Directional derivative: radial direction

[Yura]

(hmm. wasnt to sure about where to post this)

im given an equation to a 3d surface and asked to find the gradient at a certain point, in the radial direction.

my question is, what is the radial direction?

[ the equation is f(x,y) = 3*(x^2)*y + 2*y if its needed ]

 

[HallsofIvy]

The radial direction, at the point (x,y,z) is the direction of the ray from (0,0,0) through (x,y,z). In particular, a unit vector in that direction is
\frac{x}{\sqrt{x^2+ y^2+ z^2}}\vec i+ \frac{y}{\sqrt{x^2+ y^2+ z^2}}\vec j+ \frac{z}{\sqrt{x^2+ y^2+ z^2}}\vec k.

The gradient of the function f(x,y)= 3x²y+ 2y is, of course,
\nabla f= 6xy\vec i+ (3x^2+ 2)\vec j
and the derivative "in the radial direction" is the dot product
(6xy\vec i+ (3x^2+ 2)\vec j) \cdot (\frac{x}{\sqrt{x^2+ y^2+ z^2}}\vec i+ \frac{y}{\sqrt{x^2+ y^2+ z^2}}\vec j+ \frac{z}{\sqrt{x^2+ y^2+ z^2}}\vec k)
= \frac{6x^2y}{\sqrt{x^2+ y^2+ z^2}}\vec i+ \frac{y(2x^2+ 2)}{\sqrt{x^2+ y^2+ z^2}}\vec j

 

[gnomedt]

Do you mean

= \frac{6x^2y}{\sqrt{x^2+ y^2+ z^2}}+ \frac{y(2x^2+ 2)}{\sqrt{x^2+ y^2+ z^2}}

i.e. sans the i-hat and j-hat? The dot product is a scalar.

 

[HallsofIvy]

Oops! Yes, of course.

 

[Yura]

thankyou! i was stuck on that one for ages ><

thanks again for the help.