good: Multivariable Calculus: Directional Derivatives and The Gradient Study Guide

Q1. Compute the gradient of the function $f(x,y)=10x^2-8y$ at the point $(8, -5)$.

Background

Topic: Gradient of a function in two variables.

This question tests your ability to find the gradient vector of a function $f(x,y)$ at a specific point. The gradient points in the direction of greatest increase and is composed of partial derivatives.

Key Terms and Formulas:

• Gradient: $\nabla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

• Partial derivatives: $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

Step-by-Step Guidance

1. Compute $\frac{\partial f}{\partial x}$ for $f(x,y)=10x^2-8y$. What is the derivative with respect to $x$?

2. Compute $\frac{\partial f}{\partial y}$ for $f(x,y)=10x^2-8y$. What is the derivative with respect to $y$?

3. Write the gradient vector $\nabla f(x,y)$ using your results from steps 1 and 2.

4. Evaluate the gradient at the point $(8, -5)$ by substituting $x = 8$ and $y = -5$ into your gradient expression.

Try solving on your own before revealing the answer!

Final Answer: $\nabla f(8, -5) = 10\mathbf{i} - 8\mathbf{j}$

The partial derivatives are constant, so the gradient is $10\mathbf{i} - 8\mathbf{j}$ at any point, including $(8, -5)$.

Q2. Find the gradient of $f(x,y,z) = x + y - 2z + z \ln x$ at the point $(1,5,4)$.

Background

Topic: Gradient in three variables.

This question tests your ability to compute the gradient vector for a function of three variables and evaluate it at a specific point.

Key Terms and Formulas:

• Gradient: $\nabla f(x,y,z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$

• Partial derivatives: $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial z}$

Step-by-Step Guidance

1. Find $\frac{\partial f}{\partial x}$: Differentiate $f(x,y,z)$ with respect to $x$. Remember $z \ln x$ is a function of $x$.

2. Find $\frac{\partial f}{\partial y}$: Differentiate $f(x,y,z)$ with respect to $y$.

3. Find $\frac{\partial f}{\partial z}$: Differentiate $f(x,y,z)$ with respect to $z$.

4. Write the gradient vector $\nabla f(x,y,z)$ using your results from steps 1–3.

5. Substitute $x = 1$, $y = 5$, $z = 4$ into your gradient expression to evaluate at the given point.

Try solving on your own before revealing the answer!

Final Answer: $\nabla f(1,5,4) = (1 + 4 \cdot \frac{1}{1})\mathbf{i} + 1\mathbf{j} + (-2 + \ln 1)\mathbf{k} = 5\mathbf{i} + 1\mathbf{j} -2\mathbf{k}$

We computed each partial derivative and substituted the values to get the gradient vector at $(1,5,4)$.

Q3. Find the derivative of $f(x,y) = -4x + 8y$ at the point $(-6, -2)$ in the direction of $\mathbf{A} = 3\mathbf{i} - 4\mathbf{j}$.

Background

Topic: Directional derivative in two variables.

This question tests your ability to compute the directional derivative of a function at a point in a specified direction.

Key Terms and Formulas:

• Directional derivative: $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$

• Unit vector: $\mathbf{u} = \frac{\mathbf{A}}{\|\mathbf{A}\|}$

• Gradient: $\nabla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

Step-by-Step Guidance

1. Compute the gradient $\nabla f(x,y)$ for $f(x,y) = -4x + 8y$.

2. Evaluate the gradient at $(-6, -2)$.

3. Find the unit vector in the direction of $\mathbf{A} = 3\mathbf{i} - 4\mathbf{j}$ by dividing by its magnitude.

4. Set up the dot product $\nabla f(-6, -2) \cdot \mathbf{u}$ for the directional derivative.

Try solving on your own before revealing the answer!

Final Answer: $D_{\mathbf{u}} f = \frac{112}{5}$

The directional derivative is computed by taking the dot product of the gradient and the unit vector in the direction of $\mathbf{A}$.

Q4. Find the derivative of $f(x,y,z) = xy + yz + zx$ at the point $(-3,3,2)$ in the direction of $\mathbf{A} = 2\mathbf{i} + 6\mathbf{j} - 9\mathbf{k}$.

Background

Topic: Directional derivative in three variables.

This question tests your ability to compute the directional derivative of a function of three variables at a point in a specified direction.

Key Terms and Formulas:

• Gradient: $\nabla f(x,y,z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$

• Directional derivative: $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$

• Unit vector: $\mathbf{u} = \frac{\mathbf{A}}{\|\mathbf{A}\|}$

Step-by-Step Guidance

1. Compute the partial derivatives $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial z}$ for $f(x,y,z) = xy + yz + zx$.

2. Evaluate the gradient at $(-3,3,2)$.

3. Find the unit vector in the direction of $\mathbf{A}$ by dividing by its magnitude.

4. Set up the dot product $\nabla f(-3,3,2) \cdot \mathbf{u}$ for the directional derivative.

Try solving on your own before revealing the answer!

Final Answer: $D_{\mathbf{u}} f = -\frac{21}{\sqrt{121}}$

The directional derivative is the dot product of the gradient at the point and the unit vector in the direction of $\mathbf{A}$.

Q5. Find the directions in which the function $f(x,y) = x + xy + y^2$ increases and decreases most rapidly at $P_0(-4, -2)$. Then find the derivatives of the function in these directions.

Background

Topic: Gradient and directional derivatives.

This question tests your understanding of how the gradient determines the direction of greatest increase and decrease, and how to compute the corresponding directional derivatives.

