# Parametric expression

1. Apr 22, 2005

### andrewdavid

I have x(t)=t(1-t) and y(t)=t(1-t^2). As t goes from 0 to 1 in forms a loop and I need to know the point where the tangent line is vertical. I know this must be easy but I'm clueless right now. Any help?

2. Apr 22, 2005

### dextercioby

U need the derivative in that point to be infinite.That is the slope must be infinite...

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Daniel.

3. Apr 23, 2005

### Jameson

So basically you need to set $$\frac{dx}{dt}$$ equal to zero.

Jameson

4. Apr 23, 2005

### Data

solving $dx/dt=0$ is not quite enough in general. Dexter's method is fine.

Consider $x(t)=t, \ y(t)=t$.

5. Apr 24, 2005

### whozum

Data: but x = t, and y = t doesnt have a vertical asymptote?

6. Apr 24, 2005

### dextercioby

It was a counterexample to the statement of analyzing $\frac{dx}{dt}$ instead of $$\frac{\displaystyle{\frac{dy}{dt}}}{\displaystyle{\frac{dx}{dt}}}$$

Daniel.

Last edited: Apr 24, 2005
7. Apr 24, 2005

### Jameson

I still don't see how that's an example where simply setting dx/dt equal to zero will not yield the vertical asymptotes.

Jameson

8. Apr 24, 2005

### kleinwolf

This was not, because if x(t)=t, dx/dt=1

However, if you take the dumb example : x(t)=y(t)=sin(t). Then obviously, y=x, so this is just the diagonal, so there are no vertical tangents.

However, there exist some t such that dx/dt=0...but this is precisely where dy/dt=0, so that both cancel....you can imagine example where this gives other value than 1...

In fact to solve the question why you have to look at (dy/dt)/(dy/dt)...it's because you could have dy/dt=infty and dx/dt finite...or other limits such as dx/dt->0, and dy/dt->0...(indefinite)...

9. Apr 24, 2005

### Data

oops, yeah, i needed to use $x(t)=y(t)=t^2$!