to the top of the glass is... 3 - h.

    Now draw a diagonal line from the center of the sphere (which is where your vertical line crosses the top of the glass) to the place where the liquid touches the right side of the glass. This should form a right triangle. The diagonal line will have length 3 because it is from the center of the sphere to the sphere surface. The right triangle has three sides, hypotenuse length 3, the vertical leg is length 3 - h. By the Pythagorean Thm, the other leg has length
    (3^2 - (3-h)^2)^(1/2).

    If you were to look down the glass from the top, you would see the liquid's top surface in the shape of a circle. The radius,r, of that circle is what we just found,
    r = (3^2 - (3-h)^2)^(1/2)
    So the area of that circle is pi*r^2 = pi*(3^2-(3-h)^2)

    Now, if you were to add a drop of liquid, the height of the liquid would raise a small amount, call it dh. The shape of the new layer of liquid is like a disk with a thickness of dh. The radius is what we found above. The volume of that thin slice of liquid is the amount that the volume increases, call it dV.

    So, dV = pi*(3^2-(3-h)^2) * dh
    So, dV/dh = pi*(3^2-(3-h)^2) (this becomes a derivative after we let dh -> 0)

    You also know, dV/dt = 1.
    And, by the chain rule, dV/dt = dV/dh * dh/dt

    So, 1 = pi*(3^2-(3-h)^2) * dh/dt

    So, dh/dt = 1/[pi*(3^2-(3-h)^2)]

    You want to find dh/dt when h = 1, so now plug in for h...

    dh/dt (when h=1) = 1/[pi*(9-(3-1)^2)] = 1/(5*pi) cm/sec
     
  7. Apr 17, 2009 #6
    The answer is [tex]\frac{1}{4 \pi}[/tex] not [tex]\frac{1}{5 \pi}[/tex] maybe there was a mistake? and also why did u use h=1 cm thats for the cherry not the cocktail. I dont even know the height of the cocktail at the moment when the cocktail covers half the cherry. How do you find that out ?

    and did u see the picture i drew take a look at my previous post with an attachment ? here is my full link again

    http://www.mathhelpforum.com/math-help/calculus/83955-cocktail-cherry-related-rate.html
     
  8. Apr 18, 2009 #7
    Sorry, the problem i did was the basic version, no cherry, just finding the dh/dt when the height was 1 cm (which is half the size of the non existant cherry resting at the bottom.)

    If you modify my argument to include the cherry you will get it. It is tricky, because the cherry takes up space. It has a circular cross section whose area needs to be subtracted from the area of the liquid's surface, pi*(3^2-(3-h)^2). It would be messy to describe the cross-section of the cherry for all values of h. To be quick, I will just call the cross-sectional area of the cherry q(h) at a given liquid height h.

    Then, as before, dV/dh = pi*(3^2-(3-h)^2) - q(h)

    So, as before, dh/dt = 1/[pi*(3^2-(3-h)^2) - q(h)]

    Now, when h =1 we are half way up the cherry. So we need to plug in h = 1. But what is q(1)? Well, what is the cross sectional area of the middle of the cherry? It is a circle of radius 1, so the area is pi*1^2 = pi.

    Thus, dh/dt (when h=1) = 1/[pi*(9-(3-1)^2) - pi] = 1/(4*pi) cm/sec

    By the way, this is a very messy problem, I assume it is extra credit.
     
  9. Apr 18, 2009 #8
    Are u trying to say subtract the volume of the cherry from the volume of the cocktail without the cherry? but then again i dont know the radius of the cocktail........

    And in this problem do i need to introduce variables and make an equation for all values of h if the cherry is submerged inside the cocktail in a 3 dimensional way ? by a 3 dimensional way i mean using the Volume formulas and one more thing is [tex]V=\frac{2}{3} \pi r^3[/tex] the correct formula for a hemisphere ?


    Last question... how can u figure out what the height of the cocktail is when its submerging half the cherry. If i knew the height of the cocktail at the moment when half the cherry was submerged i can sub that into the hemisphere formula and differentiate. Thats what im trying to figure out and your trying to figure out something different and its extremely confusing.....
     
