Tangent Planes And Linear Approximations

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The Cross Product, Multivariable Functions & Partial Derivatives

Mechanics like the cross product, partial derivatives, and tangent planes provide the toolkit for navigating and approximating 3D space in multivariable calculus.

• The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation for two vectors in three-dimensional space. It results in a vector that is strictly perpendicular (orthogonal) to both of input vectors and has a magnitude equal to the area of the parallelogram formed by $\vec{A}$ and $\vec{B}$.

  The cross product $\vec{A}x\vec{B}$ can be computed using the following formula, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|\vec{k}$

• A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns to each ordered pair in its domain $(x, y)$ a unique real number (often represented as a $z$-coordinate), e.g., $\text{Area} = \frac{1}{2}b\cdot h$, $f(x, y) = 2x + 3y$ or $f(x, y) = x^2 + y^2$, $f(x, y) = e^{x+y}$.

• To understand how these surfaces change, we use partial derivatives. The partial derivative of $f$ with respect to $x$ measures how the function changes as $x$ shifts, while strictly locking $y$ as a constant.
  Consider a function $f(x_1, x_2, \cdots, x_n)$ that depends on n variables. The partial derivative of f with respect to one of these variables, say $x_i$, is the derivative of f while holding the other variables constant.
  For a function f(x,y), the partial derivative with respect to x at a point $(x_0, y_0)$ is defined as: $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$
  Using a different notation, $\frac{\partial f}{\partial x}(x, y) = \lim_{h \to 0} \frac{f(x+h, y)-f(x, y)}{h}$

The Tangent Plane

The tangent plane is the flat 2D plane that best fits a 3D surface at a specific point (Figure i). Just as a tangent line approximates a curve in 2D space, a tangent plane approximates a surface locally in 3D space.

🧱 1. The surface as a vector function.
We can write the equation of the surface as a vector function: $\vec{r}(x, y) = ⟨x, y, f(x, y)⟩$

🧭 2. Tangent directions come from slicing the surface.
To understand the tangent plane, you look at two special curves on the surface:
$\star$. Slice 1. Fix $x = x_0$. This gives a curve $C_1(y) = \vec{r}(x_0, y) = ⟨x_0, y, f(x_0, y)⟩$
The curve $C_1$ on S has a tangent line with a slope given by the function $f_y = \frac{\partial z}{\partial y}$
If we have a vector function of one variable we can get a tangent vector by differentiating the vector function. Differentiate with respect to 𝑦: $ \frac{\partial \vec{r}}{\partial y} = \langle 0, 1, f_y(x_0, y) \rangle$.
This is the first tangent direction: $\vec{v_1} = \langle 0, 1, f_y(x_0, y) \rangle$. It is the tangent to the vertical slice parallel to the yz‑plane.
$\star$. Slice 2. Fix $y=y_0$. This gives a curve $C_2(x) = \vec{r}(x, y_0) = ⟨x, y_0, f(x, y_0)⟩$
The curve $C_2$ on S has a tangent line with a slope given by the function $f_x = \frac{\partial z}{\partial x}$
If we have a vector function of one variable we can get a tangent vector by differentiating the vector function. Differentiate with respect to 𝑦: $ \frac{\partial \vec{r}}{\partial x} = \langle 1, 0, f_x(x, y_0) \rangle$.
This is the second tangent direction: $\vec{v_2} = \langle 1, 0, f_x(x, y_0) \rangle$. It is the tangent to the vertical slice parallel to the xz‑plane.

🧩 3. The normal vector comes from the cross product
The normal vector to our plane is $\vec{n} = \vec{v_1}x\vec{v_2} = (\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\\ 0 & 1 & f_y\\\ 1 & 0 & f_x\end{smallmatrix}) = \langle f_x, f_y, -1 \rangle$.

📐 4. Use the normal vector to write the plane equation
The general equation of a plane is given by $a(x -x_0) +b(y -y_0) +c(z -z_0) = 0$ where $P(x_0, y_0, z_0)$ is a point in the plane.
We can rewrite it as: $z -z_0 = \frac{-a}{c}(x -x_0) -\frac{b}{c}(y -y_0) = [\text{Rename the constants}] A(x -x_0) + B(y -y_0)$.
If we assume that $y = y_0$ (i.e., fixed), the equation of the tangent plane becomes, $z -z_0 = A(x -x_0)$, and this is the equation of the tangent line to the surface at $(x_0, y_0)$ assuming that $y = y_0$ -in the x-slice- (i.e., fixed), and we know that the slope of this line is given by $f_x(x_0, y_0)$, hence $A = f_x(x_0, y_0)$. Mutatis mutandis, $B = f_y(x_0, y_0)$. The equation of the tangent plane to the surface given by z = f(x, y) at $(x_0, y_0)$ is then $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)$.
Alternatively, we plug in $\vec{n} = \langle f_x, f_y, -1 \rangle$ in $a(x -x_0) +b(y -y_0) +c(z -z_0) = 0$, and we get: $f_x(x -x_0) +f_y(y -y_0) -(z -z_0) = 0$. Solve for z: $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)$.

