Michael Corral: Vector Calculus

2.3 Tangent Plane to a Surface 77
Example 2.13. Find the equation of the tangent plane to the surface z= x 2 +y 2 at the
point (1,2,5).
Solution: For the function f(x,y)= x 2 +y 2 , we have ∂f
∂x
= 2x and∂f
∂y
= 2y, so the equation
of the tangent plane at the point (1,2,5) is
2(1)(x−1)+2(2)(y−2)−z+5=0,or
2x+4y−z−5=0.
In a similar fashion, it can be shown that if a surface is defined implicitly by an
equation of the form F(x,y,z)=0, then the tangent plane to the surface at a point
(a,b,c) is given by the equation
∂F
∂x (a,b,c)(x−a)+∂F ∂y (a,b,c)(y−b)+∂F ∂z
(a,b,c)(z−c)=0. (2.7)
Note that formula (2.6) is the special case of formula (2.7) where F(x,y,z)= f(x,y)−z.
Example 2.14. Find the equation of the tangent plane to the surface x 2 +y 2 +z 2 = 9 at
the point (2,2,−1).
Solution: For the function F(x,y,z)= x 2 + y 2 + z 2 − 9, we have ∂F
∂F
∂z
= 2z, so the equation of the tangent plane at (2,2,−1) is
2(2)(x−2)+2(2)(y−2)+2(−1)(z+1)=0,or
2x+2y−z−9=0.
∂x
= 2x,
∂F
∂y
= 2y, and
☛ ✟
✡Exercises
✠
A
For Exercises 1-6, find the equation of the tangent plane to the surface z= f(x,y) at
the point P.
1. f(x,y)= x 2 +y 3 , P=(1,1,2) 2. f(x,y)= xy, P=(1,−1,−1)
3. f(x,y)= x 2 y, P=(−1,1,1) 4. f(x,y)= xe y , P=(1,0,1)
5. f(x,y)= x+2y, P=(2,1,4) 6. f(x,y)= √ x 2 +y 2 , P=(3,4,5)
For Exercises 7-10, find the equation of the tangent plane to the given surface at the
point P.
( )
x
7. 2
4 + y2
9 + z2
16 = 1, P= 1,2, 2√ 11 8. x 2 +y 2 +z 2 = 9, P=(0,0,3)
3
9. x 2 +y 2 −z 2 = 0, P=(3,4,5) 10. x 2 +y 2 = 4, P=( √ 3,1,0)