Directional derivatives

It ain’t about how hard you can hit. Its about how hard you can get hit, and how much you can take, and keep moving forward, Rocky Balboa.

And yet despite the look on my face, you’re still talking and thinking that I care, Anonymous.

Recall

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

• The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a₁+b₁), (a₂+b₂), (a₃+b₃) >.

• The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a₁-b₁), (a₂-b₂), (a₃-b₃) >.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction. It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = < ca_1, ca_2, ca_3>$.

• The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$

  It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$

• 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, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$

• Matrices offers an efficient way for solving systems of linear equations. Utilizing matrices, we can represent these equations more concisely and conveniently.

  For a system like,

  $(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}) (\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}) = (\begin{smallmatrix}u_1\\ u_2\\u_3\end{smallmatrix})$, A · X = U provides a more convenient and concise notation and an efficient way to solve system of linear equations where A is a 3 x 3 matrix representing coefficients, X is a 3 x 1 column vector representing the results where we basically do dot products between the rows of A (a 3 x 3 matrix) and the column vector of X (a 3 x 1 matrix).

• Given a square matrix A (it is a matrix that has an equal number of rows and columns), an inverse matrix A^-1 exists if and only if A is non-singular, meaning its determinant is non-zero (det(A)≠ 0).

  The inverse matrix of A, denoted as A^-1, has the property that when multiplied by A, it results in the identity matrix, i.e., $A \times A^{-1} = A^{-1} \times A = I$. Essentially, multiplying a matrix by its inverse reverses (“undoes”) the effect of the original matrix.

• Consider a system of linear equations represented in matrix form as: AX = B, where A is a n×n matrix (coefficient matrix), X is an n×1 matrix (column vector of variables), and B is an n×1 matrix (column vector of constants).

  Solving a system of linear equations expressed as AX = B involves finding or isolating the matrix X when we are given both matrices A and B. AX = B ⇒[A should be non-singular, meaning its determinant is non-zero, det(A)≠0, then we can multiply both side by A^-1]⇒ X = A^-1B.. $A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)$. adj(A) = C^T where $C_{ij} = (-1)^{i+j} \times \text{minor}(A_{ij})$ and det(A) =[A is a 3x3 matrix] a(ei - fh) - b(di - fg) + c(dh - eg)

• 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) = x² + y², e^x+y, etc.

• Partial derivatives are derivatives of a function of multiple variables, say f(x₁, x₂, ···, x_n) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x₀, y₀) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y₀. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x₀

• The Chain Rule. Let w = f(x, y, z) where x = x(t), y = y(t), and z = z(t) are functions of another variable t, then the derivate of f with respect to t is given by $\frac{dw}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}, \frac{dw}{dt} = w_x\frac{dx}{dt} +w_y\frac{dy}{dt}+w_z\frac{dz}{dt} = ∇w·\frac{d\vec{r}}{dt}~$ where ∇w = ⟨w_x, w_y, w_z⟩ and $\frac{d\vec{r}}{dt}=⟨\frac{d\vec{x}}{dt}, \frac{d\vec{y}}{dt}, \frac{d\vec{z}}{dt}⟩$

The gradient vector is defined as follows: ∇w = ⟨w_x, w_y, w_z⟩ = $⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩$. It is a vector that points in the direction of the steepest increase or ascent of a function at a given point.

Theorem. The gradient vector is perpendicular to the level surfaces of the function.

Directional derivatives

A directional derivative represents how a function w = w(x, y) changes as you move in a specific direction from a point in its domain.

In single-variable calculus, the derivative measures the rate of change of a function with respect to one variable. For a function of two variables, $\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}$ represent the derivate of f in the directions of the x-axis and y-axis, respectively.

However, in many situations, we are interested in how the function changes in an arbitrary direction, not just along the coordinate axes.

Given a function f: ℝⁿ → ℝ and a point P (x₀, y₀) in its domain, the directional derivate of f at P in the direction of a unit vector $\vec{u}$ = ⟨a, b⟩ (a vector in ℝⁿ that has a length of 1) can be found as follows.

Steps to Compute the Directional Derivative

1. Suppose we want to move linearly in the direction of $\vec{u}$. If $\vec{u}$ = ⟨a, b⟩, then we can describe the movement in the direction of $\vec{u}$ using the position vector $\vec{r}(s) = ⟨x_0 +sa, y_0 + sb⟩$, where s is a parameter representing how far we move in that direction.
2. Gradient of the function. The gradient of w, denoted by ∇w, is a vector that contains all the partial derivatives of w. For a function w(x, y), it is given by: $∇w = (\frac{∂w}{∂x}, \frac{∂w}{∂y})$.
3. Directional Derivative Formula. The directional derivative in the direction of $\vec{u}$ is given by the dot product of the gradient ∇w and the unit vector $\vec{u}$: $\frac{dw}{ds}\bigg|_{\vec{u}}$ = ∇w·$\vec{u}$, that is, the component of gradient ∇w in the direction of $\vec{u}$.

