Question

normal vectors to a surface z^2= e^sin(x)*y when x =pi and y =9

Answer 1

The equation is:\[z^2=e^{(\sin x)y}\]or,\[z^2=e^{\sin x}y\]??

Answer 2

the second z^2= y*e^sin(x)

Answer 3

ok. Well first let's re-write this as:\[F(x,y,z)=e^{\sin x}y-z^2=0\]This is a level surface of the function F(x,y,z)=0. Now take the gradient of this function (which is orthogonal to the surface):\[grad F=<\cos x e^{\sin x}y, e^{\sin x},-2z>\]at the point x=pi, y=9 z=9, we get the vector:\[grad F=<-1,1,-18>\]which is normal to the surface. Any scalar multiple of this vector is also a normal vector to the surface. Thus,\[\vec{n}=s<-1,1,-18>\]for any real number, s.

Answer 4

ah crap. forgot to take the square root when I evaluated z. z=3 NOT 9. So,
n=s<-1,1-6>

Answer 5

wouldn't z = 0?

Answer 6

why? when x=pi and y=9 we have:\[z^2=9e^{\sin \pi}\]Since sin(pi)=0,\[z^2=9\]and z=3, assuming z is greater than 0.

Answer 7

actually z could equal -3. So our normal might also be:
\[\vec{n}=s<-1,1,6>\]

Answer 8

gradient of f =<-9,1,6>?

Answer 9

I made a mistake above:
grad F=<cosxe^{sinx}y, e^{sinx}, -2z}
inputting x=pi, y=9 and z=plus or minus 3:
grad F=<-9,1,plus or minus 6>

Answer 10

how do i find the unit normal to this problem please?

Answer 11

to find a unit vector, just re-scale it to unit length. The relation is:\[\hat {n}=\frac{\vec{n}}{\left| \vec{n} \right|}\]

Answer 12

The length of your vector is sqrt(81+1+36)=sqrt (118)
So,\[\hat{n}=\frac{1}{\sqrt{118}}<-9,1,6>\]

Answer 13

thank you for your time!!

Answer 14

no problem. I'm still wrestling with understanding what to do about that damn z^2. Is the normal <-9,1,6> or <-9,1,-6>? I'm honestly a little rusty on this stuff.

Answer 15

Ok, I gave it some thought and it makes perfect sense. If x=pi and y=9 then z=3 or z=-3. If z=3 then the gradient vector is <-9,1,-6>. If z=-3 then the gradient vector is <-9,1,6>. I starting thinking about a sphere which has two halves one above the xy-plane and one below it. Given a value for x and y, you have two values for z, one above the xy-plane, and one below it. Depending on which z you want, you get a different normal vector. One gradient vector is\[<2x_{0},2y_{0},2z_{0}>\] and the other is:\[<2x_{0},2y_{0},-2z_{0}>\]my drawing is crappy though :)