Find a function F(x) whose arc length L(x) from (1,1/2) to (x,F(x)), x>1 is (1/2)x^2 + (1/4)Ln(x).(adsbygoogle = window.adsbygoogle || []).push({});

First some short hand notation.

Int[f(x),dx], means the indefinite integral of the function f(x).

Int[f(x),dx,a,b], means the definite integral where a is the lower bound and b is the upper bound.

d/(dx)[f(x)], is the derivate of f(x).

Sqrt[f(x)], is the square root of f(x).

Basically I am trying to follow Mathematica conventions and adding a touch of Leibniz to it.

Ok, now here is what I have done and where I am at thus far.

Arc length formula is Int[Sqrt[1+(d/dt)[f(t)]^2],dt,a,x]

So I have Int[Sqrt[1+((d/dt)[f(t)])^2],dt,1,x] = (1/2)x^2 + (1/4)Ln(x)

So by the FTC I can say that

d/(dx)[(1/2)x^2 + (1/4)Ln(x)] = Sqrt[1+(d/dt)[f(t)]^2]

Which is

x + 1/(4x) = Sqrt[1+(d/dt)[f(t)]^2]

Squaring both sides I have

(x + 1/(4x))^2 = abs[1+(d/dt)[f(t)]^2]

I drop the absolute value here so I have.

(x + 1/(4x))^2 = 1+(d/dt)[f(t)]^2

x^2 + 1/2 + (1/(4x))^2 = 1+(d/dt)[f(t)]^2

Subtracting 1 from both sides and factoring the RHS

(x - 1/(4x))^2 = ((d/dt)[f(t)])^2

Taking square root of both sides and discarding the absolute value I have.

(d/dt)[f(t)] = x - 1/(4x)

Integrating both sides I get.

f(x) =(1/2)x^2 - (1/4)Ln(x) +c

I use the point (1,(1/2)) that is given to find c. Thus c = 0.

Seems ok, right?

Well that means that if I put f(x) back into my arc length formula I should get back to the given function.

Ok.... Did that and now I have an extra -(1/2) on my function? What am I missing here? One thing to notice is

Int[Sqrt[1+((d/dt)[f(t)])^2],dt,1,x] = (1/2)x^2 + (1/4)Ln(x)

When x=1 you have 0 = 1/2

Of course the restriction was x>1

Any help would be great, thanks.

cheers

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# Arc Length formula

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