Unit normal problem

[mcwang719]

Suppose we have the right circular cone z=sqrt(x^2+y^2). What is the UNIT upward normal vector at the point (3,4,5).

so this is what i have done:
so i took the partial derv. at (3,4,5)
fx=-2x=-6 fy=2y=8 fz=2z=10
(-6,8,10)
so sqrt(6^2+8^2+10^2)=sqrt(200)= 10sqrt(2)
so the answer would be 1/10sqrt(2) <-6,8,10>
right?? thanks!

 

[mcwang719]

This was actually a multiple choice question
a.\(\displaystyle 1\sqrt 2 \left\langle { - {3 \over 5}, - {4 \over 5},1} \right\rangle\)
b.\(\displaystyle \left\langle { - {3 \over {\sqrt {26} }}, - {4 \over {\sqrt {26} }},{1 \over {\sqrt {26} }}} \right\rangle\)
c.\(\displaystyle \left\langle { - {3 \over 5}, - {4 \over 5},0} \right\rangle\)
d.\(\displaystyle \left\langle {{3 \over {5,}}{4 \over 5},1} \right\rangle\)
e. none of these
so the answer i came up with was \(\displaystyle {1 \over {10\sqrt 2 }}\left\langle { - 6,8,10} \right\rangle\)So the answer would be e right. Can someone please help? thanks!!!!

 

[pka]

The upward normal vector at any point \(\displaystyle \L
\left\langle { - z_x , - z_y ,1} \right\rangle = \left\langle {\frac{{ - x}}{{\sqrt {x^2 + y^2 } }},\frac{{ - y}}{{\sqrt {x^2 + y^2 } }},1} \right\rangle\)

So at (3,4,5) we have \(\displaystyle \L
\left\langle {\frac{{ - 3}}{5},\frac{{ - 4}}{5},1} \right\rangle\)

To make that a unit vector divide by its length.

 

[mcwang719]

So what would the length be? How would you finish this problem? thanks!

 

[tkhunny]

Maybe an extension to the Pythagorean Theorem?

Don't forget your basic geometry.

 

[mcwang719]

wait! so the length would be \(\displaystyle \sqrt {{3 \over 5}^2 + {4 \over 5}^2 + 1^2 } = \sqrt 2 \Rightarrow {1 \over {\sqrt 2 }}\left\langle { - {3 \over 5}. - {4 \over 5},1} \right\rangle\)right?! thanks.