Find the length of the loop of the given curve: $x=3t-t^3$ $y=3t^2$

Question

I used the arc length formula (where you take the integral of square root of x' squared + y' squared $\int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$) to get $t^3 + 3t + C$ which seems to be the wrong answer. Not sure what I did wrong. Please tell me the correct answer and why my solution was incorrect. Thanks

Answer 1

If the path described makes a loop, then there will be 2 values $t_1$ and $t_2$ such that $ \begin{bmatrix} x(t_1) \\ y(t_1) \end{bmatrix} = \begin{bmatrix} x(t_2) \\ y(t_2) \end{bmatrix} $ , so:

$$\begin{bmatrix} 3t_1 - t_1{}^3 \\ 3t_1{}^2 \end{bmatrix} = \begin{bmatrix} 3t_2 - t_2{}^3 \\ 3t_2{}^2 \end{bmatrix}$$

From $3t_1{}^2 = 3t_2{}^2$, we get $t_1 = -t_2$, and from $3t_1 - t_1{}^3 = -3t_1 + t_1{}^3$, we get that

$$t_1 = -\sqrt{3} \text{ and }t_2 = \sqrt{3}$$

Now the formula for parametrized path length is similar to the pythagorean formula, explicitly:

$$L = \int_{t_1}^{t_2} { \sqrt{ \left(\frac{{\rm d}y}{{\rm d}t} \right)^2 + \left(\frac{{\rm d}x}{{\rm d}t} \right)^2 } {\rm d}t }$$

And filling in:

$$\begin{align} % L &= \int_{-\sqrt{3}}^{\sqrt{3}} { \sqrt{ \left(6t \right)^2 + \left(3 - 3t^2 \right)^2 } {\rm d}t } % \\ &= \int_{-\sqrt{3}}^{\sqrt{3}} { \sqrt{ 9t^4 + 18t^2 + 9 } {\rm d}t } % \\ &= \int_{-\sqrt{3}}^{\sqrt{3}} { \left \vert 3t^2 + 3 \right \vert {\rm d}t } % \\ &= t^3 + 3t ~\bigg\vert_{t = -\sqrt{3}}^{t = \sqrt{3}} % \\ &= 12 \sqrt{3} % \end{align}$$

Answer 2

Too long for comment (it's not an answer) $$ x'=3(1-t^2), \enspace y'=3\cdot 2t\\ x'^2 + y'^2 = 3^2 [(1-t^2)^2 + (2t)^2] = 3^2 (1 - 2t^2 + t^4 + 4t^2) = 3^2 (1 + t^2)^2\\ \sqrt{x'^2 + y'^2} = 3(1+t^2)\\ \int \sqrt{x'^2 + y'^2}\,dt = t^3 + 3t + C $$ Why it "seems to be the wrong answer"?

Answer 3

There are three places where $x(t)=0$, namely $t\in\{0,\pm\sqrt3\}$.

Of these, only $\pm\sqrt3$ yields equal values for $y(t)$, so the bounds of the loop must be $[-\sqrt3,\sqrt3]$.

Integrating the differential arc length, you get $$t^3+3t|_{-\sqrt3}^\sqrt3=6\sqrt3-(-6\sqrt3)=12\sqrt3$$

Answer 4

First one has to determine $t_1$ and $t_2$, with $t_1<t_2$ such that $$ (x(t_1),y(t_1))=(x(t_2),y(t_2)). $$ We have: $$ (x(t),y(t))=(x(s),y(s))\iff \left\{ \begin{array}{lcl} 3t-t^3-3s+s^3&=&0\\ 3s^2-3t^2&=&0 \end{array}\right.. $$ Since we are looking for $t_1$ and $t_2$, with $t_1<t_2$, we deduce from the second equation of the system above that $s=-t$. The first equation of our system then becomes: $$ 3t-t^3=t(3-t^2)=0, $$ i.e. $$ t=0,\pm\sqrt{3}. $$ Hence $$ t_1=-\sqrt3=-t_2,\quad t_2=\sqrt3. $$ The length of the loop is then: \begin{eqnarray} \int_{t_1}^{t_2}\sqrt{(x'(t))^2+(y'(t))^2}\,dt&=&\int_{t_1}^{t_2}\sqrt{(3-3t^2)^2+(6t)^2}\,dt=\int_{-t_2}^{t_2}\sqrt{3^2[(1-t^2)^2+4t^2]}\,dt\\ &=&2\cdot3\int_0^{t_2}\sqrt{1-2t^2+t^4+4t^2}\,dt=6\int_0^{t_2}\sqrt{1+2t^2+t^4}\,dt\\ &=&6\int_0^{t_2}\sqrt{(1+t^2)^2}\,dt=6\int_0^{t_2}(1+t^2)\,dt\\ &=&6\left[t+\frac13t^3\right]_0^{t_2}=6t_2+2t_2^3=6\sqrt3+2\cdot3\sqrt3=12\sqrt3. \end{eqnarray}