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Math Help - Acceleration problem

  1. #1
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    Acceleration problem

    The position equation for the movement of a particle is given by s(t) = \sqrt{t^3 +1} where s is measured in feet and t is measured in seconds. Find the acceleration when t = 2 seconds.

    I know that I have to get y'', I got  {3t^2/2} -\sqrt{t^3+1} , but when trying to get y'' I just made a huge mess and came out to a wrong answer. Someone want to walk me through the rest of this?
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  2. #2
    Junior Member slider142's Avatar
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    Remember to keep the symbols straight; there is no y in the actual problem. Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).
    First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and t^3 + 1 is the inner function. To keep things clear, I like to use exponents in place of roots, so I would write it out:
    s(t) = (t^3 + 1)^\frac{1}{2}
    Explicitly, we can see that s(t) = f(g(t)) where g(t) = t^3 + 1 and f(t) = t^{\frac{1}{2}}. By the chain rule, s'(t) = f'(g(t))g'(t).
    s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)
    In order to get the second derivative you will have to use the product rule in addition to the chain rule. Can you find s''(t) from here?
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  3. #3
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    Quote Originally Posted by slider142 View Post
    Remember to keep the symbols straight; there is no y in the actual problem. Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).
    First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and t^3 + 1 is the inner function. To keep things clear, I like to use exponents in place of square roots, so I would write it out:
    s(t) = (t^3 + 1)^\frac{1}{2}
    Explicitly, we can see that s(t) = f(g(t)) where g(t) = t^3 + 1 and f(t) = t^{\frac{1}{2}}. By the chain rule, s'(t) = f'(g(t))g'(t).
    s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)
    Explicitly, we can see that s(t) = f(g(t)) where g(t) = t^3 + 1 and f(t) = t^(1/2)
    In order to get the second derivative you will have to use the product rule in addition to the chain rule. Did you find your problem here?
    Yes sir, whenever I integrate the power rule and the chain rule together I usually have a difficult time. So can you explain this to me so I can understand the concept better?
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  4. #4
    Junior Member slider142's Avatar
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    In the above, we have
    <br />
s'(t) = \frac{3}{2}(t^3 + 1)^{-\frac{1}{2}}(t^2)<br />
    Since we do not know its derivative offhand, we note that it is a product of two functions, each of whose derivative we do know. Specifically
    f(t) = (t^3 + 1)^{-\frac{1}{2}}
    and
    g(t) = t^2
    so that we have
    s'(t) =  \frac{3}{2}f(t)g(t)
    By the product rule, we know that
    s''(t) = \frac{3}{2}(f'(t)g(t) + f(t)g'(t))
    The only place you will need that chain rule here is in finding f'(t). Note how similar f(t) is to s(t). What do you propose s''(t) is?
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  5. #5
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    Okay I did it the way you asked me to and came out with {-9t^4/2}+{3(t^3+1)^-3/2/2}+3t^4 +3t
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  6. #6
    Junior Member slider142's Avatar
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    It looks like you're missing some small steps. Don't skip any steps until you get more experience with these rules. The first step is to get the parts of the product rule that you do not have, f'(t) and g'(t). Do not worry about how they interact with the rest of the equation yet. What did you get for f'(t)? What did you get for g'(t)?
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  7. #7
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    for f' I got 6t and for g' I got -1/2(-\sqrt{(t^3+1)}) (3t^2)
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  8. #8
    Junior Member slider142's Avatar
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    Quote Originally Posted by radioheadfan View Post
    for f' I got 6t and for g' I got -1/2(-\sqrt{(t^3+1)}) (3t^2)
    Okay, so in your case, you switched them around from my post and you let
    f(t) = 3t^2
    and
    g(t) = (t^3 + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{t^3 + 1}}.
    Your f'(t) is correct, but your g'(t) is not. g'(t) should be
    -\frac{1}{2}(t^3 + 1)^{-\frac{1}{2} - 1}(3t^2) = -\frac{1}{2}(t^3 + 1)^{-\frac{3}{2}}(3t^2)
    The outer function gets taken down one unit to the -3/2 power by the power rule of differentiation.
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  9. #9
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    Quote Originally Posted by slider142 View Post
    Okay, so in your case, you switched them around from my post and you let
    f(t) = 3t^2
    and
    g(t) = (t^3 + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{t^3 + 1}}.
    Your f'(t) is correct, but your g'(t) is not. g'(t) should be
    -\frac{1}{2}(t^3 + 1)^{-\frac{1}{2} - 1}(3t^2) = -\frac{1}{2}(t^3 + 1)^{-\frac{3}{2}}(3t^2)
    The outer function gets taken down one unit to the -3/2 power by the power rule of differentiation.

    Just to make sure I'm doing this correctly
    right now I have
    <br />
\frac{1}{2}[6t(t^3+1)^{-\frac{1}{2}}+{-\frac{9t^4}{2}}(t^3+1)^{-\frac{3}{2}}]
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  10. #10
    Junior Member slider142's Avatar
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    Quote Originally Posted by radioheadfan View Post
    Just to make sure I'm doing this correctly
    right now I have
    <br />
\frac{1}{2}[6t(t^3+1)^{-\frac{1}{2}}+{-\frac{9t^4}{2}}(t^3+1)^{-\frac{3}{2}}]
    That is exactly correct.
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  11. #11
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    Quote Originally Posted by slider142 View Post
    That is exactly correct.
    Awesome, okay so the final answer is?

    3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{2}
    Last edited by radioheadfan; August 4th 2009 at 06:01 PM.
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  12. #12
    Junior Member slider142's Avatar
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    Quote Originally Posted by radioheadfan View Post
    Awesome, okay so the final answer is?

    3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{2}
    Almost. The second term should also be multiplied by 1/2.
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  13. #13
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    3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{4}
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  14. #14
    Junior Member slider142's Avatar
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    Quote Originally Posted by radioheadfan View Post
    3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{4}
    Yep, you've got it.
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