Question

given z= (e^x)cosy find the normal line at (0,0,1)
anyone helps me, please

Answer 1

find the gradiant vector(u) of
$$f(x,y,z)=e^x\cos y-z$$
then the line will be,
$$t\vec{u}=\langle0,0,1\rangle-\langle x,y,z\rangle$$

Answer 2

what is the normal line? i know the normal vector is the vector perpendicular to the surface, i don't know the normal line

Answer 3

that's what i've mentioned in the end
if you get u as,
$$\vec{u}=\langle a,b,c \rangle$$
then,
the line will be,
$$t\langle a,b,c \rangle=\langle 0,0,1 \rangle-\langle x,y,z \rangle$$
then,
$$at=-x,\quad bt=-y,\quad ct=1-z$$
$$t=\frac{-x}{a}=\frac{-y}b=\frac{1-z}{c}$$
this is the equation of the line

Answer 4

I skip the first part in which i have to find tangent plane of the surface at that point. I got the tangent plane equation is x-z+1=0. can i find the normal line by using the vector perpendicular to tangent plane ?

Answer 5

yes you can,
we know that (0,0,1) is on the plane so the equation of the plane should be
$$a(x-0)+b(y-0)+c(z-1)=0$$
you've got the equation of the plane as x-z+1=0 which can be adjusted as,
$$1(x-0)+0(y-0)+(-1)(z-1)=0$$
so the tangent vector is (1,0,-1). If you consider the gradiant at (0,0,1) you'll get the same result

Answer 6

thanks a lot, I understand what you mean now. i am familiar with that way than the previous one. I'll send you my result to get your check after I have it done because I have to step away a while. thank you very very much,

Answer 7

will you check mine?

Answer 8

sure. you're welcome

Answer 9

I let the window here to make sure that i can contact you then.

Answer 10

x=1-z, y=0 is it right?