    Last edited: Apr 18, 2009
  10. Apr 18, 2009 #9
    If the problem does not state "the cherry remains at the bottom" or something like that, then the problem is unclear and they should rewrite it.

    But, by saying that the cherry will be half-way submerged, I am assuming that the cherry will not float and so stay at the bottom of the glass. Thus, measuring from the bottom up, the cherry's top is 2 cm from the bottom, and the cherry's middle (half-way) is 1 cm from the bottom.

    As far as using 2/3*pi*r^3, that is only useful for the volume of the entire hemisphere.

    You need a formula that shows the volume of the bottom portion of the glass that is filled with liquid, which is what I was developing before. But I did not derive the formula entirely because you said you have not done integration, which would be needed. Instead, I went after dV (small changes in the volume of the liquid) at a given height h of the liquid.

    As I said, this is NOT A BASIC related rates problem. It is better to practice a lot of basic ones than to focus on one very messy one.
     
  11. Apr 18, 2009 #10
    This is the last question on my worksheet and I've completed all other questions. There isnothing more left. Just this last question.

    Also, i think they want us to picture a cherry floating and not staying at the bottom when the cocktail pours in. It isnt realistic.
     
  12. Apr 18, 2009 #11
    I think we are supposed to assume that the cherry stays at the bottom at least until the liquid is half-way up the cherry.

    By the way, how would you describe the shape of the portion of liquid in the bottom of the glass (no cherry)? Just wondering if you think it is a "hemisphere"? Because, it is not. This might be why you are trying to use the volume of a hemisphere formula. But it does not apply.
     
  13. Apr 18, 2009 #12
    What do u mean?
     
  14. Apr 18, 2009 #13
    When you pour a little liquid in, say to a height of 1 cm, not all the way to the top, what is the shape of the space that the liquid occupies? Like, if you froze it and pulled out the ice, what shape would you call it?
     
  15. Apr 18, 2009 #14
    Okay here is what i thought in the first place. A sphere has a radius at the center and its the same distance from the center to any point at the edge of the sphere right? so a hemisphere is similiar. It has a radius at the center and it is the same going all around and down the glass. The only exception is that the radius does not go "up" from the glass which is outside the glass. Think of the radius shaping the glass itself if it was moving. Its the same everywhere shaping the glass.
     
  16. Apr 18, 2009 #15
    A hemisphere
     
  17. Apr 18, 2009 #16
    It is not.
     
  18. Apr 18, 2009 #17
    Explain why not
     
  19. Apr 18, 2009 #18
    Consider the whole sphere of radius 3.

    Now think about where that piece of ice would sit in it at the bottom.
    Now if we have another piece of ice just like it, we could picture it up at the top of the sphere. The two pieces do not make up the entire sphere. In fact, if we put them together, flat sides touching, they make a strange squashed shape, but definitely not a sphere.
     
  20. Apr 18, 2009 #19
    You are thinking in terms of a solid. What about a liquid. Remember science class? A liquid takes the shape of its container and if that is the case, the liquid is just a smaller version of the larger glass in which it is contained in.
     
  21. Apr 18, 2009 #20
    OK, I will try another way.

    You know that a sphere is the set of all points that are the same distance from a fixed point P.

    Now, a hemisphere is one half of those points (on one side) such as all points on or below the middle of the sphere (let us call the circle around the middle of the sphere the equator.)

    If we consider the bottom half hemi-sphere, then the same point P now lies at the top of that hemisphere. It lies in the center of the circle called the equator. It is still an equal distance from P to all points on the hemisphere.

    Now look at our liquid that only fills a little portion at the bottom of the glass (hemisphere). If you look down at the surface of the liquid, you see the top is a circle. We will call Q the point in the middle of that circle.

    But, Q is closer to the bottom of the glass than it is to the sides (such as directly to its left or right.) So, that bottom portion cannot be a hemisphere.

    Remember, if the shape is a hemisphere, then taking the center point on that top circle, that center point has to be an equal distance to all points on the sphere.
     
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