Solved Tangent Plane Examples

• Find the equation of the tangent plane to the surface $z = \sin(x+y)$ at $P(1, -1, 0)$

1. To find the tangent plane, we first need the partial derivatives of z with respect to x and y: $f_x = \cos(x+y)$ and $f_y = \cos(x+y)$.
2. The normal vector $\hat{\mathbf{n}}$ to the tangent plane at any point on the surface z = f(x, y) is given by: $\hat{\mathbf{n}}=⟨f_x, f_y, -1⟩ = ⟨cos(x+y), cos(x+y), -1⟩$
3. Evaluate at $P$: $\hat{\mathbf{n}}\bigg|_{⟨1, -1, 0⟩} = ⟨cos(1-1), cos(1-1), -1⟩ = ⟨1, 1, -1⟩$.
4. The equation of the tangent plane to the surface is given by $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)$
5. Simplify the equation: $x + y -z = 0.$
6. Another way to find the equation of the tangent plane is to use the general form of the plane equation with the normal vector components: $a(x −x_0) + b(y −y_0) + c(z −z_0) = 0.$
   Here $\vec{n} = \langle a, b, c \rangle = \langle 1, 1, -1 \rangle$ and $P(1, -1, 0)$, hence $1 \cdot (x -1) + 1 \cdot (y + 1) -1\cdot (z- 0) = 0.$ Therefore, the equation is $x + y -z = 0.$

• Find the equation of the tangent plane to the surface $z = \ln(2x + y)$ at $P(-1, 3)$

1. Find $z_0$: $f(-1, 3) = \ln(-2 + 3) = \ln(1) = 0$.
2. To find the tangent plane, we first need the partial derivatives of z with respect to x and y: $f_x = \frac{2}{2x+y}, f_y = \frac{1}{2x +y}.$
3. The normal vector $\hat{\mathbf{n}}$ to the tangent plane at any point on the surface z = f(x, y) is given by: $\hat{\mathbf{n}}=\langle f_x, f_y, -1\rangle⟩ = \langle\frac{2}{2x+y}, \frac{1}{2x +y}, -1\rangle$
4. Evaluate at P(-1, 3, 0): $\hat{\mathbf{n}}\bigg|_{(-1, 3, 0)} = \langle \frac{2}{2·-1+3}, \frac{1}{2·-1+3}, -1 \rangle = \langle 2, 1, -1 \rangle$.
5. The equation of the tangent plane to the surface is given by $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \implies z -0 = 2·(x +1) + 1(y -3)↭[\text{Simplify the equation}] 2x + y -1 = z.$
6. Another way to find the equation of the tangent plane is to use the general form of the plane equation with the normal vector components: $a(x −x_0) + b(y −y_0) + c(z −z_0) = 0.$
   Here $\vec{n} \langle a, b, c \rangle = \langle 2, 1, -1 \rangle$ and P(-1, 3, 0), hence $2 \cdot (x +1) + 1\cdot (y - 3) + -1 \cdot (z- 0) = 0 \implies 2x +2 + y -3 -z = 9.$ Therefore, the equation is $2x + y -1 = z.$

Summary

The tangent plane to the surface defined by the function f(x,y) at the point $(x_0, y_0)$ is given by: $\boxed{z \approx f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0)}$

This can be rewritten using increments as: $\Delta z \approx \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) =f_xΔx+f_yΔy$ (Figure B) where $\Delta z = z - f(x_0, y_0), \Delta x = x -x_0, \Delta y = y -y_0.$

This is the differential: $dz = f_xdx + f_ydy$. 🎯 Geometrically, this says that small changes in height are approximately the sum of the slopes in each direction times the small changes in those directions.

This equation states that the graph of f is close to its tangent plane in the vicinity of the point $(x_0, y_0)$. The tangent plane provides a good approximation of the surface near this point.

Let’s suppose that $\frac{\partial f}{\partial x}(x_0, y_0) = a$, then the tangent line $L_1$ in the x-direction is:

$L_1 = \begin{cases} z = f(x_0, y_0) + a(x -x_0) \\\\ y = y_0~\text{constant} \end{cases}$

If $\frac{\partial f}{\partial y}(x_0, y_0) = b$, then the tangent line $L_2$ in the y-direction is:

L₂ = $ \begin{cases} z = f(x_0, y_0) + b(y -y_0) \\\\ x = x_0~\text{constant} \end{cases}$

Together, these tangent lines determine the tangent plane given by the formula: $z = f(x_0 , y_0) + a(x -x_0) + b(y - y_0)$ where $a = \frac{\partial f}{\partial x}(x_0, y_0)$ and $b = \frac{\partial f}{\partial y}(x_0, y_0)$. This linear approximation states that the graph of f is close to its tangent plane in the intermediate vicinity of the point $(x_0, y_0)$. The tangent plane is the best linear approximation of the surface near $(x_0, y_0)$, built from the two tangent lines in the x- and y- directions.

Linear Approximation

One great use of tangent planes is that they provide a way to approximate a surface near a given point. If we are close to the point $(x_0, y_0)$, the tangent plane should closely approximate the function at that point.

