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Math Help - Differentiation: Tangent Lines and Velocity. Use position function s find velocity

  1. #1
    JDS
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    Differentiation: Tangent Lines and Velocity. Use position function s find velocity

    Use the position function s (in meters) to find the velocity at time t= a seconds.

    s(t) = 4/t, (a) a=2; (b) a=4

    I know the formula starts out something like this.....

    the limit as h approaches 0....... [s(2+h) - s(2)] / [(2+h) - 2]

    but after that I am quite lost.....I don't understand what it is I am supposed to do next. I have looked at my calculus book, looked at some of the solutions to similar problems, but I just don't get it. None of the examples deal with this exact problem.

    Any help or guidance would be greatly appreciated!!
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    fkf
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Do you have to motivate the derivative with a difference quotient or are you allowed to use the "known" rules which follows from the definition? If you know the rules you have
    s(t) = 4/t = 4*t^-1
    s'(t) = -1*4*t^-2 = -4/t^2

    If you have to use the limit of the difference quotient then we have the definition
    s'(t) = lim(h->0) (s(t+h)-s(t))/h = lim(h->0) ((4/(t+h))-4/t)/h = lim(h->0) ((4t-4(t+h))/(t(t+h)))/h = lim(h->0) ((-4h))/(t(t+h)))/h = lim(h->0) ((-4))/(t(t+h)))/1 = -4/t^2

    So we have the derivative
    s'(t) = -4/t^2

    Calculating the velocity at a = 2 were t = a
    s'(t) = -4/t^2 => s'(2) = -4/2^2 = -1 m/s

    I leave a = 4 as exercise for you.
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  3. #3
    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Thanks for the quick reply, yes I have to use the limit.....I am reading through your response now, and If I do not understand any portion of it, I will post with further questions. Thank you so much for the reply though, I really appreciate it!!
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    fkf
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Write if you don't understand, maybe I can write in LaTeX. But try to write it down with pen and paper, and you will get the idea.
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    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Okay, so as I understand it......

    A=4

    s'(t) = lim(h->0) (s(t+h)-s(t))/h

    = lim(h->0) ((4/(t+h))-4/t)/h
    = lim(h->0) ((4t-4(t+h))/(t(t+h)))/h
    = lim(h->0) ((-4h))/(t(t+h)))/h
    = lim(h->0) ((-4))/(t(t+h)))/1
    = -4/t^2
    s'(t) = -4/t^2
    => s'(4)
    = -4/4^2
    = -0.25 m/s


    By all means correct me If I am wrong.
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Quote Originally Posted by JDS View Post
    Thanks for the quick reply, yes I have to use the limit.....I am reading through your response now, and If I do not understand any portion of it.
    You don't because it is so damn hard to read.
    Why not use LaTeX coding? It is not helpful if it cannot be read.

    \frac{{\frac{4}{{t + h}} - \frac{4}{t}}}{h} = \frac{{4(t) - 4(t + h)}}{{h(t + h)t}} = \frac{{ - 4}}{{(t + h)t}}

    Now you finish the limit.
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  7. #7
    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    You are right, Yours is much cleaner and I can understand it more clearly. However, shouldnt the end function there be......

    -4/[2+(t+h)t]

    Perhaps I am looking at it wrong though?
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  8. #8
    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    nevermind, im an idiot, I see what I did, lOl
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    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    grrr, actually the more I look at this the more I confuse myself.
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    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Quote Originally Posted by Plato View Post
    You don't because it is so damn hard to read.
    Why not use LaTeX coding? It is not helpful if it cannot be read.

    \frac{{\frac{4}{{t + h}} - \frac{4}{t}}}{h} = \frac{{4(t) - 4(t + h)}}{{h(t + h)t}} = \frac{{ - 4}}{{(t + h)t}}

    Now you finish the limit.
    So I Did it with the way you have it set up here, and I still get the same answer I got above which is - 1/4 m/s or -0.25 m/s

    Hope I understood it correctly!
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  11. #11
    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    But looking at it for A=2, I am coming up with -1/2 or -0.5 m/s

    Am I correct, or am I missing something?
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    You said earlier
    s'(t) = lim(h->0) (s(t+h)-s(t))/h

    = lim(h->0) ((4/(t+h))-4/t)/h
    = lim(h->0) ((4t-4(t+h))/(t(t+h)))/h
    = lim(h->0) ((-4h))/(t(t+h)))/h
    = lim(h->0) ((-4))/(t(t+h)))/1
    = -4/t^2
    s'(t) = -4/t^2
    When t= 2, that is -4/2^2= -1, not -1/2.
    Last edited by HallsofIvy; November 13th 2012 at 10:41 AM.
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  13. #13
    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Yes you are correct but in the portion you quoted, that was based on the info I obtained from "fkf", When I reposted the answer of -1/2, I was going off of the formula version that Plato posted.....Was fkf and myself correct or is Plato correct?
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    Quote Originally Posted by JDS View Post
    Was fkf and myself correct or is Plato correct?
    \lim _{h \to 0} \left( {\frac{{ - 4}}{{(t + h)t}}} \right) = \frac{{ - 4}}{{t^2 }}

    So the derivative is \frac{-4}{t^2} which is -1 if t=2.
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  15. #15
    JDS
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    Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

    I see, So then in that case my second answer is still , -(1/4)........Correct?

    P.S. Thanks for all the help!!
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