Find the length of the curve $\mathcal C$, $x(t)=2\cos^3(t)$, $y(t)=2\sin^3(t)$ for $0 ≤ t ≤ 2π$.

Question

Find the length of the curve $\mathcal C$, $x(t)=2\cos^3(t)$, $y(t)=2\sin^3(t)$ for $0 ≤ t ≤ 2π$.

I got a bit stuck trying to do this. I started off by finding $dx/dt$ and $dy/dt$ ($-6\cos^2(t)\sin(t)$ and $6\sin^2(t)\cos(t)$ respectively). Then, I found the modulus for this which I found to be $6\sin t\cos t$.

Then, I integrated by substituting $u=\sin t$ and got $\sin t$. However, I am unsure which limits to chose as if I use $2π$ and $0$, then the answer returned is $0$, which I am pretty sure is not the answer. Any help on how to choose the limits would be appreciated, or pointing out if I've made any mistakes.

Answer 1

Well, the formula is given by:

$$\mathcal{I}\left(\alpha,\beta\right):=\int_\alpha^\beta\sqrt{\left(\frac{\text{d}x}{\text{d}t}\right)^2+\left(\frac{\text{d}y}{\text{d}t}\right)^2}\space\text{d}t\tag1$$

So, in your case we get:

$$\mathcal{I}\left(0,2\pi\right)=\int_0^{2\pi}\sqrt{\left(-6\sin\left(t\right)\cos^2\left(t\right)\right)^2+\left(6\sin^2\left(t\right)\cos\left(t\right)\right)^2}\space\text{d}t=$$ $$\int_0^{2\pi}\sqrt{9\sin^2\left(2t\right)}\space\text{d}t=3\int_0^{2\pi}\left|\sin\left(2t\right)\right|\space\text{d}t=3\cdot4\int_0^\frac{\pi}{2}\sin\left(2t\right)\space\text{d}t=3\cdot4\cdot1=12\tag2$$

Answer 2

I want to add to Jan's solution what happens with the absolute value of the direction vector of the curve: $$\sqrt{\Big(-6 \sin{(t)}\cos^2{(t)}\Big)^2 + \Big(6\sin^2{(t)}\cos{(t)}\Big)^2} = 6 \sqrt{\Big(\sin^2{(t)} + \cos^2{(t)}\Big) \cdot \cos^2{(t)}\sin^2{(t)}}$$

Now use the trigonometric pythagorean theorem, that $\cos^2{(t)} + \sin^2{(t)} = 1$ and use the trignometric theorems for adding: $$\sin(a + b) = \mathfrak{Im}\big(\exp{\big(i(a + b)\big)}\big) = \mathfrak{Im}\big(\exp{\big(ia\big)}\exp{\big(ib\big)}\big) = \mathfrak{Im}\Big((\cos{a} + i\sin{a})(\cos{b} + i\sin{b})\Big) = \cos{a}\sin{b} + \cos{b}\sin{a}$$ which in reverse means that: $$\cos{a}\sin{b} = \frac12\big(\cos{a}\sin{b} + \cos{a}\sin{b} + \cos{b}\sin{a} - \cos{b}\sin{a}\big) = \frac12\big(\cos{(a)}\sin{(b)} + \cos{(-a)}\sin{(b)} + \cos{(b)}\sin{(a)} + \cos{(b)}\sin{(-a)}\big) = \frac12\big(\sin{(a+b)} + \sin{(b-a)}\big)$$

Thus the square root from above which we got as absolute value of the curves direction vector results in $$... = 6\ \Big|{\cos{(t)}\sin{(t)}}\Big| = 6\ \Big|\frac12\Big(\sin{(t+t)} + \sin{(t-t)}\Big)\Big| = 3 |\sin{(2t)}|$$

Always pay attention when integrating over an absolute value! Identify the zero roots of the absolute value and split the integral for each differently signed part. When integrating $|\sin{(2t)}|$ from $0$ to $2\pi$, the parts with alternating sign go from $0$ to $\frac{\pi}{2}$, to $\pi$, to $\frac{3\pi}{2}$ and finally to $2\pi$ (4 parts). Since each of the 4 parts are equal, you only need to calculate $$3\cdot 4 \int^{\frac{\pi}{2}}_{0} \sin{(2t)} dt$$

Most important are three things:

• never forget taking the absolute value when applying the square root to a squared expression
• use $\sin$/$\cos$ laws to simplify products of trignometric functions to sums
• use trignometric pythagorean theorem for sum of $\cos$ and $\sin$ squares

If you keep all these things in mind you will succeed well with integration over trigonometric courses. If you want to check the answer you can also enter the integral in Wolfram Alpha:

integrate 6*sqrt(cos^2(x)*sin^4(x) + sin^2(x)*cos^4(x)) from 0 to 2pi

Wolfram Alpha query

Answer 3

I prefer to solve problems like this in the complex plane. Let me show you why. First of all, the arc length in the complex plane is given by

$$s=\int |\dot z|dt$$

Now, we have

$$ \begin{align} z&=x+iy\\ &=2\cos^3t+i\sin^3t\\ &=\frac{1}{4}\big(\cos3t+3\cos t+i(-\sin3t+3\sin t) \big)\\ &=\frac{1}{2}(\cos3t-i\sin3t)+\frac{3}{2}(\cos t+i\sin t)\\ &=\frac{1}{2}(3e^{it}-e^{-i3t}) \end{align} $$

Then

$$ \begin{align} \dot z&=\frac{3i}{2}(e^{it}+3e^{it})\\ &=-6e^{it}\sin t \cos t\quad\quad (\text{thank you, WoflramAlpha!})\\ &=-3e^{it}\sin2t \end{align} $$

Finally,

$$ \begin{align} s&=\int |\dot z|dt\\ &=3\int_0^{2\pi}|\sin2t|dt\\ &=3\cdot 4\int_0^{\pi/2}\sin2t\ dt\\ &=12 \end{align} $$