Length of the curve $x(t)=4\sin(t/4), y(t)=1-2\cos(t/4)^2$

Question

I struggle for a while with the following exercise:

What is the length of the curve determined by $x(t) = 4\sin(t/4)$ and $y(t)=1-2\cos(t/4)^2$?

I played around a little bit with the hyperbolic substitution, but no luck with arriving at the, hopefully, correct answer: $\DeclareMathOperator{\arcsinh}{arcsinh} 4(\sqrt{2}+\arcsinh(1))$.

I'd appreciate if someone could show me the right way to do this :)

Answer 1

This curve is a parabola, which we can prove by eliminating the parameter.

$$y = 1-2\cos^2\left(\frac{t}{4}\right) = 2\sin^2\left(\frac{t}{4}\right) - 1 = \frac{1}{8}x^2-1$$ $x = 4\sin\left(\frac{t}{4}\right)\implies -4\leq x \leq 4$, so we can apply the arclength formula: $$L = \int_{-4}^4\sqrt{1+\frac{x^2}{16}}dx = 2\int_{0}^4\sqrt{1+\frac{x^2}{16}}dx$$ Next, we can use the substitution $x = 4\sinh t$: $$L = \int_0^{\sinh^{-1}(1)} 8\cosh t\sqrt{1+\sinh^2 t} dt = \int_0^{\sinh^{-1}(1)} 8\cosh^2 t dt= \int_0^{\sinh^{-1}(1)} 4+4\cosh 2t dt$$ $$= 4t+4\sinh t \cosh t \Bigr |_0^{\sinh^{-1}(1)} = 4t+4\sinh t \sqrt{1+\sinh^2 t} \Bigr |_0^{\sinh^{-1}(1)} = 4\sinh^{-1}(1) + 4\sqrt{2}$$

Answer 2

By eliminating $t$, you get the curve in $xy$-coordinates

$$y=\frac{1}{8}x^2-1$$

and its derivative,

$$\frac{dy}{dx} = \frac{1}{4}x$$Then, use the standard integral for length,

$$L=\int_{-4}^4 \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

to get

$$L=\int_{-4}^4 \sqrt{1+\frac{1}{16}x^2}dx= 8\int_0^1\sqrt{1+t^2}dt=4\left[\sqrt{2}+\ln(1+\sqrt{2})\right]$$

Note that $\ln(1+\sqrt{2}) = \sinh^{-1}(1)$.