# 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

### HallsofIvy

Staff Emeritus
If y= Y(x), you can write the "position vector" of a point on the curve as $\vec{r}= x\vec{i}+ y\vec{j}= x\vec{i}+ Y(x)\vec{j}$. Differentiating that vector with respect to x gives $\vec{r}'= \vec{i}+ Y'(x)\vec{j}$ as a vector tangent to the curve. It's length is, of course, $|\vec{r}'|= \sqrt{1^2+ Y'(x)^2}= (1+ Y'(x)^2)^{1/2}$. Dividing $\vec{i}+ Y'(x)\vec{j}$ by that, which is the denominator in what you give, makes it a unit tangent vector.

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