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# arclength watch

1. length of a curve y=f(x) with f(1)=0 from (1,0) to (x,f(x)) x>1 is
(x^2-1)/2.
find f(x) defined on[1, ∞ ) assuming that it is an increasing function.

I think i have done this, but am confused as to where to use the limits. should i solve

∫√(1+(f'(x))2 dx=(x2 -1)/2 (taking x=1,x as the bounds of integration)

by differentiating both sides etc?
2. (Original post by realicetic)
length of a curve y=f(x) with f(1)=0 from (1,0) to (x,f(x)) x>1 is
(x^2-1)/2.
find f(x) defined on[1, ∞ ) assuming that it is an increasing function.

I think i have done this, but am confused as to where to use the limits. should i solve

∫√(1+(f'(x))2 dx=(x2 -1)/2 (taking x=1,x as the bounds of integration)

by differentiating both sides etc?
then i end up having

f(x)=∫√(x2 +2x) dx

i dont know how i can solve this...
3. sqrt(1 + f '(t)^2) dt = (1/2)(x^2 - 1)

Differentiating wrt x,

sqrt(1 + f '(x)^2) = x
1 + f '(x)^2 = x^2
f '(x) = sqrt(x^2 - 1)
f(x) = (1/2)x sqrt(x^2 - 1) - (1/2)arccosh(x) + c . . . . . see the other thread
f(x) = (1/2)x sqrt(x^2 - 1) - (1/2)arccosh(x) . . . . . since f(1) = 0
4. (Original post by Jonny W)
sqrt(1 + f '(t)^2) dt = (1/2)(x^2 - 1)

Differentiating wrt x,

sqrt(1 + f '(x)^2) = x
1 + f '(x)^2 = x^2
f '(x) = sqrt(x^2 - 1)
f(x) = (1/2)x sqrt(x^2 - 1) - (1/2)arccosh(x) + c . . . . . see the other thread
f(x) = (1/2)x sqrt(x^2 - 1) - (1/2)arccosh(x) . . . . . since f(1) = 0
this pleases me, it is what i got cheers xxx

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Updated: January 29, 2006
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