Equation for the Tangent Plane

PROBLEM:
Give an equation for the tangent plane to x^2+y^2+z^2=209 at the point (-9,8,8).

a. _________=0

Find equations for the normal line to the surface through that point.
b. x(t)= ___________
c y(t)= ___________
d. z(t)= ___________

MY WORK:
f(-9,8,8)=209 [81+64+64]
d/dx=2x d/dy= 2y d/dz=2z
= -18 =16 =16

plug into formula...
f(x,y)+d/dx(x-x0)+d/dy(y-y0)+d/dz(z-z0…

209-18(x-9)+16(y-8)+16(z-8)


Thanks for your help. Also can you offer insight for what it is looking for for the normal lines?

Write F(x,y,z) = x^2 + y^2 + z^2 = 209.

Then grad F (gradient) is a vector normal to F at any given point.

In this case, grad F = <2x, 2y, 2z>
==> grad F(-9, 8, 8) = <-18, 16, 16>.

So, the equation of the tangent plane is
-18(x - (-9)) + 16(y - 8) + 16(z - 8) = 0,

and the equation of the normal line (in vector form) is
r(t) = <-9, 8, 8> + t<-18, 16, 16).
==> x = -9 - 18t, y = 8 + 16t, z = 8 + 16t.

I hope this helps!

Remember that F(-9,8,8) = 209 by definition.

Report Abuse