What is the Point Where the Tangent Line is Vertical for x(t)=y(t)=t^2?

[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?

 

[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.

 

[Jameson]

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

Jameson

 

[Data]

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

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

 

[whozum]

Data: but x = t, and y = t doesn't have a vertical asymptote?

 

[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.

 

[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

 

[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)...

 

[Data]

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