Length of a Parabolic Curve

I just wanted to know how I can find the length of a curve given by $f(x) = x^2$ from $x=0$ to $x=1$.

For appproximation, the length is a bit larger than the hypotenuse of isosceles right triangle with the shorter side being 1 long. It's definitely less than the sum of two shorter sides. Thus, if we represent the length by $L$, the following relationship is expected: $\sqrt 2 < L < 2$

I now regard $L$ as the accumulation of hypotenuses of infinitestimally small right triangles around $f(x)$. Since $f'(x)=2x$, the general right triangle is something like this: If $x$ goes very slightly down the $x$-axis ($\Delta x$), the the $y$ value goes upwards for $2x\Delta x$.

Thus the hypontenuse is the square root of the following: $(\Delta x)^2+(2x\Delta x)^2$. The hypotenuse is: $(\Delta x) {( 4x^2 + 1)^{1/2}}$

Since $L$ has been defined as the accumulation of these hypotenuses, it is: $L = \int_0^1 ( 4x^2 + 1)^{1/2} dx$.

I am stuck just here. Could someone tell me if my chain of thoughts so far is right and how I can go from here? I don't know how to calculate the integral of a function that contains another function in it.

Thanks!!

Solution 1:

Your train of thought is exactly right; you've single-handedly rederived the formula for the length of a curve given by $y=f(x)$ :-) This can be written as

$$ L=\int_a^b\sqrt{1+f'(x)^2}\,\mathrm dx $$

in general. In your case, as you rightly determined, $f'(x)=2x$, and we want the length from $a=0$ to $b=1$, so we have

$$ L=\int_0^1\sqrt{1+4x^2}\,\mathrm dx\;. $$

To solve this integral, you can use the substitution $\sqrt{1+4x^2}=\cosh u$, so $x=\frac12\sinh u$ and $\mathrm dx=\frac12\cosh u\,\mathrm du$, to get

$$ L=\frac12\int_0^{\operatorname{arcosh}\sqrt5}\cosh^2 u\,\mathrm du=\frac14\left[x+\sinh x\cosh x\right]_0^{\operatorname{arcosh}\sqrt5}=\frac14\left(\operatorname{arcosh}\sqrt5+2\sqrt5\right)\approx1.479$$

If you don't know how to solve such an integral, you can always ask Wolfram|Alpha; it will usually know the answer and can often tell you the steps to get there if you click on "step-by-step solution"; though the solution will sometimes, as in this case, not be the most simple one.

You might also be interested in the question Intuition behind arc length formula.