# Finding the initial direction of a parametric curve?

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This is done by computing the derivative vector and seeing how it behaves near $0$.

That is to say what can we say about

$${d\over dt}\begin{pmatrix} x(t) \\ y(t)\end{pmatrix}$$

when $0<t<\epsilon$?

This is just

$$\begin{pmatrix}-2\sin t\cos t \\ -2\cos t\sin t\end{pmatrix}=-\begin{pmatrix}\sin(2t) \\ \sin(2t)\end{pmatrix}$$

for small $t$. As $t$ increases from $0$, $\sin(2t)$ increases, so both of these coordinates are decreasing because of the minus sign.

In fact, since the two entries are the same, we can say that it is headed in the direction of the angle $\theta = {5\pi\over 4}$.

Alternatively, if you do not know calculus

Let

$$0<t_1<t_2<\epsilon <{\pi\over 2}$$

be small. Then by examining the unit circle, we see that

$$2+\cos^2 t_2<2+\cos^2t_2$$

and the $y$-coordinate of our vector is decreasing, as a result.

Similarly $x(t)$ has that for

$$0<t_1<t_2<\epsilon<{\pi\over 2}$$

that $\sin t$ is increasing (again by considering the unit circle) and we have that

$$1-\sin^2t_2<1-\sin^2t_1$$

so that $x(t)$ is also decreasing. We no longer can easily tell they are headed in the same direction like with the calculus approach, though you can convince yourself of it by using

$$\cos^2t+\sin^2t=1.$$

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### Calculistening helplessly

Updated on October 15, 2020

• Calculistening helplessly almost 2 years

With the parameters: $$x(t)=1-\sin^2t$$ $$y(t)=2+\cos^2t$$

It starts at (1,3) and when t=pi/2 it's at (0,2), so I'm tempted to say it's going down to the right; is this correct?

In general, is there a way to find the direction of a curve besides just plotting points?

• DSinghvi almost 8 years
Do you know calculus and in particular" derivative "
• Calculistening helplessly almost 8 years
Forgive me for being a little slow on the uptake, but why $\frac{5pi}{4}$?
• Adam Hughes almost 8 years
@Calculisteninghelplessly because that is the angle of the vector $\langle -1,-1\rangle$. Remember that the angle of a vector $\langle a,b\rangle$ is just $\theta =\arctan\left({b\over a}\right)$ if the vector is in quadrants 1 or 4, and $\theta =\arctan\left({b\over a}\right)+\pi$ if in quadrants 2 or 3.