Arc length Integral from 2 point

Question

For $x = \frac{1}{3}y^{3} + \frac{1}{4}y$, the arc length from $(\frac{7}{12},1)$ to $(a,b)$ is $\frac{53}{6}$. What is the value of $a + b$?

I've tried to solve this but I can't find the solution to because I didn't really understand what it means

Answer 1

You know that $$a=\frac 13 b^3+\frac 1 4b$$ The formula for this specific arclength is

$$L=\int_1^b \sqrt{1+\left(y^2+\frac{1}{4}\right)^2}\,dy$$ which is really messy (it involves elliptic functions of the first and second kinds).

Let us assume (this looks reasonable taking into account the target value) that $b$ is not too large. So, using series $$\sqrt{1+\left(y^2+\frac{1}{4}\right)^2}=\frac{\sqrt{41}}{4}+\frac{10 (y-1)}{\sqrt{41}}+\frac{333 (y-1)^2}{41 \sqrt{41}}+O\left((y-1)^3\right)$$ and integrating termwise $$L\sim -\frac{1305}{164 \sqrt{41}}+\frac{1373 b}{164 \sqrt{41}}-\frac{128 b^2}{41 \sqrt{41}}+\frac{111 b^3}{41 \sqrt{41}}+O\left(b^4\right)$$

Now, use $L=\frac{53}6$. Solving the cubic gives, as an estimate, $b=2.90794$.

Repeat the process around $y=3$ $$\sqrt{1+\left(y^2+\frac{1}{4}\right)^2}=\frac{\sqrt{1385}}{4}+\frac{222 (y-3)}{\sqrt{1385}}+\frac{52397 (y-3)^2}{1385 \sqrt{1385}}+O\left((y-3)^3\right)$$

$$L\sim -\frac{512747}{16620 \sqrt{1385}}+\frac{114877 b}{5540 \sqrt{1385}}-\frac{3456 b^2}{1385 \sqrt{1385}}+\frac{52397 b^3}{4155 \sqrt{1385}}+O\left(b^4\right)$$ Solving this new cubic gives $b=2.93919$.

At this point, I stop ! The exact value of $b$ was given by @DavidP.