Answer to Question #123846 in Calculus for Tau

Question

Find the length of the arc of the parabola y2 = x from (0, 0) to (1, 1)

Answer 1

We have an explicitly defined function $f(y) = y^2$.

So the length of this curve on the segment $y \in [0, 1]$ will be calculated using the formula:

$L = \intop_0^1\sqrt{1 + {f'(y)}^2}dy$ , then $f'(y) = 2y$ $\implies L = \intop_0^1\sqrt{1 + 4y^2}dy$.

$\intop_0^1\sqrt{1 + 4y^2}dy =$

$= \begin{bmatrix} u=\sqrt{1 + 4y^2} & du=\frac{4y}{\sqrt{1 + 4y^2} }\\ dv = dy & d = y \end{bmatrix} = y\sqrt{1+4y^2}\mid_0^1-$

$- \int_0^1\frac{4y^2}{\sqrt{1+4y^2}}dy$ .

$\int_0^1\frac{4y^2}{\sqrt{1+4y^2}}dy = \int_0^1\frac{1+4y^2-1}{\sqrt{1+4y^2}}dy =\int_0^1\sqrt{1+4y^2}dy -$

$-\int_0^1\frac{dy}{\sqrt{1+4y^2}}$ .

So, $L= y\sqrt{1+4y^2}\mid_0^1-L +\int_0^1\frac{dy}{\sqrt{1+4y^2}}$.

$2L = y\sqrt{1+4y^2}\mid_0^1+\int_0^1\frac{dy}{\sqrt{1+4y^2}}$.

$L = \frac{\sqrt{5}}{2} + \frac{1}{2}\times\frac{1}{2}\int_0^1\frac{d(2y)}{\sqrt{1+(2y)^2}}$.

$L = \frac{\sqrt{5}}{2} + \frac{1}{4}ln\mid2y + \sqrt{1+4y^2}\mid\vert_0^1$ .

$L = \frac{\sqrt{5}}{2} + \frac{1}{4}ln\mid2+\sqrt{5}\mid - \frac{1}{4}ln\mid1\mid$.

But $ln\mid1\mid = 0$, then

$L = \frac{\sqrt{5}}{2} + \frac{1}{4}ln\mid2+\sqrt{5}\mid \approx1.4789$.

Answer: 1.4789