Recipe

Find a tangent plane and use it for linear approximation

Use when: you have z=f(x,y), differentiable at a point (a,b), and need either the tangent plane's equation or a quick numeric estimate of f at a nearby point

Goal: the tangent plane's equation, and, if a nearby point is given, a linear-approximation estimate of f there

Taught in Tangent planes · a later section extends this recipe with more cases

1. 1

   Compute f(a,b), the exact height the plane must pass through.

   Example

   For f(x,y)=x^2+y^3 at (2,1): f(2,1)=2^2+1^3=4+1=5.

   Why: "tangent" means the plane touches the surface exactly here, so matching this one height is the first non-negotiable condition.

2. 2

   Find f_x and f_y as fresh functions of x and y, then substitute (a,b) into each to get two plain numbers.

   Example

   f_x=2x \Rightarrow f_x(2,1)=4, and f_y=3y^2 \Rightarrow f_y(2,1)=3.

   Why: a plane needs an independent tilt in two directions, holding y fixed for one and x fixed for the other. A single derivative only pins down a line.

3. 3

   Substitute f(a,b), f_x(a,b), and f_y(a,b) into z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b), then distribute and simplify.

   Example

   z=5+4(x-2)+3(y-1)=5+4x-8+3y-3=4x+3y-6.

   Why: this is the one plane matching the surface's height and both slopes at (a,b). No other plane satisfies all three conditions at once.

4. 4

   Reread the question to see whether it stops at the plane's equation or also asks for a value at some nearby point.

   Example

   Estimating f(2.02,0.97): the plane gives z=4(2.02)+3(0.97)-6=8.08+2.91-6=4.99. The true value is f(2.02,0.97)=2.02^2+0.97^3\approx4.9931, matching to two decimal places.

   • only the equation was asked for — the answer from the previous step is complete; stop here.
   • a nearby point (x_0,y_0) is given — substitute it into the PLANE, not into f, for a cheap estimate of f(x_0,y_0).
   • Not yet: bounding how much error an estimate like this one can carry, from known uncertainties in the inputs themselves — covered in Linear approximation and differentials

   Why: near the point of tangency the plane and the surface stay close together. Differentiability guarantees their difference shrinks faster than the distance moved away from (a,b).