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Math Help - [SOLVED] Differentiation / Integration

  1. #1
    Ife
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    [SOLVED] Differentiation / Integration

    Given the velocity and initial postion of a body moving along a coordinate line at time t, find the body's postion at time t.
    v = cos \frac{ \pi}{2}t, s(0)=1

    I am sure that my answer is wrong, i am thinking that if s' = cos \frac{ \pi}{2}t, then s = sin \frac{ \pi}{2}t+c, and if s(0)=1, then c=1 so s = sin \frac{ \pi}{2}t +1

    But i know that if i were to differentiate that same expression, i would get something else not so? (chain rule)... so what should I do here?
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Ife View Post
    Given the velocity and initial postion of a body moving along a coordinate line at time t, find the body's postion at time t.
    v = cos \frac{ \pi}{2}t, s(0)=1

    I am sure that my answer is wrong, i am thinking that if s' = cos \frac{ \pi}{2}t, then s = sin \frac{ \pi}{2}t+c, and if s(0)=1, then c=1 so s = sin \frac{ \pi}{2}t +1

    But i know that if i were to differentiate that same expression, i would get something else not so? (chain rule)... so what should I do here?
    Your integral should be

    s = \frac{2\sin{(\frac{\pi t}{2}})}{\pi} + C
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  3. #3
    Ife
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    Quote Originally Posted by 11rdc11 View Post
    Your integral should be

    s = \frac{2\sin{(\frac{\pi t}{2}})}{\pi} + C
    ohhhhh so that i can multiply by the differential of the inner term to get 1... ok.. thanks
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Ife View Post
    Given the velocity and initial postion of a body moving along a coordinate line at time t, find the body's postion at time t.
    v = cos \frac{ \pi}{2}t, s(0)=1

    I am sure that my answer is wrong, i am thinking that if s' = cos \frac{ \pi}{2}t, then s = sin \frac{ \pi}{2}t+c, and if s(0)=1, then c=1 so s = sin \frac{ \pi}{2}t +1

    But i know that if i were to differentiate that same expression, i would get something else not so? (chain rule)... so what should I do here?
    Hey Ife

    s(t)=\int{v}(t)dt=\int\cos{(\frac{\pi}{2}t)}dt

    Let u=\frac{\pi}{2}t, then \frac{2}{\pi}du=dt .

    \Rightarrow{s}(t)=\frac{2}{\pi}\int\cos{u}du=\frac  {2}{\pi}\sin{u}+C=\frac{2}{\pi}\sin(\frac{\pi}{2}t  )+C

    Now

    s(0)=1\Rightarrow1=\frac{2}{\pi}\sin(\frac{\pi}{2}  \cdot0)+C\Rightarrow{C}=1
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