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Math Help - Differential Word Problem

  1. #1
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    Post Differential Word Problem

    I'm trying to solve this problem, but I don't know how to set up the problem:

    A golf ball consists of a spherical cover over a solid core. Let "t" denote the thickness of the cover and "r" denote the radius of the core.

    a) Determine a formula for the exact volume of the cover

    b) Use differentials to approximate the volume of the cover at "r"

    c) If r = 0.8 inches and t = 0.04 inches, determine the exact volume of the cover

    d) If r = 0.8 inches and t = 0.04 inches, approximate the volume of the cover

    e) Compute the error ( true value - approximate value)

    f) Compute relative error (error/actual value)


    I know it seems like a lot to ask for, and I'm not asking for the answers, or all of the answers; I just need some help so that I can work through the problem.

    Thank you in advance.
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  2. #2
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    Re: Differential Word Problem

    Quote Originally Posted by OmegaJudicator View Post
    I'm trying to solve this problem, but I don't know how to set up the problem:

    A golf ball consists of a spherical cover over a solid core. Let "t" denote the thickness of the cover and "r" denote the radius of the core.

    a) Determine a formula for the exact volume of the cover

    b) Use differentials to approximate the volume of the cover at "r"

    c) If r = 0.8 inches and t = 0.04 inches, determine the exact volume of the cover

    d) If r = 0.8 inches and t = 0.04 inches, approximate the volume of the cover

    e) Compute the error ( true value - approximate value)

    f) Compute relative error (error/actual value)


    I know it seems like a lot to ask for, and I'm not asking for the answers, or all of the answers; I just need some help so that I can work through the problem.

    Thank you in advance.
    (a) cover volume = outer volume - inner volume = \frac{4}{3}\pi[(r+t)^3 - r^3]

    (b) V = \frac{4}{3}\pi r^3

    determine \frac{dV}{dr} and solve for dV, the approximate volume of the cover at "r"

    you should have enough info to finish the problem
    Thanks from OmegaJudicator
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  3. #3
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    Re: Differential Word Problem

    Quote Originally Posted by skeeter View Post
    (a) cover volume = outer volume - inner volume = \frac{4}{3}\pi[(r+t)^3 - r^3]

    (b) V = \frac{4}{3}\pi r^3

    determine \frac{dV}{dr} and solve for dV, the approximate volume of the cover at "r"

    you should have enough info to finish the problem
    I don't understand what part b is trying to ask for. As far as I know, when there is a problem involving differentials, I use the linear approximation formula. I think I'm having issues with the variables.

    For b, if I'm approximating the volume of the cover, wouldn't I be using the formula from part a, then take the derivative?

    Side note:

    I asked my professor for some help, and she told me that the volume of the cover, which can be called f, will be the difference of the volume of the two concentric spheres. She also menioned that f = ΔV, where V is the volume of a sphere of radius r, which would be the sphere on the inside. She told me to approximate f, not V.

    This confuses me more.
    Last edited by OmegaJudicator; March 25th 2012 at 12:56 PM.
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  4. #4
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    Re: Differential Word Problem

    Quote Originally Posted by OmegaJudicator View Post
    I don't understand what part b is trying to ask for. As far as I know, when there is a problem involving differentials, I use the linear approximation formula. I think I'm having issues with the variables.

    For b, if I'm approximating the volume of the cover, wouldn't I be using the formula from part a, then take the derivative?
    for (b) ...

    \frac{dV}{dr} = 4\pi r^2

    dV = 4\pi r^2 \cdot dr

    for r = 0.8 , t = 0.04 \approx dr

    dV = 4\pi (0.8)^2 \cdot (0.04)
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    Re: Differential Word Problem

    Quote Originally Posted by skeeter View Post
    for (b) ...

    \frac{dV}{dr} = 4\pi r^2

    dV = 4\pi r^2 \cdot dr

    for r = 0.8 , t = 0.04 \approx dr

    dV = 4\pi (0.8)^2 \cdot (0.04)
    I'm confused, where are you getting those values, when those are for part c and d?

    And if we are approximating the volume of the cover, shouldn't we be using the first formula you presented? I'm sorry I an trying to understand where you are approaching this problem.
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  6. #6
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    Re: Differential Word Problem

    Quote Originally Posted by OmegaJudicator View Post
    I'm confused, where are you getting those values, when those are for part c and d?

    And if we are approximating the volume of the cover, shouldn't we be using the first formula you presented? I'm sorry I an trying to understand where you are approaching this problem.
    the first formula gets you the exact volume of the cover, the differential equation approximates it

    Last edited by skeeter; March 25th 2012 at 01:14 PM.
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    Re: Differential Word Problem

    Quote Originally Posted by skeeter View Post
    the first formula gets you the exact volume of the cover, the differential equation approximates it
    But we are not given any values on b. Part a and part c are related, and part b and part d are related as well. The problem is I don't know what to do with part b; shouldn't I use: the formula for the cover volume, and take the derivative of it, with respect to r?
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  8. #8
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    Re: Differential Word Problem

    go to my previous post and watch the video
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    Re: Differential Word Problem

    Quote Originally Posted by skeeter View Post
    go to my previous post and watch the video
    I'm still confused in part b, the video helped clarify some parts, but I'm still lost.

    I'm sorry if I am not with you on the same page, I'm trying to work into it.
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  10. #10
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    Re: Differential Word Problem

    (b) ... the differential, dV = 4\pi r^2 \cdot dr , is the approximation for for the volume of the cover.

    I just put the numbers given in parts (c) and (d) to show you how the approximation works. If you put the same numbers in the equation written for part (a), that gets you the exact volume of the cover.

    calculate both, then finish (e) and (f).
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    Re: Differential Word Problem

    Quote Originally Posted by skeeter View Post
    (b) ... the differential, dV = 4\pi r^2 \cdot dr , is the approximation for for the volume of the cover.

    I just put the numbers given in parts (c) and (d) to show you how the approximation works. If you put the same numbers in the equation written for part (a), that gets you the exact volume of the cover.

    calculate both, then finish (e) and (f).
    I might come off as sounding hard-headed, but I and getting confused with what you're telling me and what my professor told me.

    she told me that the volume of the cover, which can be called f, will be the difference of the volume of the two concentric spheres. She also menioned that f = ΔV, where V is the volume of a sphere of radius r, which would be the sphere on the inside. She told me to approximate f, not V.
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  12. #12
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    Re: Differential Word Problem

    Quote Originally Posted by OmegaJudicator View Post
    I might come off as sounding hard-headed, but I and getting confused with what you're telling me and what my professor told me.

    she told me that the volume of the cover, which can be called f, will be the difference of the volume of the two concentric spheres. She also menioned that f = ΔV, where V is the volume of a sphere of radius r, which would be the sphere on the inside. She told me to approximate f, not V.
    what you are calling f is the formula I wrote for part (a).

    f = \Delta V \approx dV = 4\pi r^2 \cdot dr ...

    ... speak to your prof if this last statement is not clear.
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  13. #13
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    Re: Differential Word Problem

    Quote Originally Posted by skeeter View Post
    what you are calling f is the formula I wrote for part (a).

    f = \Delta V \approx dV = 4\pi r^2 \cdot dr ...

    ... speak to your prof if this last statement is not clear.
    Thank you so much for your help; I will definitely go over this with my professor tomorrow.

    Have a great day
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