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Math Help - I need HELP!

  1. #1
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    I need HELP!

    I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by 10164064 View Post
    I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
    Note that f(t)=\int f'(t)\,dt

    Integrating, we get

    \int t \,dt = \frac{1}{2}t^2+C

    However, you're given an initial condition. Use it to find the value of C.

    \because f(0)=3, \ 3=\frac{1}{2}(0)^2+C\implies \color{blue}\boxed{C=3}

    Thus, \color{blue}\boxed{f(t)=\frac{1}{2}t^2+3}

    Hope that this makes sense!

    --Chris
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by 10164064 View Post
    I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
    f'(t)=t, means that f(t)=\int_0^t u \ du +c. Do the integral then put t=0 into the resulting expression for f(x) and solve for c.

    RonL
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  4. #4
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    got it

    YES thank you that makes sense
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 10164064 View Post
    I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3
    Haha, maybe this will be fun

    Let f'(t)=at+b

    So then

    \int{f'(t)}dt=f(t)=\frac{at^2}{2}+bt+C

    Now seeing that

    f(0)=3\Rightarrow{C=3}

    So we have that

    f(t)=\frac{at^2}{2}+bt+3

    ...you guys should thank me for doing all the heavy lifting around here.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    Haha, maybe this will be fun

    Let f'(t)=at+b

    So then

    \int{f'(t)}dt=f(t)=\frac{at^2}{2}+bt+C

    Now seeing that

    f(0)=3\Rightarrow{C=3}

    So we have that

    f(t)=\frac{at^2}{2}+bt+3

    ...you guys should thank me for doing all the heavy lifting around here.
    That is not the question asked, and is a pointless extension why not f'(x)=P_n(x) where P_n(x) is an n-th order polynomial in x?

    Also, the idea is to give the minimum help to allow the original poster to complete the solution themselves.

    RonL
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    That is not the question asked, and is a pointless extension why not f'(x)=P_n(x) where P_n(x) is an n-th order polynomial in x?

    Also, the idea is to give the minimum help to allow the original poster to complete the solution themselves.

    RonL
    Let f'(x)=P_n(x) where....just kidding
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