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Math Help - Antiderivative problems!!

  1. #1
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    Antiderivative problems!!

    1. Find the particular antiderivative that satisfies the following conditions:


    ??

    2. Find the particular antiderivative that satisfies the following conditions:



    ???

    3. A particle is moving as given by the data below:


    where is the position, and is the velocity.
    ????
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  2. #2
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    Quote Originally Posted by Kayla_N View Post
    1. Find the particular antiderivative that satisfies the following conditions:


    ??

    2. Find the particular antiderivative that satisfies the following conditions:



    ???

    3. A particle is moving as given by the data below:


    where is the position, and is the velocity.
    ????
    They are all done the same ways so here is the first

    Since we know it is a derivative we can integrate both sides to get

    \int dR=\int \frac{40}{t^2}dt \iff R=-\frac{40}{t}+C

    Now we use the inital condition

    R(1)=40=-\frac{40}{1}+C \iff 80=C

    R(t)=-\frac{40}{t}+80
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  3. #3
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    Quote Originally Posted by Kayla_N View Post
    1. Find the particular antiderivative that satisfies the following conditions:


    ??

    2. Find the particular antiderivative that satisfies the following conditions:



    ???

    3. A particle is moving as given by the data below:


    where is the position, and is the velocity.
    ????

    CAn Someone please lead me into number 2. Where do i start? I look at number 1 and still bit confuse. Please help
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  4. #4
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    1.

    Integrate both sides.

    The integral of dR/dt is R. The integral of 40/t^2 is \frac{-40}{t} + C. Now you know that R = -40/t + C And when t = 1, R = 40. So, it follows that C = 40 + 40 = 80. Therefore, the particular integral (or antiderivative) is R(t) = -40/t + 80.

    2.

    Same thing. Integrate both sides.

    p'(x) is really dp/dx. Its integral is p. You can find out the integral of 40/x^3. Do NOT forget the +C (there are no limits), or else you won't get a 'particular' integral. You know that p(4) = 4. That means, when x = 4, p = 4 as well. Find C, then get the particular integral.

    I hope that helps.

    ILoveMaths07.

    P.S. Give it a try, show how you worked it out, and we'll help you out.
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  5. #5
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    so the integral i got was -(20/x^2)+C. Now p=4 means x=4.

    C= 21/4 and final answer i got is -20/x^2 +21/4.

    Thanks for helping me...

    CAN SOMEONE HELP ME WITH PLEASEEEEEEEE
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  6. #6
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    Quote Originally Posted by Kayla_N View Post
    so the integral i got was -(20/x^2)+C. Now p=4 means x=4.

    C= 21/4 and final answer i got is -20/x^2 +21/4.

    Thanks for helping me...

    CAN SOMEONE HELP ME WITH PLEASEEEEEEEE
    Please .... less of the hysterical pleading.

    Your value for C is wrong, I have no idea how you got it. Check your arithmetic.

    Q3 is done like the others. Note that by definition \frac{ds}{dt} = v.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Please .... less of the hysterical pleading.

    Your value for C is wrong, I have no idea how you got it. Check your arithmetic.

    Q3 is done like the others. Note that by definition \frac{ds}{dt} = v.

    Sorry but my C value is correct. Thanks for helping me on 3, i got (-3sin(t)-3cos(t))+3.
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  8. #8
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    Quote Originally Posted by Kayla_N View Post
    Sorry but my C value is correct. [snip]
    You're right. My mistake.
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