Multivariable Calculus: Directional Derivatives and The Gradient Study Guide

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

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 with respect to $x$ and $y$.

Key Terms and Formulas:

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

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

Step-by-Step Guidance

1. Find $\frac{\partial f}{\partial x}$ for $f(x,y) = 10x - 8y$.

2. Find $\frac{\partial f}{\partial y}$ for $f(x,y) = 10x - 8y$.

3. Write the gradient vector $\nabla f(x,y)$ using these partial derivatives.

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

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 $\frac{\partial f}{\partial x} = 10$ and $\frac{\partial f}{\partial y} = -8$. The gradient at $(8, -5)$ is $10\mathbf{i} - 8\mathbf{j}$.

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

Background

Topic: Gradient of a function in three variables.

This question tests your ability to compute the gradient vector for a function $f(x,y,z)$ at a specific point. The gradient in three variables is a vector of partial derivatives with respect to $x$, $y$, and $z$.

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. Compute $\frac{\partial f}{\partial x}$ for $f(x,y,z) = x + y - 2z + z \ln x$.

2. Compute $\frac{\partial f}{\partial y}$ for the same function.

3. Compute $\frac{\partial f}{\partial z}$ for the same function.

4. Substitute $x = 1$, $y = 5$, $z = 4$ into each partial derivative to evaluate the gradient at $(1,5,4)$.

Try solving on your own before revealing the answer!

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

We calculated each partial derivative and substituted the values. Note that $\ln 1 = 0$.

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

Background

Topic: Directional derivative in two variables.

This question tests your ability to compute the directional derivative of a function at a point in the direction of a given vector. The directional derivative measures the rate of change of the function in the specified direction.

Key Terms and Formulas:

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

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

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

Step-by-Step Guidance

1. Find the gradient $\nabla f$ by computing the partial derivatives of $f(x,y)$.

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

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

4. Set up the dot product $\nabla f \cdot \mathbf{u}$ to compute the directional derivative.

Try solving on your own before revealing the answer!

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

The gradient at $(-6, -2)$ is $-4\mathbf{i} + 8\mathbf{j}$, and the unit vector is $\frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}$. The dot product gives $\frac{112}{5}$.

Q4. Find the derivative of the function $f(x,y,z) = xy + yz + zx$ 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 $f(x,y,z)$ at a point in the direction of a given vector. The process involves finding the gradient and taking the dot product with a unit vector in the 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 of $f(x,y,z)$ to find the gradient.

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

3. Find the unit vector in the direction of $\mathbf{A}$.

4. Set up the dot product $\nabla f \cdot \mathbf{u}$ for the directional derivative.

Try solving on your own before revealing the answer!

Final Answer: $D_{\mathbf{u}} f = \frac{-21}{11}$

The gradient at $(-3,3,2)$ is $(5, -1, 0)$, and the unit vector is $\frac{2}{11}\mathbf{i} + \frac{6}{11}\mathbf{j} - \frac{9}{11}\mathbf{k}$. The dot product gives $\frac{-21}{11}$.

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 in these directions.

Background

Topic: Gradient and directional derivatives.

This question tests your understanding of how the gradient vector determines the direction of greatest increase and decrease for a function, 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 of $f(x,y)$ to find the gradient.

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

3. Normalize the gradient to get the unit vector $\mathbf{u}$ in the direction of greatest increase.

4. The direction of greatest decrease is $-\mathbf{u}$.

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: $\mathbf{u} = \frac{1}{\sqrt{17}}\mathbf{i} + \frac{4}{\sqrt{17}}\mathbf{j}$

Direction of greatest decrease: $\mathbf{v} = -\frac{1}{\sqrt{17}}\mathbf{i} - \frac{4}{\sqrt{17}}\mathbf{j}$

Derivative in direction of increase: $D_{\mathbf{u}} f = \sqrt{17}$

Derivative in direction of decrease: $D_{\mathbf{v}} f = -\sqrt{17}$

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: $D_{\mathbf{u}} f$

• Partial derivatives: $f_x$, $f_y$, $f_z$

• Unit vectors: $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$

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 the definition of the directional derivative: $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$.

4. For $\mathbf{u} = \mathbf{i}$, $\mathbf{j}$, or $\mathbf{k}$, relate the dot product to the partial derivatives.

Try solving on your own before revealing the answer!

Final Answer:

$D_{\mathbf{i}} f = f_x$, $D_{\mathbf{j}} f = f_y$, $D_{\mathbf{k}} f = f_z$.

This is because the directional derivative in the direction of a coordinate axis is simply the partial derivative with respect to that variable.

Q7. Find the gradient of the function $f(x,y) = 4y - 10x$ 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 interpretation with respect 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. Identify 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: $\nabla f(1,9) = -10\mathbf{i} + 4\mathbf{j}$

The gradient is perpendicular to the level curve at $(1,9)$. The level curve is $4y - 10x = 26$.

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

Background

Topic: Tangent lines and gradients.

This question tests your ability to find the tangent line to a curve defined implicitly, using the gradient and the point of tangency.

Key Terms and Formulas:

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

• Tangent line: line passing through the point with slope determined by the gradient

Step-by-Step Guidance

1. Compute the gradient of $f(x,y) = x^2 + y^2$ at $(1,3)$.

2. Find the direction vector for the tangent line to $x + y = 4$ at $(1,3)$.

3. Set up the equation for the tangent line using the point and direction.

4. Write the equation in standard form.

Try solving on your own before revealing the answer!

Final Answer: Equation for the tangent line is $x - y = -2$

The tangent line at $(1,3)$ to $x + y = 4$ is $x - y = -2$.

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 dimensions.

This question tests your ability to find the normal line to a surface at a given point using the gradient, and to write its parametric equation.

Key Terms and Formulas:

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

• Parametric equation: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{n}$

Step-by-Step Guidance

1. Compute the gradient of $F(x,y,z) = 3x^2 + 4y^2 + z^2$ at $(1,3,2)$.

2. Use the gradient as the direction vector for the normal line.

3. Write the parametric equation for the line using the point and the direction 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 gradient at $(1,3,2)$ is $(6,24,4)$, so the normal line is $\mathbf{r}(t) = (1 + 6t)\mathbf{i} + (3 + 24t)\mathbf{j} + (2 + 4t)\mathbf{k}$.