Converting parametric equations with trigonometric functions into Cartesian form

Question

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

Answer 1

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 $$\left(\frac{x}{2}\right)^{2 / 5} = \cos^2 t \qquad \text{and} \qquad \left(\frac{y}{2}\right)^{2 / 5} = \sin^2 t.$$

Answer 2

As Travis wrote in the last part of his answer, a convenient of writing in cartesian form $$x = a\cos^m (t), \quad y = b \sin^n (t)$$ is to extract $\cos^2(t)$ from $x$ and $\sin^2(t)$ from $y$ and then to apply $\sin^2(t)+\cos^2(t)=1$.

In the case, $$\Big(\frac xa\Big)^{2/m}+\Big(\frac yb\Big)^{2/n}=1$$ but, as Travis answered, there are restrictions depending on the considered quadrant.