Say I have a curve defined as: $$ x = a cos^3t\\ y = a sin^3t\\ 0 ≤ t ≤ 2\pi$$
And I wish to find its length. $$\int_{A}^{B}\{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2\}^{1/2}$$
Finding $dx/dt$ by way of the chain rule: $$ x = acos^3t \\ u = cost \\ x = au^3 \\ \frac{dx}{du} = 3au^2 \\ \frac{du}{dt} = -sint \frac{dx}{dt} = \frac{dx}{du} * \frac{du}{dt} = -3acos^2tsint $$
Similarly (work omitted for terseness) $\frac{dy}{dt} = 3asin^2tcost$
Plugging into the length formula:
$$\int_{0}^{2\pi}\{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2\}^{1/2} = \\ \int_{0}^{2\pi}\{(-3acos^2tsint)^2 + (3asin^2tcost)^2\}^{1/2} = \\ \int_{0}^{2\pi}\{9a^2cos^4tsin^2t + 9a^2sin^4tcos^2t\}^{1/2} = \\ \int_{0}^{2\pi}\{9a^2cos^2tsin^2t(cos^2t + sin^2t)\}^{1/2}$$
Since $cos^2t + sin^2t = 1$ that simplifies to
$$\int_{0}^{2\pi}\sqrt[2]{9a^2cos^2tsin^2t} = \\ \int_{0}^{2\pi}{3a\cos{t}\sin{t}}$$
Since $costsint = \frac{1}{2}sin2t$ the above simplfies further to: $$\frac{3a}{2}\int_{0}^{2\pi}{sin2t}$$
which integrates as $$ \frac{3a}{2} * \left[-cos2t\right]_0^{2\pi} $$ Since $\cos{4\pi} = 1$ and $\cos{0} = 1$ that expands to $\frac{3a}{2}((-1) + 1)$ resulting in a length of 0, which is.... clearly impossible.
My real question is, could someone please help me identify where I went wrong?
Thanks!
You simply forgot that $$\sqrt{9a^2\cos^2 t\sin^2t}=\lvert3a\sin t\cos t\rvert.$$
You can use that this curve (an astroid) has many symmetries: w.r.t. the $x$-axis, the $y$-axis and the first bissectrix $y=x$, since $$\Bigl(x\bigl(\tfrac\pi2-t\bigr), y\bigl(\tfrac\pi2-t\bigr)\Bigr)=\bigl(x(t),y(t)\bigr)$$ So all you have to do is compute the integral on the interval $\;\bigl[0,\frac\pi4\bigr]$ and multiply the resulting length by $8$.