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Another Related Rates

  1. Aug 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A light shines from the top of a pole 50 ft high. A ball is dropped from the same height at a point 30 ft away. (see thumbnail). How fast is the shadow of the ball moving along the ground .5 seconds later assuming ball falls a distance of [tex] s=16t^2[/tex]

    2. Relevant equations
    Chain Rule/ Implicit Differentiation

    3. The attempt at a solution
    I have labeled the height of the ball after 1/2 second as [tex]h=50-16t^2[/tex] which=46.
    I have assumed (possibly incorrectly) that the base of the triangle larger is [tex]x[/tex] and the smaller base is [tex]x-30[/tex].
    I calculate [tex]\frac{dh}{dt}=-32t\left|_t_=_.5=-16[/tex]
    I know that[tex] \frac{dx}{dt}=\frac{dh}{dt}*\frac{dx}{dh}[/tex] ...but I am having trouble writing x as a function of h.

    I think it has to do with similar triangles. But I can't seem to make it work.

  2. jcsd
  3. Aug 2, 2007 #2


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    If I'm reading that very small picture correctly,
    one of the triangles has sides of height 50 and length x,
    and the other has sides of height h and length x-30.
    The ratios of these lengths are the same since the triangles are
  4. Aug 2, 2007 #3
    That is what I had been trying, but am failing miserably. I do want to write [tex]x(h)[/tex] right?

    BTW~Click on the picture to enlarge it!

    so [tex]\frac{X}{50}=\frac{X-30}{h}[/tex]

    I don't know why I am having so much trouble solving this for X? I do need to solve this for X right?
    Last edited: Aug 2, 2007
  5. Aug 2, 2007 #4


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    Yes, you do need to solve for X. It's just a little tiny algebra problem:

    Multiply both sides by 50,

    cancel the 50 in the denominator on the LHS,

    multiply both sides by h,

    cancel the h in the denominator on the RHS,

    isolate all the terms involving X onto one side of the equation,

    isolate all the terms not involving X onto the other side of the equation,

    write the side of the equation involving X as a product of X times one other factor,

    divide both sides by that factor.
  6. Aug 2, 2007 #5
    Thanks for the algebra lesson, I had never encoutered that before:wink:

    \Rightarrow h=\frac{50x-1500}{x}\leftarrow[/tex] for some reason I go retarded
    here....why? why can't I seem to isolate x?
  7. Aug 2, 2007 #6


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    If you leave it like

    xh = 50x - 1500

    does it look any easier? Hint: group like terms.
  8. Aug 2, 2007 #7
    How about this:
    [tex] \frac{x}{50}= \frac{x}{h}-\frac{30}{h}[/tex]
    Can you rearrange for x from there?
  9. Aug 2, 2007 #8
    I don't know....I got [tex] xh=50x-1500\Rightarrow xh-50x=-1500\Rightarrow x(h-50)=-1500\Rightarrow x=\frac{-1500}{h-50}[/tex]

    Thanks:redface:but now I am having trouble differentiating for [tex]\frac{dx}{dh}[/tex] I am getting [tex]\frac{1500}{h-50}^2}[/tex] is that correct? Thatis supposed to be just the denominator squared...I cannot fig how to keep it down there
  10. Aug 2, 2007 #9


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    yes, just the denominator.

    you have:
    the (-1500) just sits out front. why would it get squared?
  11. Aug 3, 2007 #10
    So dh/dt at .5 seconds is -16 and dx/dh at h=46 is -1500/16[tex]^2[/tex] so their product is dx/dt correct? so dx/dt at .5 seconds is 1500/16 correct?
    Last edited: Aug 3, 2007
  12. Aug 3, 2007 #11


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    dx/dh at h=46 is +1500/(16^2).
  13. Aug 3, 2007 #12
    so I need to change it to -1500/16 yes?
    The reason I am having so much trouble is that the Text gives an answer of dx/dt=-1500.
  14. Aug 3, 2007 #13


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    yes -1500/16 ft per second
  15. Aug 3, 2007 #14
    Well then. Thank you for all of your time and patience. Most of my trouble, has, however, stemmed from the classic "incorrect textbook solution." The book insists that the answer is -1500 ft/sec. It must be a typo.

    Thanks again for the help,
  16. Aug 4, 2007 #15
    Not so fast, check dx/dh at h=46 again. You squared twice, leaving an extra 16
  17. Aug 4, 2007 #16


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    oopsy daisy, turdferguson is right
    sorry about not catching that
  18. Aug 4, 2007 #17
    Ah ha! Thanks again guys.

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