# Homework Help: Tangent to Parametric Equations

1. Oct 29, 2014

### Gabble1

1. The problem statement, all variables and given/known data
Consider the curve with parametric equations: x = t - cos t, y = sin t.
Determine exactly the equation of the tangent to the curve at the point where t=-0.5pi.

2. Relevant equations

3. The attempt at a solution
The equation of a line is y - y1 = m ( x - x1 )
I substituted t = -pi/2 into x and y to get x = -pi/2 and y = -1
Differentiating dy/dx as (dy/dt)/(dx/dt) came out as cos t/1 + sin t
When substituting in for t = -pi/2, I was left with 0/0
From this point, I am unsure what the nature and equation of the tangent would be, as a horizontal line would have 0 as the numerator and a non-0 value as the denominator, and vice versa for a vertical line.
Any help would be greatly appreciated.

2. Oct 29, 2014

### PeroK

3. Oct 29, 2014

### Gabble1

I have to solve it algebraically. Plotted the graph on WolframAlpha, doesn't seem to be of any use.

4. Oct 29, 2014

### BvU

If you get 0/0 algebraically, it's still worth looking at the picture (e.g. on) to find out why you are in trouble...
Perhaps then you can algebraically compare the limit of the tangent line when coming from the left with idem when coming from the right.

5. Oct 29, 2014

### Gabble1

Cheers for the link, just had a look at the article. Would this mean that at the cusp, a tangent would not exist or would it simply be a vertical line at x = -pi/2 ?

6. Oct 29, 2014

### BvU

I have no idea of the formal definition of "tangent line" in your context. As a physicist, I would sign off for $x = -{\pi\over 2}$: the fact that the motion changes direction at that point doesn't bother me.

This link digs in a bit deeper, but I don't get any wiser from it.

Last edited: Oct 29, 2014
7. Oct 29, 2014

### Gabble1

Ah fair enough, thank you very much for your help

8. Oct 29, 2014

### RUber

To get a different perspective, you can always apply L'Hopiatal's rule to your 0/0 limit. If you needed to discuss the limits of the derivatives as they approach the time -pi/2.
However, there are infinitely many choices of lines that are tangent to a cusp like that.

9. Oct 29, 2014

### BvU

Beg to differ. Left derivative and right derivative "go to" $-\infty$ and $+\infty$, respectively, but that happens to give the same tangent line.
In the second link I gave (definition 1 page 136 ) that line actually does give the best linear approximation near the cusp. So -- unlike for y = |x| -- there is an argument to claim that $x=-{\pi\over 2}$ is the tangent line.