I am not bossy, I am blessed with an administrative gift.
The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the input vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
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: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x^{2} + y^{2}, e^{x+y}, etc.
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 is the plane that best fits the surface (Figure i). We can write the equation of the surface as a vector function: $\vec{r}(x, y) = ⟨x, y, f(x, y)⟩$
Let x = x_{0}, the curve C_{1} on S has a tangent line with a slope given by the following 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. The same will hold true here. If we differentiate with respect to y we will get a tangent vector to traces for the plane x = x_{0}. Hence, the tangent (or the direction) vector for traces with fixed x: $\vec{v_1} = ⟨0, 1, f_y⟩$.
Similarly, let y = y_{0}, the curve C_{2} on S has a tangent line with a slope given by the following f_{x} = $\frac{\partial z}{\partial x}$, the tangent (or the direction) vector for traces with fixed y: $\vec{v_2} = ⟨1, 0, f_x⟩$.
The normal vector to our plane is $\vec{n} = \vec{v_1}x\vec{v_2} = (\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ 0 & 1 & f_y\\ 1 & 0 & f_x\end{smallmatrix}) = ⟨f_x, f_y, -1⟩$.
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} (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 $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)$,
To find the tangent plane, we first need the partial derivatives of z with respect to x and y: $f_x = cos(x+y), f_y = cos(x+y).$
The normal vector $\hat{\mathbf{n}}$ to the tangent plane at any point on the surface z = f(x, y) is given by:
$\hat{\mathbf{n}}=⟨f_x, f_y, -1⟩ = ⟨cos(x+y), cos(x+y), -1⟩$
$\hat{\mathbf{n}}\bigg|_{⟨1, -1, 0⟩} = ⟨cos(1-1), cos(1-1), -1⟩ = ⟨1, 1, -1⟩$.
The equation of the tangent plane to the surface is given by $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)$ ↭ z -0 = 1·(x -1) + 1(y -1)↭[Simplify the equation] x + y -z = 0.
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 $\vec{n} = ⟨a, b, c⟩ = ⟨1, 1, -1⟩$ and P(1, -1, 0), hence 1·(x -1) + 1·(y + 1) -1·(z- 0) = 0⇒ the equation is x + y - z = 0.
z_{0} = f(-1, 3) = ln(2·-1 +3) = ln(1) = 0.
To find the tangent plane, we first need the partial derivatives of z with respect to x and y: $f_x = \frac{2}{2x+y}, f_y = \frac{1}{2x +y}.$
The normal vector $\hat{\mathbf{n}}$ to the tangent plane at any point on the surface z = f(x, y) is given by:
$\hat{\mathbf{n}}=⟨f_x, f_y, -1⟩ = ⟨\frac{2}{2x+y}, \frac{1}{2x +y}, -1⟩$
$\hat{\mathbf{n}}\bigg|_{⟨-1, 3, 0⟩} = ⟨\frac{2}{2·-1+3}, \frac{1}{2·-1+3}, -1⟩ = ⟨2, 1, -1⟩$.
The equation of the tangent plane to the surface is given by $z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0)$ ↭ z -0 = 2·(x +1) + 1(y -3)↭[Simplify the equation] 2x + y -1 = z.
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 $\vec{n} = ⟨a, b, c⟩ = ⟨2, 1, -1⟩$ and P(-1, 3, 0), hence 2·(x +1) + 1·(y - 3) + -1·(z- 0) = 0 ⇒ 2x +2 + y -3 -z = 0⇒ the equation is 2x + y - 1 = z.
The tangent plane to the surface defined by the function f(x,y) at the point (x_{0}, y_{0}) is given by: $z ≈ f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0)$
This can be rewritten as:
$Δz ≈ \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) =f_xΔx+f_yΔy$ (Figure B) where $Δz = z - f(x_0, y_0), Δx = x -x_0, Δy = y -y_0.$
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:
L_{2} = $ \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}).
It helps approximate the value of a function near a specific point using partial derivatives.
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) ≈ $f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0)$. Here, z_{0} = f(x_{0}, y_{0}) represents the value of the function at the point (x_{0}, y_{9})
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.
Find the value at the point: f(1, 1) = 1^{2} + 1^{2} = 2.
Compute the partial derivatives: $\frac{\partial f}{\partial x}(x_0, y_0) = 2x\bigg|_{(1, 1)} = 2$.
Similarly, $\frac{\partial f}{\partial y}(x_0, y_0) = 2y\bigg|_{(1, 1)} = 2$
Use the linear approximation formula: f(x, y) ≈ $f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) = 2 + 2(x -1) + 2(y -1) = 2x + 2y -2$.
So, So, near the point (1,1), the function f(x,y) ≈ 2x +2y −2 gives us an approximate value of f.
Find the value at the point: f(-4, 3) = $3 + \frac{(-4)^2}{16} + \frac{3^2}{9} = 3 + 1 + 1 = 5.$
Compute the partial derivatives: $\frac{\partial f}{\partial x}(x_0, y_0) = \frac{2x}{16} = \frac{x}{8}\bigg|_{(-4, 3)} = \frac{-1}{2}$.
Similarly, $\frac{\partial f}{\partial y}(x_0, y_0) = \frac{2y}{9}\bigg|_{(-4, 3)} = \frac{2}{3}$
Use the linear approximation formula: f(x, y) ≈ $f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) = 5 + \frac{-1}{2}(x +4) + \frac{2}{3}(y -3)$.
So, near the point (-4, 3), the function f(x,y) ≈ $5 + \frac{-1}{2}(x +4) + \frac{2}{3}(y -3)$ gives us an approximate value of f.
Find the value at the point: f(1, 2) = $1^2 -2·1·2 + 3·2 = 3.$
Compute the partial derivatives: $\frac{\partial f}{\partial x}(x_0, y_0) = 2x -2y\bigg|_{(1, 2)} = -2$.
Similarly, $\frac{\partial f}{\partial y}(x_0, y_0) = -2x+3\bigg|_{(1, 2)} = 1$
Use the linear approximation formula: f(x, y) ≈ $f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x -x_0)+ \frac{\partial f}{\partial y}(x_0, y_0)(y -y_0) = 3 -2(x -1) + 1·(y -2)$.
Approximate f(1.03, 1.99): f(1.03, 1.99) ≈ 3 -2(1.03-1) + (1.99-2) = 3-2·0.03 -0.01 = 3 -0.06 -0.01 = 2.93