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Tough derivative problem

  1. Feb 19, 2006 #1

    k = 3/2*w
    y=1/w^2 <- w is the x component of y = 1/x^2

    w is increasing at a constant rate of 7 units per second. When w = 5, what is the rate of change of k with respect to time?
    dw/dt = 7

    I'm not sure but can I do this? It seems like this is wrong or that I'm skipping a step.
    dk/dt = 3/2 * dw/dt
    dk/dt = 10.5 ??
    Last edited: Feb 19, 2006
  2. jcsd
  3. Feb 19, 2006 #2


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    Dearly Missed

    Fix up your expressions.
    w=1/w^2 means that the only real number w can be is 1.
  4. Feb 19, 2006 #3


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    Your missing k now: Post the problem as given if need be, or the source if it is a commom text.
  5. Feb 19, 2006 #4


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    "w is the x component of y= 1/x^2" doesn't make much sense. I think you just mean y= 1/w^2. In fact, I don't see reason to mention y or x. I also assume you mean k= (3/2)w rather than k= 3/(2w)

    dk/dt= (dw/dk)(dk/dt)

    dw/dk= (3/2) and you are told that dw/dt= 7 so dk/dt= (3/2)(7)= 10.5 just as you say.

    But are you sure you've copied the problem correctly? There is no need to know that w= 5 if dk/dw is a constant.
  6. Feb 19, 2006 #5
    Thanks for all your help... I'm pretty sure that it must be 10.5 now, I don't see how any other relationship with time could be formed, although the w = 5 is a bit suspicious.

    Here's the full picture, sorry I should have posted it first but I thought I could condense it.
  7. Feb 19, 2006 #6


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    it seems that [tex]\ell[/tex] is the tangent line to the curve y=1/w^2 at the point w, and k is the x-intercept thereof. Hence

    [tex]\ell : y-\frac{1}{w^2}=-\frac{2}{w^3}\frac{dw}{dt}(x-w)[/tex]

    is the equation of the tangent line.
  8. Feb 19, 2006 #7


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    the x-intercept of the above line is k.

    Since w=5 and [tex]\frac{dw}{dt}=7\frac{units}{sec}[/tex] we have...

    Can you get it from here?
  9. Feb 19, 2006 #8
    Hmm, I see where you got the equation for the line but now I'm a bit confused...

    Would I use the equation of the line to find k and then take the derivative of k?( i get 7.5 that way as opposed to 10.5)
    Last edited: Feb 19, 2006
  10. Feb 20, 2006 #9


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    If you are a little bit confused, then you should check it up in your textbook, there should be a part that says:
    The equation of the tangent line of the function f(x) at the point (x0, f(x0)) is:
    y - f(x0) = f'(x0) (x - x0)
    Yes, you need to express k in terms of w, then use the chain rule:
    [tex]\frac{dk}{dt} = \frac{dk}{dw} \times \frac{dw}{dt}[/tex]
    You were told that:
    [tex]\frac{dw}{dt} = 5[/tex], so what you have to do now is to find k in terms of w, then differentiate k with respect to w to get:
    Do you know how to find k in terms of w?
    Hint: k is the x-intercept of the tangent line at the point P(w, 1 / w2). And the tangent line is:
    [tex]y - \frac{1}{w ^ 2} = \frac{-2}{w ^ 3} (x - w)[/tex]
    [tex]k'_t(5) = k'_w(5) \times w'_t(5) = 7 k'_w(5) = ?[/tex]
    Can you go from here? :)
    I don't really understand this.
    [tex]\ell : y-\frac{1}{w^2}=-\frac{2}{w^3}\frac{dw}{dt}(x-w)[/tex]
    Why is there dw / dt in the tangent line equation???
    Am I missing something?
    Last edited: Feb 20, 2006
  11. Feb 21, 2006 #10


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    No, you're not missing anything, I just goofed. It should be:

    [tex]\ell : y-\frac{1}{w^2}=-\frac{2}{w^3}(x-w)[/tex]

    as you pointed out.
  12. Feb 21, 2006 #11


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    related rates

    To finish setting-up the related rates problem:

    note that since k is the x-intercept of [tex]\ell[/tex], one may put y=0 and x=k into the above equation of the tangent line (a.k.a. [tex]\ell[/tex],) to get the following implicit relationship between w and k:

    [tex] 0-\frac{1}{w^2}=-\frac{2}{w^3}(k-w)[/tex]

    multiply by [itex]-w^3[/itex] to get

    [tex] w=2(k-w)[/tex]

    which simplifies to... :rolleyes: eh, you'll get it from here.
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