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Math Help - particle motion

  1. #1
    Senior Member euclid2's Avatar
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    particle motion

    a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
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  2. #2
    Senior Member eumyang's Avatar
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    Again, if you know derivatives, then you can take the derivative of d(t) to get
    d(t) = 2t - 8
    ... and then plug in t = 4 to see if the particle is indeed at rest.

    Or, mimicking your previous post, find d(4) and d(4 + h), and plug into
    \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}
    and take the limit.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by euclid2 View Post
    a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
    The graph of position against time is a parabola, and the particle is stationary when the tangent to this parabola is horizontal, which is the vertex of the parabola. So you need to show that the point on the parabola corresponding to t=4 is the vertex.

    CB
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    Senior Member euclid2's Avatar
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    Quote Originally Posted by eumyang View Post
    Again, if you know derivatives, then you can take the derivative of d(t) to get
    d(t) = 2t - 8
    ... and then plug in t = 4 to see if the particle is indeed at rest.

    Or, mimicking your previous post, find d(4) and d(4 + h), and plug into
    \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}
    and take the limit.
    i got f(4)=7 and f(4+h)=h^2+7 so limh-->0 h^2+7-7/h limh-->0 h^2/h = 0 so the answer to this one is also 0?
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  5. #5
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    What function are you using? You said that f(t)= t^2-8t+15 so that f(4)= 16- 32+ 15= 1, not 7.
    And f(4+h) is NOT h^2+ 7. Where did you get that?

    Since f(t)= t^2-8t+15, f(4+h)= (4+ h)^2- 8(4+ h)+ 15
    f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1

    \frac{f(4+h)- f(4)}{h}= \frac{h^2+ 8h+ 1- 1}{h}= \frac{h^2+ 8h}{h}
    and, for h\ne 0, that is h+ 8. Take the limit of that as h goes to 0.
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    Grand Panjandrum
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    Quote Originally Posted by euclid2 View Post
    i got f(4)=7 and f(4+h)=h^2+7 so limh-->0 h^2+7-7/h limh-->0 h^2/h = 0 so the answer to this one is also 0?
    So is this a calculus question after all?

    Please tell, so I can move it to the appropriate forum.

    CB
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  7. #7
    Member grgrsanjay's Avatar
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    for a particle to be in rest its velocity should be zero for finding velocity we need to integrate velocity which comes out to be 2t-8
    v=2t-8
    v(4)=2(4)-8
    =8-8
    =0
    hence proved
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by grgrsanjay View Post
    for a particle to be in rest its velocity should be zero for finding velocity we need to integrate velocity which comes out to be 2t-8
    v=2t-8
    v(4)=2(4)-8
    =8-8
    =0
    hence proved
    "Need" is wrong here, "can" is appropriate, in fact we don't need calculus at all.

    Also velocity is the derivative of position

    CB
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  9. #9
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    Quote Originally Posted by grgrsanjay View Post
    for a particle to be in rest its velocity should be zero for finding velocity
    No, you need to differentiate the position function to get velocity.

    we need to integrate velocity which comes out to be 2t-8
    v=2t-8
    v(4)=2(4)-8
    =8-8
    =0
    hence proved
    And it was not clear whether the OP could use "rules" of differentiation like that. Since this was posted in "Precalculus" it is likely that he knows only the "limit of the difference quotient" definition of derivative- or that he could find where the slope of the tangent line is 0 by completing the square as Captain Black suggested.
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  10. #10
    Senior Member euclid2's Avatar
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    Technically, I am not sure if it should be moved to the calculus section. I haven't yet learned derivatives so i thought pre-calculus was appropriate. Although it is a "calculus course". Thank you everyone for your help!
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  11. #11
    Senior Member euclid2's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    What function are you using? You said that f(t)= t^2-8t+15 so that f(4)= 16- 32+ 15= 1, not 7.
    And f(4+h) is NOT h^2+ 7. Where did you get that?

    Since f(t)= t^2-8t+15, f(4+h)= (4+ h)^2- 8(4+ h)+ 15
    f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1

    \frac{f(4+h)- f(4)}{h}= \frac{h^2+ 8h+ 1- 1}{h}= \frac{h^2+ 8h}{h}
    and, for h\ne 0, that is h+ 8. Take the limit of that as h goes to 0.
    Maybe I am missing something but how does f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1? wouldn't the 8h's cancel and leave us with h^2-1
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  12. #12
    Senior Member eumyang's Avatar
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    \begin{aligned}<br />
f(4 + h) &= (4 + h)^2 - 8(4 + h) + 15 \\<br />
&= 16 + 8h + h^2 - 32 - 8h + 15 \\<br />
&= h^2 + 8h - 8h + 16 - 32 + 15 \\<br />
&= h^2 - 1<br />
\end{aligned}
    You're right, looks like a typo.

    Furthermore, f(4) = -1, not 1.
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by euclid2 View Post
    a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
    As mentioned elsewhere the the curve represented by the equation of displacement against time is a parabola and the particle is at rest when the tangent to the parabola is horizontal and this occurs at the vetex of the parabola.

    The value of t corresponding to the vertex is the mid point between the roots of the quadratic obtained by setting displacement to zero:

    t^2-8t+15=0

    Applying the quadratic formula to this quadratic give the roots as t=4 \pm 1

    QED
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  14. #14
    Senior Member euclid2's Avatar
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    Quote Originally Posted by eumyang View Post
    \begin{aligned}<br />
f(4 + h) &= (4 + h)^2 - 8(4 + h) + 15 \\<br />
&= 16 + 8h + h^2 - 32 - 8h + 15 \\<br />
&= h^2 + 8h - 8h + 16 - 32 + 15 \\<br />
&= h^2 - 1<br />
\end{aligned}
    You're right, looks like a typo.

    Furthermore, f(4) = -1, not 1.
    but then we get

    limh-->0 h(4+h)-h(4)/h

    limh-->0 h^2-1-1/h

    limh--> h-2

    limh--> (0)-2

    =-2

    which doesn't make sense?? because it's not at rest??
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  15. #15
    Senior Member eumyang's Avatar
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    Nope. Careful with your signs...
    \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}

    = \lim_{h \to 0} \dfrac{h^2 - 1 - (-1)}{h}

    = \lim_{h \to 0} \dfrac{h^2 - 1 + 1}{h}

    = \lim_{h \to 0} \dfrac{h^2}{h}

    = \lim_{h \to 0} h

    = 0
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