# equation of the tangent plane and normal

• Dec 15th 2010, 06:09 AM
snopy
equation of the tangent plane and normal
hello, i have task Z= y*cos*(x-y) in point T (π; π/2; ?) .....is this derivation all right?

Z´x = y*sin*(x-y)/T1= - π/2*sin*( π- π/2) = - π/2

Z´y = y*sin*(x-y) + cos*(x-y)/T1 = π/2 * sin*( π – π/2) + cos*( π – π/2) = π/2

Z0 = π/2*cos*(2 π – π) = π/2*cos* π/2 = π/2
2

• Dec 15th 2010, 06:14 AM
snopy
• Dec 15th 2010, 11:08 AM
SammyS
Quote:

Originally Posted by snopy

I have edited your post as I understand it, & re-done the mathematical statements in LaTeX.

From the title of the original post, I understand that you need to:

Find the equation of the tangent plane and normal to the surface:
$\displaystyle \ z= y\cos(x-y)$ at point T: $\displaystyle (\pi,\ \pi/2,\ z_0)$

$\displaystyle \displaystyle {{\partial z}\over{\partial x}} = -y\sin(x-y)$. Evaluated at point T, this is $\displaystyle - {{\pi}\over{2}}\sin( \pi- {{\pi}\over{2}}) = - {{\pi}\over{2}}$

$\displaystyle \displaystyle {{\partial z}\over{\partial y}} = y\sin(x-y)+\cos(x-y)$. Evaluated at point T, this is $\displaystyle {{\pi}\over{2}}\sin( \pi- {{\pi}\over{2}})+\cos(\pi- {{\pi}\over{2}})={{\pi}\over{2}}$

$\displaystyle z_0={{\pi}\over{2}}\cos({{2\pi-\pi}\over{2}})={{\pi}\over{2}}\cos({{\pi}\over{2}} )={{\pi}\over{2}}$

Quote:

.....is this derivation all right?

$\displaystyle \displaystyle z_0=0$ because $\displaystyle \displaystyle \cos(\pi/2)=0$.