Why dot product has functions in it when it is for vectors?

Vol

New member
TeX
[FONT=&quot] [/FONT] Hi. I ran into the above studying method of characteristics. You can see that the first dot product has two partial derivatives in it. But derivatives are scalar functions. I thought dot products were for vectors only. How can you have dot products of functions? There are no coordinates to calculate. How will you know if it equals zero or not? What I not getting? Thanks. :roll:

Dr.Peterson

Elite Member
TeX Hi. I ran into the above studying method of characteristics. You can see that the first dot product has two partial derivatives in it. But derivatives are scalar functions. I thought dot products were for vectors only. How can you have dot products of functions? There are no coordinates to calculate. How will you know if it equals zero or not? What I not getting? Thanks. :roll:
Just as a and b are scalars, but (a, b) is a vector with those as components, ux and uy are scalars but (ux, uy) is a vector. In fact, the dot product is equal to aux + buy.

If there is more to your question, you'll have to show us the whole problem.

Vol

New member
Just as a and b are scalars, but (a, b) is a vector with those as components, ux and uy are scalars but (ux, uy) is a vector. In fact, the dot product is equal to aux + buy.

If there is more to your question, you'll have to show us the whole problem.
Oh, that's right. I forgot about the brackets. Thanks. :roll: