# Unit tangent

1. Jan 6, 2009

### coverband

Hi

In my lecturer's notes he describes the unit tangent to a curve y=Y(X) as

(i + Y'(X)j)/[(1+[Y'(X)]^2)^(0.5)]

in an introduction to second order PDEs

I'm a bit confused by this. Where did it come from?

Can anyone explain

Thanks

2. Jan 6, 2009

### OrderOfThings

A curve $(x, y(x))$ parameterized by $x$ has tangent vectors $(1, y'(x))$ with lengths $\sqrt{1+(y')^2}$.

3. Jan 6, 2009

### flatmaster

Also, this function would give you a unit vector pointing in the positive x direction. Notice that the function will tell you only the tangent vector. It will not tell you at what x and y coordinates the actual tangent touches the curve. You need the origional function for that.

4. Jan 6, 2009

### tiny-tim

Hi coverband!

It looks more logical if you parametrise the curve: X = x(t), Y = y(t).

Then your Y'(X) is dY/dX = y'/x', and the formula (after multiplying top and bottom by x') becomes …

(x',y')/√(x'2 + y'2) …

so the tangent is parallel to the gradient (x',y'),

and to make it a unit vector you divide by its magnitude.

5. Jan 6, 2009

### flatmaster

Let's clarify the analogy with the gradient operator. This would actually be a one dimentional case of the gradient operator, not a 2D case correct?

y(x) = Grad[Y(x)] / Abs[Grad[Y(x)]]

Where in general, x is a position vector of arbitrary dimention. This of course goes much beyond the origional question, but I want to check for myself.

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