Ahoy,
I am having trouble with a computer-based assignment and the question is as follows:
$$x = 2\cos^5 t, \quad y = 2 \sin^5 t$$
Write these in Cartesian form, $F(x,y) = c$.
I understand how to perform this operation for simpler examples but so far this is what I have done.
I have solved $x$ for $t$, which I have said is $(\cos^{-1}\frac{x}{2})^{1/5}$.
Once this is done I then plug in this value for $t$ into my $y$ equation and I believe this gives me my $c$ value.
I am not really sure if this is right or how I am supposed to present my answer here. I guess I am failing to understand the process on how to deal with these trig functions with powers as well as how to present my answer.
Thank you,
Sean
Best Answer
Hint First, note that solving the equation for $x$ gives $$t = \arccos \left[\left(\frac{x}{2}\right)^{1 / 5}\right]$$ (for appropriate values of the argument of $\arccos$), rather than with $\arccos$ and $\cdot^{1 / 5}$ in the reverse order.
Now, one can simplify the expression produced by substituting this expression for $t$ into the formula for $y$ by using the identity $$\sin \arccos t = \sqrt{1 - t^2}$$ (again for appropriate $t$). We can derive this identity by drawing a right triangle with leg lengths $1$ and $t$ and applying the usual definitions of trig and inverse trig functions.
Alternatively, (at least for $(x, y)$ in the first quadrant,) rearranging the original parametric equations gives $$\left(\frac{x}{2}\right)^{2 / 5} = \cos^2 t \qquad \text{and} \qquad \left(\frac{y}{2}\right)^{2 / 5} = \sin^2 t.$$