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Math Help - Differential Analysis

  1. #1
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    Differential Analysis

    For some reason, I'm not getting this....

    The volume of an object is given by V=[pi(4.5(t)+1)^2(t)] If t changes from 4 to 5:

    Find dV and delta(V)


    Heres how I am solving it:
    Vprime= pi(4.5t+1)^2 +pi*t(2)(4.5t+1)(4.5)
    dV= [Vprime(t)]*[dt].. dt=1 and t=4 so:

    [pi(4.5(4)+1)^2 +pi*(4)(2)(4.5(4)+1)(4.5)][1]= 3282.96

    deltaV=f(t+delta(t))-f(t) so:
    deltaV=f(4+1)-f(4)=
    [pi(4.5(5)+1)^2(5)]-[pi(4.5(4)+1)^2(4)]=4138.26

    dV approximates delta(V) so the numbers should be very close but as you can see, they're not.

    Can anyone help?
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  2. #2
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    Quote Originally Posted by Matt85 View Post

    If t changes from 4 to 5

    Vprime= pi(4.5t+1)^2 +pi*t(2)(4.5t+1)(4.5)
    dV= [Vprime(t)]*[dt].. dt=1 and t=4 so:

    [pi(4.5(4)+1)^2 +pi*(4)(2)(4.5(4)+1)(4.5)][1]= 3282.96

    deltaV=f(t+delta(t))-f(t) so:
    deltaV=f(4+1)-f(4)=
    [pi(4.5(5)+1)^2(5)]-[pi(4.5(4)+1)^2(4)]=4138.26

    dV approximates delta(V) so the numbers should be very close but as you can see, they're not.
    The dV is correct. The delta(V) is correct.

    The size of the change of V works with the size of the change of t. t increased by 25%, so of course V increased by a lot. If t went from 4 to 4.001, then you would see a more obvious connection.

    EDIT: Thanks for the explanation
    Last edited by Turiski; June 8th 2010 at 11:17 PM.
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  3. #3
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    Quote Originally Posted by Turiski View Post
    The derivative is correct. The delta(V) is correct

    You assumed that dV was intended to be evaluated at t=4. You didn't seem to have any motivation for that, but perhaps there was some. In general though dV shouldn't be evaluated numerically.

    Also, the size of the change of V works with the size of the change of t. t increased by 25%, so of course they are not particularly close. If t went from 4 to 4.001, then you would see a more obvious connection.
    4 was the starting point, and the change in t (5-4) is 1. So, I was evaluating the change in the tangent line, dV, against the change in the actual equation, delta(V). Throughout this section, we've been comparing values like this.

    Ahh, I get what you're saying about the change being more significant than from 4 to 4.001. And, it sounds like you came up with the same thing I did.
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