Tangent planes and linear approximations to surfaces

WHAT are we approximating?

We have a surface $z = f(x,y)$ . Near a point $(a,b)$ we want a plane that:

Passes through the surface point $(a,b,f(a,b))$ .
Has the same slope as the surface in every direction at that point.

HOW to derive it (from first principles)

Step 1 — A general plane through $(a,b,z_0)$ . Any non-vertical plane can be written $z = z_0 + A(x-a) + B(y-b).$ Why this form? It guarantees passage through $(a,b,z_0)$ , and $A,B$ are the slopes in the $x$ - and $y$ -directions.

Step 2 — Match the height. At $(a,b)$ we need $z = f(a,b)$ , so $z_0 = f(a,b)$ .

Step 3 — Match the $x$ -slope. Hold $y=b$ . Then $z = z_0 + A(x-a)$ , slope $A$ . The surface's slope here is $f_x(a,b)$ . So $A = f_x(a,b)$ .

Step 4 — Match the $y$ -slope. Hold $x=a$ similarly: $B = f_y(a,b)$ .

Figure — Tangent planes and linear approximations to surfaces

WHY does this work? (Differentiability)

The error $E(x,y) = f(x,y) - L(x,y)$ must shrink faster than the distance to $(a,b)$ : $\lim_{(x,y)\to(a,b)} \frac{f(x,y)-L(x,y)}{\sqrt{(x-a)^2+(y-b)^2}} = 0.$ This is the precise meaning of differentiable. Just having partials exist is not enough — they must also be continuous (a sufficient condition) for the plane to truly "hug" the surface.

Gradient / normal-vector view

Define $F(x,y,z) = f(x,y) - z = 0$ as a level surface. Then $\nabla F = (f_x, f_y, -1)$ is normal to the surface. The tangent plane is everything perpendicular to this normal: $f_x(a,b)(x-a) + f_y(a,b)(y-b) - 1\cdot(z-f(a,b)) = 0,$ which rearranges to the same formula. The normal vector is $\mathbf{n} = \langle f_x, f_y, -1\rangle$ .

Worked Examples

Common Mistakes

Recall Feynman: explain to a 12-year-old

Imagine a hilly chocolate-coated surface. If you put a tiny flat sticker right on one spot, the sticker touches the chocolate perfectly and tilts the same way the hill tilts. That sticker is the tangent plane. To guess the height of a nearby point, you just walk along the flat sticker instead of the bumpy chocolate — close points give nearly the right answer, far points are off. The tilt of the sticker left-right is $f_x$ and front-back is $f_y$ .

Connections

Partial derivatives — supply the slopes $f_x, f_y$ .
The gradient vector — gives the normal $\langle f_x, f_y, -1\rangle$ .
Differentiability of multivariable functions — the condition the plane needs.
Tangent line and linear approximation (single variable) — the 1D parent idea.
The chain rule (multivariable) — built on the linear-approximation engine.
Directional derivatives — slope in any direction, also lives in the tangent plane.

#flashcards/maths

What is the tangent plane equation to

z=f(x,y)

(a,b)

z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)

Why must

f

be differentiable (not just have partials) for a tangent plane?

Partials only test axis directions; differentiability requires the error

f-L

to vanish faster than the distance to

(a,b)

, ensuring the plane fits in every direction. :::

What is the normal vector to the surface

z=f(x,y)

\mathbf{n} = \langle f_x, f_y, -1 \rangle

What is the linear approximation

L(x,y)

L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)

What is the differential

dz

and what does it estimate?

dz = f_x\,dx + f_y\,dy

; it estimates the true change

\Delta z = f(x,y)-f(a,b)

for small steps.

Sufficient condition for differentiability at a point?

f_x

and

f_y

exist and are continuous near the point.

How accurate is the linear approximation as you move away?

Error grows roughly like the square of the distance from

(a,b)

; only reliable nearby.

Estimate area error of

A=xy

with

dx,dy

errors.

dA = y\,dx + x\,dy

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek pahaadi surface hai, $z=f(x,y)$ . Kisi ek point pe agar tum bahut zoom-in karo, toh surface flat dikhne lagta hai — bilkul ek samtal sticker ki tarah. Yahi flat sticker hota hai tangent plane. Iska kaam hai: us point ke aas-paas curvy surface ki jagah ek seedhi plane use karke height ka andaaza lagana. Yeh single-variable wale $y \approx f(a)+f'(a)(x-a)$ ka hi 2D version hai.

Formula banta kaise hai? Plane ko likho $z = z_0 + A(x-a) + B(y-b)$ . Point pe height match karo toh $z_0 = f(a,b)$ . $x$ -direction ki slope match karo toh $A = f_x(a,b)$ , aur $y$ -direction se $B = f_y(a,b)$ . Bas, mil gaya: $z = f(a,b) + f_x(x-a) + f_y(y-b)$ . Yaad rakhne ka tareeka — height + slope × step, do baar.

Ek important baat: sirf partial derivatives exist kar jaayein, isse plane guarantee nahi hota. Surface ko har direction mein smooth hona chahiye — isko differentiability kehte hain, aur agar $f_x, f_y$ continuous hain toh kaam ho jaata hai. Yeh exam mein favourite trap hai.

Practical use: chote-chote changes (errors) estimate karne ke liye differential $dz = f_x\,dx + f_y\,dy$ lagao. Jaise rectangle area $A=xy$ mein measurement error nikaalna. Bas dhyaan rakho — yeh approximation sirf point ke paas accurate hai; door jaake error tezi se badhta hai (distance ke square ke barabar).

Tangent planes and linear approximations to surfaces

WHAT are we approximating?

HOW to derive it (from first principles)

WHY does this work? (Differentiability)

Gradient / normal-vector view

Worked Examples

Common Mistakes

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections