tangent plane

tangent plane

1. What a Tangent Plane Is

For a surface $S$ given by $F(x,y,z)=0$ (e.g., $z = f(x,y)$ ⇒ $F= z - f(x,y)=0$) at a regular point $P_0(x_0,y_0,z_0)$ where $∇F(P_0)≠0$, the tangent plane is the plane through $P_0$ whose normal is $∇F(P_0)$:

Equivalently for explicit $z = f(x,y)$:

is the tangent plane (the linear approximation from differentiability). The normal line at $P_0$ is:

with direction vector $∇F(P_0)$.

2. Derivation Sketch

Take a curve $r(t)$ lying on $S$ through $P_0$; differentiate $F(r(t))=0$ ⇒ $∇F· r'(t)=0$ ⇒ every tangent vector $r'(t)$ is perpendicular to $∇F$ ⇒ all tangents lie in plane with normal $∇F$.

3. Worked Examples

(a) Plane? Actually surface S: x² + y² + z² =14 at P(1,2,3):
    F= x²+y²+z²-14, F_x=2x=2, F_y=4, F_z=6 ⇒ plane: 2(x-1)+4(y-2)+6(z-3)=0 ⇒ x+2y+3z=14 ✓
    Normal line: (x-1)/1 = (y-2)/2 = (z-3)/3 ✓

(b) z = f(x,y)= x² + y² at (1,1,2): f_x=2x=2, f_y=2 ⇒ plane: z-2=2(x-1)+2(y-1) ⇒2x+2y -z=2 ✓
    Normal line? Implicit F= z -x² -y²: ∇F= (-2x,-2y,1) ⇒ at (1,1): (-2,-2,1) ⇒ (x-1)/-2=(y-1)/-2=(z-2)/1 ✓

(c) Implicit: z² = x² + y² (cone) at (3,4,5): F= x²+y² -z², ∇F=(2x,2y,-2z)=(6,8,-10) ⇒ plane:6(x-3)+8(y-4)-10(z-5)=0 ⇒3x+4y-5z=0 ✓ (touches cone along generator? Check).

(d) Find tangent plane to S: x y z =6 at (1,2,3): F= x y z-6, ∇F=(y z, x z, x y)=(6,3,2) ⇒ plane:6(x-1)+3(y-2)+2(z-3)=0 ⇒6x+3y+2z=18 ✓

(e) Show plane is horizontal exactly when F_x=F_y=0 ⇒ normal vertical (like at maxima from earlier module).

4. Relation to Earlier Topics

• This plane is exactly the linear part from differentiability / Taylor first-order for $z=f(x,y)$.
• Gradient $∇F$ was normal to level surface $F=c$ — here $S$ is level surface.

5. Exam Points

• Write tangent plane $F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0$ for $F(x,y,z)=0$ — 5 marks, very frequent.
• For $z=f(x,y)$, use $z-z_0= f_x(x-x_0)+f_y(y-y_0)$ — 5 marks, frequent.
• Find normal line symmetric form — 3-5 marks, frequent.
• Show $∇F$ is normal — 3 marks.