Key Terms and Formulas:

• Gradient: $\nabla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

• Direction of greatest increase: direction of $\nabla f$

• Direction of greatest decrease: direction of $-\nabla f$

• Directional derivative: $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$

Step-by-Step Guidance

1. Compute the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for $f(x,y) = x + xy + y^2$.

2. Evaluate the gradient at $P_0(-4, -2)$.

3. Find the unit vector in the direction of the gradient for the direction of greatest increase.

4. Find the unit vector in the direction of $-\nabla f$ for the direction of greatest decrease.

5. Set up the directional derivatives in these directions using the dot product.

Try solving on your own before revealing the answer!

Final Answer:

Direction of greatest increase: $\mathbf{u} = \frac{\nabla f(-4, -2)}{\|\nabla f(-4, -2)\|}$

Direction of greatest decrease: $\mathbf{v} = -\frac{\nabla f(-4, -2)}{\|\nabla f(-4, -2)\|}$

Derivative in direction of increase: $D_{\mathbf{u}} f = \|\nabla f(-4, -2)\|$

Derivative in direction of decrease: $D_{\mathbf{v}} f = -\|\nabla f(-4, -2)\|$

Q6. How are $D_i f$, $D_j f$, and $D_k f$ related to $f_x$, $f_y$, and $f_z$? Give reasons for your answer.

Background

Topic: Directional derivatives and partial derivatives.

This question tests your understanding of the relationship between directional derivatives in the coordinate directions and partial derivatives.

Key Terms and Formulas:

• Directional derivative: $D_{\mathbf{u}} f$ is the rate of change of $f$ in the direction of $\mathbf{u}$.

• Partial derivatives: $f_x$, $f_y$, $f_z$ are derivatives in the $x$, $y$, $z$ directions.

Step-by-Step Guidance

1. Recall that the directional derivative in the direction of $\mathbf{i}$ is $D_{\mathbf{i}} f$.

2. Recall that $f_x$ is the partial derivative with respect to $x$.

3. Consider what happens when the direction vector is $\mathbf{i}$, $\mathbf{j}$, or $\mathbf{k$.

4. Compare the definitions of directional derivatives and partial derivatives in these directions.

Try solving on your own before revealing the answer!

Final Answer:

$D_{\mathbf{i}} f$, $D_{\mathbf{j}} f$, and $D_{\mathbf{k}} f$ are equal to $f_x$, $f_y$, and $f_z$ respectively, because the directional derivative in the coordinate direction is the partial derivative in that direction.

Q7. Find the gradient of $f(x,y) = 4y - 10x$ at the point $(1,9)$. Then sketch the gradient and the level curve that passes through that point.

Background

Topic: Gradient and level curves.

This question tests your ability to compute the gradient and understand its geometric relationship to level curves.

Key Terms and Formulas:

• Gradient: $\nabla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

• Level curve: Set of points where $f(x,y)$ is constant.

Step-by-Step Guidance

1. Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for $f(x,y) = 4y - 10x$.

2. Write the gradient vector $\nabla f(x,y)$.

3. Evaluate the gradient at $(1,9)$.

4. Find the equation of the level curve passing through $(1,9)$ by setting $f(x,y) = f(1,9)$.

Try solving on your own before revealing the answer!

Final Answer: $\nabla f(1,9) = -10\mathbf{i} + 4\mathbf{j}$; Level curve: $4y - 10x = 26$

The gradient is perpendicular to the level curve at the point $(1,9)$.

Q8. Sketch the curve $x^2 + y^2 = 4$ together with $\nabla f$ and the tangent line at the point $(1,3)$. Then write an equation for the tangent line.

Background

Topic: Tangent lines and gradients for implicit curves.

This question tests your ability to find the tangent line to a curve defined implicitly and relate it to the gradient.

Key Terms and Formulas:

• Implicit curve: $x^2 + y^2 = 4$

• Gradient: $\nabla f(x,y) = (2x, 2y)$ for $f(x,y) = x^2 + y^2$

• Tangent line: Use the gradient to find the normal, then find the tangent direction.

Step-by-Step Guidance

1. Compute the gradient $\nabla f(x,y)$ for $f(x,y) = x^2 + y^2$.

2. Evaluate the gradient at $(1,3)$.

3. Find the equation of the tangent line at $(1,3)$ using the gradient (normal vector).

4. Set up the tangent line equation in the form $a(x - x_0) + b(y - y_0) = 0$ where $(a,b)$ is the gradient.

Try solving on your own before revealing the answer!

Final Answer: Tangent line equation: $2(x - 1) + 6(y - 3) = 0$ or $x - 1 + 3(y - 3) = 0$

The tangent line is perpendicular to the gradient at $(1,3)$.

Q9. Find a parametric equation for the line that is perpendicular to the graph of $3x^2 + 4y^2 + z^2 = 43$ at the point $(1,3,2)$.

Background

Topic: Parametric equations and gradients in three variables.

This question tests your ability to find the normal line to a surface at a given point using the gradient.

Key Terms and Formulas:

• Gradient: $\nabla f(x,y,z) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$

• Parametric line: $\mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{n}$, where $\mathbf{n}$ is the normal vector.

Step-by-Step Guidance

1. Compute the gradient $\nabla f(x,y,z)$ for $f(x,y,z) = 3x^2 + 4y^2 + z^2$.

2. Evaluate the gradient at $(1,3,2)$ to get the normal vector.

3. Write the parametric equation for the line using the point $(1,3,2)$ and the normal vector.

Try solving on your own before revealing the answer!

Final Answer: $\mathbf{r}(t) = (1 + 6t)\mathbf{i} + (3 + 24t)\mathbf{j} + (2 + 4t)\mathbf{k}$

The normal vector is $\nabla f(1,3,2) = (6,24,4)$, so the line is $\mathbf{r}(t) = (1,3,2) + t(6,24,4)$.