Parametrised curves.

Question

I've been working through the following question:

Q1= What points on the parameterised curve $x(\theta)=\cos^2{\theta}, y(\theta)=\sin{\theta}\cos{\theta}$ correspond to the parameter values ${\theta}=-\frac{\pi}{4}, {\theta}=\frac{\pi}{4}.$

My answers for that question were $x=\frac{1}{2},y=-\frac{1}{2}$ and $x=\frac{1}{2}, y=\frac{1}{2}$.

Q2 Find the tangent lines at these points.

So, when it says ''at these points'' does it mean that I calculate the slope using $$\frac{dy}{dx}=\frac{\frac{dy}{d{\theta}}}{\frac{dx}{d{\theta}}}$$ and plug in the parameter values? $${\theta}=-\frac{\pi}{4}, {\theta}=\frac{\pi}{4}$$

I've done this and I encounter two horizontal tangents at $y=-\frac{1}{2}$ and $y=\frac{1}{2}$.

Does this seem reasonable?

The third part of the question is:

Q3-Draw this curve on a set of labelled axes.

To do this I simply input numerous values for $\theta$ and plotted the resultant $x,y $ values.

Is this a reasonable method, or would it be easier to have $y$ in terms of $x$, i.e. remove the $\theta$ parameter?

The graph looks like an ellipse of sorts.

Answer 1

Your answer for Q1 and Q2 look fine.

For Q3, since we have $$x=\frac{\cos(2\theta)+1}{2}\iff \cos(2\theta)=2x-1$$ $$y=\frac{\sin(2\theta)}{2}\iff \sin(2\theta)=2y,$$ we have $$(2x-1)^2+(2y)^2=1\iff \left(x-\frac 12\right)^2+y^2=\frac 14,$$ which is a circle whose center is $(1/2,0)$ with the radius $1/2$.

Answer 2

You've done fine, all the way around, so to speak. When first encountering parametric equations, working with them takes a shift in thinking from what we're used to.

So I think your method of plotting the curve is a nice way to immerse yourself in how to work with parametrized values of $x, y$, rather than leaping back to the familiar $(x, y)$ mode of operating.

But it would also be a nice exercise, after plotting as you did, to find the equation of the curve in terms of $x, y$, to develop facility in going back and forth between the two forms of equations. That way, you can also check your work.