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Math Help - Misc Calculus questions

  1. #1
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    Misc Calculus questions

    Hi, im having trouble with some questions, here's one, i will post more as i run into more problems.

    The acceleration of an object, a m/s^2 moving in a line at t seconds is given by a = sin(2t) if the object has an initial velocity of 4 m/s then v is equal to?

    Now you'd think i would just have to integrate a and find the c constant, so i got: -\frac{1}{2}\cos{2t}+\frac{9}{2}
    but the answer was this:
    \int_{0}^{t} sin(2x)dx + 4

    can someone explain why this is correct?
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  2. #2
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    another similar question. I basically am having a problem with the form they give the answer in. I can find the exact answer result, but they always have it in this annoying form.

    Given \frac{dy}{dx}=\frac{2-y}{4}, x=3 when y=1, then when y=1/2, x is equal to?

    i got x= 4ln(\frac{3}{2})+3
    they wanted: \int_{1}^{\frac{1}{2}} \frac{4}{2-t}dt + 3

    im confused...
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  3. #3
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    this one is seemingly easier, i probably made a silly mistake somewhere, but I'd appreciate some help anyway:

    \frac{dy}{dx}=\frac{1}{5}(y-1)^2 and y=0 when x=0, then y is equal to?

    apparently the answer is \frac{x}{x+5}

    why?

    ps: i think i was overcomplicating something...
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  4. #4
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    Quote Originally Posted by scorpion007
    Hi, im having trouble with some questions, here's one, i will post more as i run into more problems.

    The acceleration of an object, a m/s^2 moving in a line at t seconds is given by a = sin(2t) if the object has an initial velocity of 4 m/s then v is equal to?

    Now you'd think i would just have to integrate a and find the c constant, so i got: -\frac{1}{2}\cos{2t}+\frac{9}{2}
    but the answer was this:
    \int_{0}^{t} sin(2x)dx + 4

    can someone explain why this is correct?
    They are the same thing.

    <br />
\int_0^t \sin(2x)\ dx=\left[\frac{-\cos(2x)}{2}\right]_0^t=\frac{-\cos(2t)}{2}+\frac{1}{2}<br />

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by scorpion007
    another similar question. I basically am having a problem with the form they give the answer in. I can find the exact answer result, but they always have it in this annoying form.

    Given \frac{dy}{dx}=\frac{2-y}{4}, x=3 when y=1, then when y=1/2, x is equal to?

    i got x= 4ln(\frac{3}{2})+3
    they wanted: \int_{1}^{\frac{1}{2}} \frac{4}{2-t}dt + 3

    im confused...

    The result they are using is:

    if: \frac{dx}{dt}=f(x), then:

    x(t)=\int_0^t\ f(\xi)\ d\xi\ + \ x(0)

    Which is just the fundamental theorem of calculus in disguise.

    RonL
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by scorpion007
    this one is seemingly easier, i probably made a silly mistake somewhere, but I'd appreciate some help anyway:

    \frac{dy}{dx}=\frac{1}{5}(y-1)^2 and y=0 when x=0, then y is equal to?

    apparently the answer is \frac{x}{x+5}

    why?

    ps: i think i was overcomplicating something...
    Its a variables seperable differential equation, and has solution:

    <br />
\int\ \frac{5}{(y-1)^2}\ dy=\int\ dx +c<br />

    and the initial conditions will determin c.

    RonL
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