   $∇w·\hat{\mathbf{i}} = f_x$ represents the rate of change of the function f(x, y) as we vary x and hold y fixed. $∇w·\hat{\mathbf{j}} = f_y$ represents the rate of change of the function f(x, y) as we vary y and hold x fixed. This enables us to calculate the directional derivative in an arbitrary direction, by taking the dot product of ∇f with a unit vector, $\vec{u}$, in the desired direction.

In other words, we are asking what is going on if we move linearly in the direction of $\vec{u}$ (Figures A and B), say a position vector $\vec{r(s)}$ where $\frac{d\vec{r}}{ds}=\vec{u}$, that is, $\frac{dw}{ds}\bigg|_{\vec{u}}$. This is the directional derivative in the direction of ${\vec{u}}$, the rate of change of the function w as we move in the direction of $\vec{u}$, the slope of a slice (blue) of the graph by a vertical plane parallel to ${\vec{u}}$

Geometrically, $\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.

• The directional derivative is maximized when cos(θ) = 1 ↭ θ = 0, that means, $\vec{u}$ is in the same direction as the gradient ∇w, $\vec{u}=dir(∇w).$

The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.

• Conversely, the directional derivative $\frac{dw}{ds}\bigg|_{\vec{u}}$ is minimized when cos(θ) = -1 ↭ θ = 180, that is, $\vec{u}$ is in the opposite direction of the gradient ∇w.

Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$,the function w does not change. In other words, we are moving along a surface level or contour line of w ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w, that is, the direction of movement ($\vec{u}$) is perpendicular to the direction of maximum change (∇w).

Formal Definition

Definition. Suppose z = f(x, y) is a function of two variables with a domain D. Let (a, b) ∈ D and define the unit vector $\vec{u} = cos(θ)\hat{\mathbf{i}}+sin(θ)\hat{\mathbf{j}}$. The directional derivate of f in the direction of $\vec{u}$ is given by $D_{\vec{u}}f(a, b) = \lim_{h \to 0} \frac{f(a + hcos(θ), b + hsin(θ)-f(a, b))}{h}$ provided the limit exists. Source: Calculus (OpenStack) by Gilbert Strang & Edwin “Jed” Herman.

A similar definition is given by YouTube’s Dr. Trefor Bazett, Calculus III: Multivariable Calculus.

Alternative Definition. Suppose z = f(x, y) is a function of two variables with a domain D and define the unit vector $\vec{u} = ⟨u_1, u_2⟩, where ||\vec{u}|| = 1$. The directional derivate of f in the direction of $\vec{u}$ is given by or defined as $D_{\vec{u}}f(x_0, y_0) = \lim_{s \to 0} \frac{f(x_0 + su_1, y_0 + su_2)-f(x_0, y_0)}{s}$ provided the limit exists.

To compute the directional derivative using the chain rule, consider:

$x(s) = x_0 + su_1, y(s)= y_0+su_2.$

Thus, $D_{\vec{u}}f(x_0, y_0) = \lim_{s \to 0} \frac{f(x_0 + su_1, y_0 + su_2)-f(x_0, y_0)}{s} = \frac{d}{ds}[f(x(s),y(s))]\bigg|_{s=0}$

Applying the Multi-variable Chain Rule: $D_{\vec{u}}f(x_0, y_0) = (\left.\frac{∂f}{∂x}\right|_{(x_0, y_0)}\frac{dx}{ds})+$

$(\frac{∂f}{∂y}\Bigg|_{(x_0, y_0)}\frac{dy}{ds}) = $ [Since x(s)=x₀+su₁]

$(\frac{∂f}{∂x}\bigg|_{(x_0, y_0)}u_1)$ + [Since y(s)=y₀+su₂]

$(\frac{∂f}{∂y}\bigg|_{(x_0, y_0)}u_2)$. Let ∇f = $⟨\frac{∂f}{∂x},\frac{∂f}{∂y}⟩$ be the gradient of f, then we can conclude that:

$D_{\vec{u}}f(x_0, y_0) = ∇f\bigg|_{(x_0,y_0)}·\vec{u}$.

