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  1. #1
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    intergration by substitution

    OK I am getting really close now...the book shows a negative answer, which I am unsure of.

    The velocity v(t) = x'(t) at time t of an object moving along the x axis is given, along with the initial position x(0) of the object.
    Find:
    the position x(t) at time t
    the position of the object at time t=4
    the time when the object is at x=3

    x'(t) = -2(3t+1)^1/2, x(0) = 4

    y=u^1/2
    dy = 1/2u^3/2

    x=3t+1
    dx = 3
    x(t) = 1/2 u^3/2 (3t+1)^3/2 dx
    x(t) = 4/9(3t+1)^3/2 + C

    book says 4/9 should be negative.
    Unsure of how to do the position of the object at t=4 and time of x=3. The way I have tried to substitute isn't coming up with the books answer.
    Last edited by startingover; June 8th 2007 at 10:17 AM. Reason: left out sentence
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post

    x'(t) = -2(3t+1)^1/2, x(0) = 4
    where did you get that function from? was it a part of the question? if so, it should be negative, since we have a -2 in front
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  3. #3
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    Quote Originally Posted by startingover View Post
    OK I am getting really close now...the book shows a negative answer, which I am unsure of.

    The velocity v(t) = x'(t) at time t of an object moving along the x axis is given, along with the initial position x(0) of the object.
    Find:
    the position x(t) at time t
    the position of the object at time t=4
    the time when the object is at x=3

    x'(t) = -2(3t+1)^1/2, x(0) = 4

    y=u^1/2
    dy = 1/2u^3/2du

    x=3t+1
    dx = 3dt
    x(t) = 1/2 u^3/2 (3t+1)^3/2 dx
    x(t) = 4/9(3t+1)^3/2 + C

    book says 4/9 should be negative.
    Unsure of how to do the position of the object at t=4 and time of x=3. The way I have tried to substitute isn't coming up with the books answer.
    I didn't check the work above, but I did highlight some notation you are incorrect with.

    You need to integrate
    x \prime (t) = -2\sqrt{3t+1}, x(0) = 4

    So
    x(t) = \int dt -2\sqrt{3t+1}

    x(t) = -2 \int dt \sqrt{3t + 1}

    Let u = 3t + 1, then du = 3 dt

    x(u) = -2\int \left ( \frac{1}{3} du \right ) \sqrt{u}

    x(u) = -\frac{2}{3} \frac{1}{ \frac{3}{2} }u^{3/2} + C

    Where C is some constant. And since u = 3t + 1...

    x(t) = -\frac{2}{3} \cdot \frac{2}{3} (3t + 1)^{3/2} + C

    x(t) = -\frac{4}{9} (3t + 1)^{3/2} + C

    Now, we know that x(0) = 4 so

    x(0) = -\frac{4}{9} (3 \cdot 0 + 1)^{3/2} + C = 4

    -\frac{4}{9} + C = 4

    C = 4 + \frac{4}{9} = \frac{36}{9} + \frac{4}{9}

    C = \frac{40}{9}

    So finally
    x(t) = -\frac{4}{9} (3t + 1)^{3/2} + \frac{40}{9}

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by startingover View Post
    the position of the object at time t=4
    the time when the object is at x=3
    x(t) = -\frac{4}{9} (3t + 1)^{3/2} + \frac{40}{9}

    The first is easy: t = 4

    x(4) = -\frac{4}{9} (3 \cdot 4 + 1)^{3/2} + \frac{40}{9}

    x(4) = -\frac{4}{9} (13)^{3/2} + \frac{40}{9} \approx -16.3876

    The next problem is what is t when x = 3?

    x(t) = -\frac{4}{9} (3t + 1)^{3/2} + \frac{40}{9} = 3

    -\frac{4}{9} (3t + 1)^{3/2} = 3 - \frac{40}{9} = -\frac{13}{9}

    (3t + 1)^{3/2} = -\frac{9}{4} \cdot -\frac{13}{9} = \frac{13}{4}

    3t+ 1 = \left ( \frac{13}{4} \right )^{2/3}

    We can do little to improve the RHS so I'm just going to leave it this way.

    3t = \left ( \frac{13}{4} \right )^{2/3} - 1

    t = \frac{1}{3} \left ( \frac{13}{4} \right )^{2/3} - \frac{1}{3} \approx 0.398032

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    OK I am getting really close now...the book shows a negative answer, which I am unsure of.

    The velocity v(t) = x'(t) at time t of an object moving along the x axis is given, along with the initial position x(0) of the object.
    Find:
    the position x(t) at time t
    the position of the object at time t=4
    the time when the object is at x=3

    x'(t) = -2(3t+1)^1/2, x(0) = 4
    for questions like this, where the composite function consists of a LINEAR function plugged into some other function, i dont go through the whole process of substitution. i do something analogous to the chain rule of differentiation--yes, integration by substitution is analogous to the chain rule of differentiation, but i'm talking about something more direct. in this case, i'd integrate by taking the integral of the composite function and DIVIDE (as opposed to multiply if we were doing integration) by the derivative of what's inside the brackets.

    x'(t) = -2 (3t + 1)^{ \frac {1}{2}}

    \Rightarrow x(t) = -2 \int (3t + 1)^{ \frac {1}{2}} dt

    \Rightarrow x(t) = -2 \cdot \frac {2}{3} (3t + 1)^{ \frac {3}{2}} \cdot \frac {1}{3} + C

    \Rightarrow x(t) = - \frac {4}{9}(3t + 1)^{ \frac {3}{2}} + C

    and Dan took care of the rest
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