calculus

At age 15, Nan is twice as tall as her 5-year-old brother Dan, but on Dan's 21st birthday, they find that he is 6 inches taller. Explain why there must have been a time when they were exactly the same height

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  1. So you are studying the mean value theorem....

    If growth is continual, even at different rates, if Dan starts out shorter than her, and ends up taller, at some time he must have been the same height...how else can he pass her?

  2. I believe we have to express it as a function using f(x) and g(x)....not sure.

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  3. let N(t) be nan's height
    and D(t) be dan's height.

    N(a)>D(a) and
    D(b)>N(b) given

    Let N'be Nan's growth rate and D' be Dans growth rate.

    N(t)=N(a)+N'*(t-a)so
    N(b)=N(a)+N'*(b-a). Simillary,
    D(b)=D(a)+D'*(b-a)

    N(a)-N(b)>D(a)-D(b)
    N(a)-N(a)-N'(b-a)>D(a)-D(a)-D'(b-a
    or N'<D' which says that Dan's growth rate is greater than Nans.

    Is there a time which N(t)=D(t)?

    iF so, then
    D(a)+D'*(t-a)=N(a)+N'(t-a)

    (t-a)= (N(a)-D(a))/((D'-N')
    and t= ( ) + a

    Since this time computes for all N', D', such that D'>N', this time must exist.

  4. take time 0 when Dan is 5, then t = 16 when Dan is 21
    D(t) = Dan height (always positive)
    N(t) = Nan height (always positive)

    N(0) = 2D(0)
    N(16) = D(16)-6
    These two curves cross between t = 0 and t = 16. The heights are the same at the intersection.

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  5. Of course there may have been more than one time when they were equal height.

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