Theorem. Let z = f(x,y) be a function of two variables x and y, and assume that the partial derivatives f_x and f_y exists. Then, the directional derivative of f in the direction of $\vec{u} = cos(θ)\hat{\mathbf{i}}+sin(θ)\hat{\mathbf{j}}$ is given by $D_{\vec{u}}f(a, b)=f_x(x, y)cos(θ) + f_y(x, y)sin(θ)$.

Source: Calculus (OpenStack) by Gilbert Strang & Edwin “Jed” Herman.

Proof.

By definition, the directional derivate of f in the direction of $\vec{u} = cos(θ)\hat{\mathbf{i}}+sin(θ)\hat{\mathbf{j}}$ at a point (x₀, y₀) is given by: $D_{\vec{u}}f(x_0, y_0) = \lim_{t \to 0} \frac{f(x_0 + tcos(θ), y_0 + tsin(θ)-f(x_0, y_0))}{t}$ 🚀

Parameterization: Let x = x₀ + tcos(θ), y = y₀ + tsin(θ). Thus, $\frac{dx}{dt} = cos(θ), \frac{dy}{dt} = sin(θ)$ (*).

Define an intermediate function: Define a new function g(t) = f(x(t), y(t)) where x(t) = x₀ + tcos(θ), y(t) = y₀ + tsin(θ).

Apply the Chain Rule: Since, by assumption, we assume that f_x and f_y exists, we can use the chain rule to calculate g’(t):

$g’(t)= \frac{∂f}{∂x}\frac{dx}{dt} + \frac{∂f}{∂y}\frac{dy}{dt} =[*] f_x(x, y)cos(θ)+f_y(x, y)sin(θ)$

Evaluate at t = 0: At t = 0, we have x = x₀, y = y₀, and g’(0) = $f_x(x_0, y_0)cos(θ)+f_y(x_0, y_0)sin(θ)$ 🚀

By definition of the derivative g’(0): g’(0) = $\lim_{t \to 0} \frac{g(t)-g(0)}{t} = \lim_{t \to 0} \frac{f(x_0+tcos(θ), y_0+tsin(θ))-f(x_0, y_0)}{t}$ 🚀

Combining results, the directional derivative $D_{\vec{u}}f(x_0, y_0) = \lim_{t \to 0} \frac{f(x_0 + tcos(θ), y_0 + tsin(θ)-f(x_0, y_0))}{t} = f_x(x_0, y_0)cos(θ)+f_y(x_0, y_0)sin(θ)$.

Since the point (x₀, y₀) is an arbitrary point within the domain of f where the partials f_x and f_y exist, the result holds true for all points in the domain of f.∎

Solved exercises

• The directional derivative of a function w = w(x,y,z) in the direction of the unit vector $\hat{\mathbf{i}}$ is given by: $\frac{dw}{ds}\bigg|_{\vec{i}} = ∇w·\vec{i} = ⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩·\vec{i} = \frac{\partial w}{\partial x}$.

  This tells us that the rate of change of the function w in the direction of $\hat{\mathbf{i}}$ (the x-direction) is given by the partial derivative with respect to x.

• Directional Derivative of $f(x, y) = 2-x^2-y^2 ~\text{at}~ (\frac{1}{2}, \frac{-1}{2}), ~\text{in the direction of}~ \vec{u} = ⟨\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}⟩$

First, find the gradient ∇f: $∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨-2x, -2y⟩$.

At $(\frac{1}{2}, \frac{-1}{2}), D_{\vec{u}}f(x_0, y_0) = ⟨-2x, -2y⟩\bigg|_{(\frac{1}{2}, \frac{-1}{2})}·⟨\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}⟩ = $

$⟨-1, 1⟩·⟨\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}⟩ = -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} = 0$.

This result 0 indicates that at the point $(\frac{1}{2}, \frac{-1}{2})$, the function does not change as we move in the direction of $\vec{u}$. this implies there is an horizontal tangent line in this direction.

• Directional Derivative $\frac{dw}{ds}\bigg|_{\vec{u}}$ of w = f(x, y, z) = x²z + y³z² -xyz, in the direction of $\vec{v} = ⟨-1, 0, 3⟩$

First, we normalize $\vec{v}$ to get a unit vector $\vec{u}.$ $||\vec{v}|| = \sqrt{(-1)^2+0^2+3^2} = \sqrt{10} ≠ 1$

$\vec{u} = \frac{1}{\sqrt{10}}⟨-1, 0, 3⟩$

Now, find the gradient ∇f: $∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}⟩ = ⟨2xz-yz, 3y^2z^2-xz, x^2+2y^3z -xy⟩$.

