Asked by Carly
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
Answers
Answered by
bobpursley
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?
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?
Answered by
Carly
I believe we have to express it as a function using f(x) and g(x)....not sure.
Answered by
bobpursley
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.
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.
Answered by
Damon
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.
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.
Answered by
Damon
Of course there may have been more than one time when they were equal height.
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