Expert: Scotto - 3/29/2009

Question
Compute the arc length function for the curve y=f(x)=e^x relative to it's y intercept.

Answer
To compute the arc length, you need to integrat from the y inercept to the point on the graph √(1 + [f'(x)]²).

Now since the function f(x) = e^x, f'(x)=e^x.
This means that [f'(x)]² = 2^(2x).

The y intercept is when x is 0, so the y intercept is a x=0, y=1.

What we want to find, then, is
⌠x
⌡0 (1 + e^(2t))^0.5 dt.

Multiply by e^x/e^x, giving
⌠x
⌡0 (1 + e^(2t))^0.5*e^t/e^t dx

Let u=e^t, then e^(tx) = u² and du=e^t dt.  Our integral becomes
⌠ln(x)
⌡1   (1+u²)/u du.

This smells like a trig substituion to me.

Take a right triangle.  
Let the far side be u, the near side be 1,
and the hypoteneuse be √(1+u²) with an angle Θ.

What we have is tan(Θ) = u, so sec²(Θ)dΘ = du.
√(1+u²) is also sec(Θ), so (1+u²) is sec²(Θ).

This transforms the integral into
⌠ln(x)
⌡π/2 sec^4(Θ) dΘ.

Now we know that 1 + tan²(Θ) = sec²(Θ), so make that substituion into what we're integrating.  The result is
⌠ln(x)
⌡π/2  sec²(Θ)(1+tan²(Θ))dΘ.

This becomes, by multiplying it out,
⌠ln(x)
⌡π/2  sec²(Θ) + sec²(Θ)tan²(Θ)) dΘ.

Now the integral of sec²(Θ) is tan(Θ),
but we still have to integrate sec²(Θ)tan²(Θ).

First, we'll use u-v substitution.
Let dv=sec²(Θ)dΘ and u=tan²(Θ).
This makes v=tan(Θ) and du=2tan(Θ)tan(Θ)sec(Θ)=2tan²(Θ)sec(Θ).

Now the integral that arise is v du, which is 2tan^3(Θ)sec(Θ),
so here we let w=tan(Θ), so dw=tan(Θ)sec(Θ)dΘ,
and the integral becomes
⌠ln(x)
⌡1  2w² dx = 2w²/3 from(ln(x) downto 1) = 2ln²(x)/3 - 2/3.

Lets remember  that we dropped a tan() just a little while ago,
and when that was evaluated we got tan(ln(x))-tan(π/2).

Adding that to the current result gives
tan(ln(x)) - tan(π/2) + 2ln²(x)/3 - 2/3.
The tan(π/2) is 1, so this is
tan(ln(x)) - 1 + 2ln²(x)/3 - 2/3.