The directional derivative is: $\frac{dw}{ds}\bigg|_{\frac{1}{\sqrt{10}}⟨-1, 0, 3⟩⟩} = ∇w·\frac{1}{\sqrt{10}}⟨-1, 0, 3⟩ = ⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩·\frac{1}{\sqrt{10}}⟨-1, 0, 3⟩ = \frac{1}{\sqrt{10}}⟨2xz-yz, 3y^2z^2-xz, x^2+2y^3z -xy⟩·⟨-1, 0, 3⟩ = \frac{1}{\sqrt{10}}(-2xz +yz + 3x^2+6y^3z-3xy)$. This expression gives the rate of change of w in the direction of $\vec{u}$.

• Calculate the directional derivative $D_{\vec{u}}f(x, y)$ of the function f(x, y) = 3x²y -4xy³ +3y²-4x at the point (3,4) in the direction of $\vec{u} = cos(\frac{π}{3})\hat{\mathbf{i}}+sin(\frac{π}{3})\hat{\mathbf{j}}$

The directional derivative of f at a point (x, y) in the direction of a unit vector $\vec{u} = cos(θ)\hat{\mathbf{i}}+sin(θ)\hat{\mathbf{j}}$ is given by $D_{\vec{u}}f(a, b)=f_x(x, y)cos(θ) + f_y(x, y)sin(θ)$. Here, f_x and f_y are the partial derivatives of f with respect to x and y, respectively.

Wwe need to find the partial derivatives f_x and f_y: $f_x = 6xy-4y^3-4, f_y = 3x^2-12xy^2+6y.$

$D_{\vec{u}}f(3, 4)= (6xy-4y^3-4)cos(\frac{π}{3})+(3x^2-12xy^2+6y)sin(\frac{π}{3}) = \frac{1}{2}·(6xy-4y^3-4) + \frac{\sqrt{3}}{2}(3x^2-12xy^2+6y)=[\text{Substituting the values}]\frac{1}{2}·(6·3·4-4·4^3-4) + \frac{\sqrt{3}}{2}(3·3^2-12·3·4^2+6·4) = \frac{1}{2}·(72-256-4)+ \frac{\sqrt{3}}{2}(27-576+24)=-94-\frac{525\sqrt{3}}{2}$.

• Calculate the directional derivative $D_{\vec{u}}f(x, y)$ of the function f(x, y) = xe^xyy + y at the point (2, 0) and $\vec{u}$ is the unit vector in the direction of θ = $\frac{2π}{3}$

The unit vector $\vec{u}$ given the direction of θ = $\frac{2π}{3}$ is $\vec{u} = ⟨cos(\frac{2π}{3}), sin(\frac{2π}{3})⟩ = ⟨\frac{-1}{2}, \frac{\sqrt{3}}{2}⟩$

The directional derivative of f at a point (x, y) in the direction of a unit vector $\vec{u} = cos(θ)\hat{\mathbf{i}}+sin(θ)\hat{\mathbf{j}}$ is given by $D_{\vec{u}}f(a, b)=f_x(x, y)cos(θ) + f_y(x, y)sin(θ)$. Here, f_x and f_y are the partial derivatives of f with respect to x and y, respectively.

Wwe need to find the partial derivatives f_x and f_y: $f_x = e^{xy} +xye^{xy}, f_y = x^2e^{xy}+1.$

Evaluate the Partial Derivatives at the Point (2, 0) and calculate the directional derivative:

$D_{\vec{u}}f(2, 0)= (e^{xy} +xye^{xy})cos(\frac{2π}{3})+(4+1)sin(\frac{2π}{3}) = \frac{-1}{2}·(1+0) + 5\frac{\sqrt{3}}{2}⟩=[\text{Substituting the values}]\frac{-1}{2}= \frac{5\sqrt{3}-1}{2}$.

• Calculate the directional derivative $D_{\vec{u}}f(x, y)$ in the direction of the unit vector $\vec{u} = ⟨a, b⟩.$

To find the directional derivative, we start by defining a new function g(z) that captures how f(x,y) changes as we move in the direction of $\vec{u}$: g(z) = f(x₀ + az, y₀ + bz). Here (x₀, y₀) is the point from which we start moving, and $\vec{u} = ⟨a, b⟩$ is the direction of movement.

Next, we need to find the derivative of g(z): $g’(z) = \lim_{h \to 0}\frac{g(z+h)-g(z)}{h}, g’(0) = \lim_{h \to 0}\frac{g(h)-g(0)}{h} = \lim_{h \to 0}\frac{f(x_0 + ah, y_0 + bh)-f(x_0, y_0)}{h} = D_{\vec{u}}f(x_0, y_0)$. This limit gives the directional derivative $D_{\vec{u}}f(x_0, y_0)$.

