Understanding Unit Tangent to a Curve in PDEs

[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

 

[HallsofIvy]

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.