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Where did this came from?

  1. Nov 17, 2005 #1
    Hi,
    we were given this problem in the class:
    Volume of cylinder is [itex] V = \pi r^2h[/itex]
    v = 5 cm (with precision of 0.005 cm)
    r = 3 cm (with precision of 0.01 cm)
    What maximum flaw was the volume of the cylinder measured with?
    Well, I didn't know how to do it, so I just wrote teacher's solution. Now I'm trying to understand it, but without success. Here it is:

    [tex]
    h = 5 cm,\ dh = 0.0005\ cm
    [/tex]
    [tex]
    r = 3 cm,\ dr = 0.01\ cm
    [/tex]
    [tex]
    dV =\ ?
    [/tex]

    [tex]
    dV = \frac{\partial V}{\partial h} dh\ +\ \frac{\partial V}{\partial r}dr = \pi r^2 dh\ +\ 2\pi rh dr\ =\ 0.345\ \pi cm^3
    [/tex]


    I just don't get the idea at all...Could someone explain it to me?
    Thank you.
     
    Last edited: Nov 17, 2005
  2. jcsd
  3. Nov 17, 2005 #2

    HallsofIvy

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    Recall that the derivative, [itex]\frac{dy}{dx}[/itex] is the limit of the "difference quotient", [itex]\frac{y(x+h)- y(x)}{h}[/itex], as h goes to 0. If h is small, then [itex]\frac{dy}{dx}[/itex] is approximately equal to the difference quotient. Of course, then, y(x+h)- y(x) is approximately equal to [itex]\frac{dy}{dx}h[/itex]. Often we write that as if h were equal to "dx": [itex]\frac{dy}{dx}dx= dy[/itex]. That's not true but as long as h is small, it is approximately correct.
    Now, suppose y= y(x) is a function of x and, in measuring x, we make a slight error and get x+h. Calculating y with that incorrect value of x we get y(x+h) when the true value should be y(x). The error is y(x+h)- y(x) which, as above, is approximately dy= y'(x)dx. That is, although technically dx and dy only have meaning in the derivative, we can approximate dx by the "error in x" and dy by the "error in y".
    Here, [itex]V= \pi r^2h[/itex] and so, as you have,
    [tex]dV= 2\pi rh dr+ \pi r^2 dh[/itex]
    Now we approximate dr by the error in r: dr= 0.005 and dh by the error in h: dh= 0.01. setting r= 3 and h= 5 gives the result you state.
    Again, that is approximate. One can calculate the exact value by:
    Since h was measured as 5cm and the maximum error in h is 0.005 cm, h could be as large as 5.005 cm. Since r was measured as 3 cm and the maximum error in r is 0.01, r could be as large as 3.01 cm. That means that the volume of the cylinder could be as large as [itex]\pi (3.01^2)(5.005)= 142.458[/itex].
    On the other hand, h could be as small as 5-0.005= 4.995 and r could be as small as 3-0.01= 2.99. That means that the volume of the cylnder could be as small as [itex]\pi(2.99^2)(4.995)[/itex]=140.290.
    Since [itex]\pi(3^2)(5)= 141.371[/itex] you can see that [itex]142.458-141.371= 1.087[/itex] and [itex]141.371- 140.290= 1.081[/itex] so that the "possible error" of [itex]0.345\pi[/itex] = 1.0838 from the approximation is correct to 2 decimal places.
     
    Last edited: Nov 17, 2005
  4. Nov 17, 2005 #3

    benorin

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    The last line in HallsofIvy's post above with the non-displayed latex graphic is:

    "Since [itex]\pi(3^2)(5)= 141.371[/itex] you can see that 142.458-141.371= 1.087 and 141.371- 140.290= 1.081 so that the "possible error" of [itex]0.345\pi=1.0838[/itex] from the approximation is correct to 2 decimal places."
     
  5. Nov 17, 2005 #4

    HallsofIvy

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    Thanks, benorin, I've fixed it. I left out a [/itex]!
     
  6. Nov 17, 2005 #5
    Thank you HallsoftIvy for your comprehensive explanation, now I get it, excepting this step:

    If V were function of single variable, I would write

    [tex]
    dV = V'(x)\ dx
    [/tex]

    But there are two variables..

    I see that the expression is equal to:

    [tex]
    dV = \frac{\partial V}{\partial r}\ dr + \frac{\partial V}{\partial h}\ dh
    [/tex]

    Anyway, I don't know why. Is it some use of chain rule? Or something different?

    Thank you.
     
  7. Nov 17, 2005 #6

    Tom Mattson

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    It's not the chain rule. The chain rule is a theorem. What you have cited above is the definition of the total differential of a function of two variables.
     
  8. Nov 17, 2005 #7
    Well, probably you're right, we didn't any definition involving this expression for total differential.

    Anyway, total differential is always in some particular point, isn't it? (I mean the linear transformation, not the value in some point itself)

    We had defined total differential as linear transformation which satisfies certain zero limit.

    Then after few theorems it was clear that in point h the value of total differential of function f in point a is equal to

    [tex]
    Df(a)(h) = \bigtriangledown f(a) .h
    [/tex]

    where

    [tex]
    \bigtriangledown f(a) := \left(\frac{\partial f}{\partial x_1}(a), ... , \frac{\partial f}{\partial x_n}(a)\right)
    [/tex]

    Could you please show me the connection?

    Thank you.
     
    Last edited: Nov 17, 2005
  9. Nov 17, 2005 #8

    Tom Mattson

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    Sure, the 2 dimensional analog of your equation [itex]dV(x)=V'(x)dx[/itex] is:
    [tex]dV(x,y)=\vec{\nabla}V(x,y)\cdot d\vec{r}[/tex],
    where [itex]d\vec{r}=dx\hat{i}+dy\hat{j}[/itex].
    This gives us:
    [tex]dV(x,y)=\frac{\partial V}{\partial x}dx+\frac{\partial V}{\partial y}dy[/tex]
    Now let [itex]x=r[/itex] and [itex]y=h[/itex] and you've got it.
     
    Last edited: Nov 18, 2005
  10. Nov 18, 2005 #9
    Thank you Tom! I had little problems with the way of symbolics used, but now I have it and I'm happy. I hate solving problems just with learned approach but not understanding the idea.

    Thank you all, HallsoftIvy, Tom, benorin!
     
  11. Nov 18, 2005 #10

    Tom Mattson

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    Just a heads up: I had a typo in my post, which I have just corrected.

    This:

    [tex]dV(x,y)=\vec{\nabla}\cdot d\vec{r}[/tex]


    should have been this:

    [tex]dV(x,y)=\vec{\nabla}V(x,y)\cdot d\vec{r}[/tex]
     
  12. Nov 18, 2005 #11
    Yes, that's what I was referring to when I wrote "I had little problems with the way of symbolics used" :smile:
     
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