The linear approximation formula is a powerful tool in multivariable calculus. It allows us to approximate the value of a function f(x, y) near a specific point $(x_0, y_0)$ using the partial derivatives at that point.

This approximation helps us understand the local behavior of the function and is very useful in various applications, such as optimization and numerical analysis.

The linear approximation formula for a function f(x,y) around the point $(x_0, y_0)$ is given by: $f(x, y) \approx f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0)$ where $z_0 = f(x_0, y_0)$ represents the value of the function at the point $(x_0, y_0)$. This is the first-order Taylor polynomial of f at $(x_0, y_0)$.

Imagine you are standing on a surface that represents the graph of the function f(x, y). If you look around, the surface might be curved, but if you focus on a very small region around your feet, that small region will appear almost flat. This flat region is the tangent plane. The linear approximation formula essentially tells you how to move from the point $(x_0, y_0)$ to a nearby point (x, y) on the surface by moving along this flat plane.

The partial derivatives of the function with respect to x and y evaluated at the point $(x_0, y_0)$ represent the rates of change of the function in the x- and y-directions at that point. In other words, $\frac{\partial f}{\partial x}(x_0, y_0)$ measures how f changes as x changes while keeping y constant, and $\frac{\partial f}{\partial y}(x_0, y_0)$ measures how f changes as y changes while keeping x constant.

Examples

• A Standard Paraboloid. Suppose we have the function $f(x, y) = x^2 + y^2$ and we want to approximate its value near the anchor point $(1, 1)$.

1. Evaluate the function at the point:: $f(1, 1) = 1^2 + 1^2 = 2$
2. Compute the partial derivatives and evaluate them at $(1, 1)$: $\frac{\partial f}{\partial x}(x_0, y_0) = 2x\bigg|_{(1, 1)} = 2$.
   Similarly, $\frac{\partial f}{\partial y}(x_0, y_0) = 2y\bigg|_{(1, 1)} = 2$
3. Use the linear approximation formula: $f(x, y) \approx f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) = 2 + 2(x -1) + 2(y -1) = 2x + 2y -2$.
4. Conclusion. So, So, near the point (1, 1), the simplified plane equation L(x, y) = 2x + 2y − 2 gives us a highly accurate approximate value of f(x, y).

• An Elliptic Paraboloid. Suppose we have a function $f(x, y) = 3 + \frac{x^2}{16} + \frac{y^2}{9}$ and we want to approximate its value near the point $(-4, 3)$.

1. Find the value at the point: $f(-4, 3) = 3 + \frac{(-4)^2}{16} + \frac{3^2}{9} = 3 + \frac{16}{16} + \frac{9}{9} = 3 + 1 + 1 = 5$
2. Compute the partial derivatives and evaluate them at $(-4, 3)$: $\frac{\partial f}{\partial x}(x_0, y_0) = \frac{2x}{16} = \frac{x}{8}\bigg|_{(-4, 3)} = \frac{-1}{2}$.
   Similarly, $\frac{\partial f}{\partial y}(x_0, y_0) = \frac{2y}{9}\bigg|_{(-4, 3)} = \frac{2}{3}$
3. Use the linear approximation formula: $f(x, y) \approx f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) = 5 + \frac{-1}{2}(x +4) + \frac{2}{3}(y -3)$.
4. Conclusion. So, near the point $(-4, 3)$, the function $f(x,y) \approx 5 + \frac{-1}{2}(x +4) + \frac{2}{3}(y -3)$ gives us an approximate value of f.

• Estimating a Complex Decimal. This example demonstrates why linear approximation is useful: calculating decimals by hand is tedious, but tangent planes make it simple arithmetic.

Suppose we have a function $f(x, y) = x^2 - 2xy + 3y$ and we want to approximate $f(1.03, 1.99)$.

1. Because $(1.03, 1.99)$ is very close to $(1, 2)$, we will use $(1, 2)$ as our anchor point $(x_0, y_0)$.
2. Evaluate the function at the clean anchor point $(1, 2)$: $f(1, 2) = 1^2 - 2(1)(2) + 3(2) = 1 - 4 + 6 = 3$
3. Compute the partial derivatives at $(1, 2)$: $\frac{\partial f}{\partial x}(x_0, y_0) = 2x -2y\bigg|_{(1, 2)} = -2$.
   Similarly, $\frac{\partial f}{\partial y}(x_0, y_0) = -2x+3\bigg|_{(1, 2)} = 1$
4. Use the linear approximation formula: $f(x, y) \approx f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) = 3 -2(x -1) + 1·(y -2)$.
5. Approximate f(1.03, 1.99): $f(1.03, 1.99) \approx 3 -2(1.03-1) + (1.99-2) = 3-2·0.03 -0.01 = 3 -0.06 -0.01 = 2.93$
6. 🌄 The Reality Check: If you plug $f(1.03, 1.99)$ directly into a calculator, the exact value is $2.9315$. Our tangent plane approximation of $2.93$ was incredibly close, completely bypassing the need to calculate $1.03^2$ or multiply $1.03 \times 1.99$ by hand.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.