Let’s rewrite g(z) as follows g(z) = f(x, y) where x = x₀ + az, and y = y₀ + bz. g’(z) = [Differentiating g(z) with respect to z using the chain rule, we get:] $\frac{dg}{dz} = \frac{∂f}{∂x}\frac{dx}{dz} + \frac{∂f}{∂y}\frac{dy}{dz}$ =[Since x = x₀ +az, y = y₀ + bz] $f_x(x, y)a + f_y(x, y)b$ ⇒[Evaluating at z = 0, which corresponds to the original point x = x₀, y = y₀] $g’(0) = f_x(x_0, y_0)a + f_y(x_0, y_0)b$.

Combining both results, we find that the directional derivative $D_{\vec{u}}f(x_0, y_0) = g’(0) = f_x(x_0, y_0)a + f_y(x_0, y_0)b$. In general, $D_{\vec{u}}f(x, y) = f_x(x, y)a + f_y(x, y)b$.

• Calculate the direction derivate $D_{\vec{u}}f(x, y)$ =f(x, y) = xcos(y) in the direction $\vec{v} = ⟨2, 1⟩.$ In which direction does the temperature increase fastest at P? What is the rate of increase.

First, We need to normalize the vector, $\vec{u} = ⟨\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}⟩$.

Next, find the partial derivatives of f: f_x(x, y) = cos(y), f_y(x, y) = -zsin(y).

Now, calculate the directional derivative $D_{\vec{u}}f(x, y) = f_x(x, y)a + f_y(x, y)b =[\text{Substitute the partial derivatives}] cos(y)\frac{2}{\sqrt{5}} - xsin(y)\frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}}(2cos(y)-xin(y)).$

• The temperature at a point (x, y, z) is given by the function T(x, y, z) = 2xy -yz where T is measured in degrees Celsius and x, y, and z are measured in meters.

1. Find the rate of change in temperature at the point P(2, -1, 2) in the direction towards the point Q(3, -3, 3).
2. Calculate in which direction the temperature increase fastest at P? What is this rate of increase?
3. Determine if there is a direction at point P where the rate of change in temperature is −5∘ C/m. If so, find the direction; if not, explain why.

First, we calculate the vector from point P to point Q: $\vec{PQ} = ⟨3−2,−3−(−1),3−2⟩ = ⟨1, -2, 1⟩$.

To find the direction vector $\vec{u}$, we need to normalize the vector, $\vec{PQ}, \vec{u} = \frac{1}{\sqrt{6}}⟨1, -2, 1⟩$.

Next, we find the gradient of the temperature function ∇T: $∇T = ⟨\frac{∂T}{∂x}, \frac{∂T}{∂y}, \frac{∂T}{∂z}⟩ = ⟨2y, 2x -z, -y⟩$.

Evaluate the Gradient at P$(2, -1, 2), ∇T(2, −1, 2)= ⟨2(−1), 2(2)−2, −(−1)⟩ = ⟨−2, 4−2, 1⟩ = ⟨−2, 2, 1⟩

$D_{\vec{u}}f(2, -1, 2) = ⟨2y, 2x -z, -y⟩\bigg|_{(2, -1, 2)}·\frac{1}{\sqrt{6}}⟨1, -2, 1⟩$

The directional derivative at P in the direction of $\vec{u}$ is given by: $D_{\vec{u}}f(2, -1, 2) = ⟨-2, 2, 1⟩·\frac{1}{\sqrt{6}}⟨1, -2, 1⟩ = \frac{1}{\sqrt{6}}(-2-4+1) = \frac{-5}{\sqrt{6}}$. So, the rate of temperature at P(2, -1, 2) in the direction towards the point Q(3, -3, 3) is $\frac{-5}{\sqrt{6}}\frac{°C}{m}$.

The temperature increases fastest when the unit vector is parallel to the gradient vector. In other words, it changes most rapidly in the direction of the gradient vector $∇T(2, -1, 2) = ⟨2y, 2x -z, -y⟩\bigg|_{(2, -1, 2)} = ⟨-2, 2, 1⟩$.

The rate of increase is the magnitude of the gradient: |∇T(2, -1, 2)| = |⟨-2, 2, 1⟩| = $\sqrt{4 + 4 + 1} = 3°C/m$. This means the maximum rate of temperature increase is 3 °C/m

This also means that the maximum rate of temperature decrease is -3°C/m, and this is the case when the unit vector is in the opposite direction of the gradient. Since the maximum possible rate of decrease is −3 °C/m, it is impossible to have a rate of change of −5 °C/m.

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