Find the unit normal vector to the surface at the given point.

z=(x^2+y^2)^1/2 (3,4,5)

Question

Find the unit normal vector to the surface at the given point. 

z=(x^2+y^2)^1/2  (3,4,5)

anonymous · Answer

$$z=\sqrt{x^2+y^2}, at the point (3,4,5)$$

anonymous · Answer

the same of finding gradient at the given point

anonymous · Answer

find partial derivatives, and take their value at your point. That's the vector normal to the surface

anonymous · Answer

$$Gradient( z)= (x/\sqrt{x ^{2}+y ^{2}},y/\sqrt{x ^{2}+y ^{2}})$$

anonymous · Answer

Calculating the partials, I got : $$0=z-\sqrt{x^2+y^2}$$
$$\Delta F(x,y,z) = x/\sqrt{x^2+y^2}i+y/\sqrt{x^2+y^2}j-1k$$

anonymous · Answer

^ yeah thats what I got.

anonymous · Answer

so now just put your values in

anonymous · Answer

Then when I evaluated the gradient I get: 

$$\Delta F(3,4,5) = <3\sqrt{5}/5,4\sqrt{5}/5>$$

anonymous · Answer

i guess, i didn't make the calculation

anonymous · Answer

ΔF(3,4,5)=<35  √ /5,45  √ /5,1>

anonymous · Answer

how are you getting 35 and 45?

amistre64 · Answer

gradient isnt \Delta, its \nabla$$\nabla F ...$$