\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \)

You CAN Ace Calculus

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

how to read math books

The directional derivative and gradient are very closely related. So we cover both on this page. First, a couple of quick definitions.

Directional Derivative - - You know from single variable calculus that the derivative is just the slope of the tangent line. If you are given a point, you can find the slope of the specific tangent line at that specific point. Similarly, the directional derivative is the slope of a tangent line but it applies to three dimensions where you are given a point and a unit vector. This unit vector tells you which specific tangent line you are talking about at that point. In summary, a directional derivative is the slope of a tangent line in a specific direction.

Gradient - - The gradient is a vector (which is why its also called the gradient vector) calculated using the partial derivative. It contains information about the slope at every point. We can use the gradient to get the directional derivative.

Remember that in multi-variable calculus most concepts use special notation. The directional derivative and gradient are no exception.

Here is a great video explaining the directional derivative and gradient in detail and the associated notation, including some examples. It is a long video, so we separate it into video clips on this page as the discussion continues.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 1) [19mins-21secs]

video by Dr Chris Tisdell

Gradient

\(\displaystyle{ \nabla g(x,y) = \frac{\partial g}{\partial x} \hat{i} + \frac{\partial g}{ \partial y}\hat{j} }\) where \(\displaystyle{ \nabla }\) is the 'del' operator \(\displaystyle{ \nabla = \frac{\partial }{\partial x}\hat{i} + \frac{\partial }{ \partial y}\hat{j} }\)

The gradient can also be indicated by grad(g) and can be called the grad of g. Notice that we do not use vector notation on \(\nabla g(x,y)\). We could write \( \vec{ \nabla } g(x,y)\) to clearly indicate that the result is a vector and some instructors do this. However, many of them don't and we won't either. So you just need to remember that \(\nabla g(x,y)\) is a vector (or, more precisely, a vector field).

Before we go on, let's make sure we understand this by working a few practice problems.

If \( f(x,y,z) = x^3+y+z^2 \), calculate \( \nabla f \) and \( \nabla f(1,2,3) \).

Problem Statement

If \( f(x,y,z) = x^3+y+z^2 \), calculate \( \nabla f \) and \( \nabla f(1,2,3) \).

Solution

804 solution video

video by Dr Chris Tisdell

close solution

Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)

Problem Statement

Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)

Hint

This problem uses unusual variables \( \tau \) and \( \kappa \). However, they work just as if you had t and x.

Problem Statement

Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)

Hint

This problem uses unusual variables \( \tau \) and \( \kappa \). However, they work just as if you had t and x.

Solution

2293 solution video

close solution

Calculate the gradient vector field of \( f(x,y) = \ln(2x+5y) \).

Problem Statement

Calculate the gradient vector field of \( f(x,y) = \ln(2x+5y) \).

Solution

2294 solution video

video by MIP4U

close solution

Given \( f(x,y) = xy \), determine \( \nabla f(-3,2) \).

Problem Statement

Given \( f(x,y) = xy \), determine \( \nabla f(-3,2) \).

Solution

2295 solution video

video by MIP4U

close solution

Directional Derivative

\(\displaystyle{ D_{\vec{u}}f = \nabla f \cdot \vec{u} }\)
We read this as the directional derivative of f in the direction of \(\vec{u}\).

The directional derivative can be thought of as the magnitude of the gradient at a certain point in a specific direction. To calculate the directional derivative, we take the dot product of the gradient with the unit vector ( \(\vec{u}\) ) in the direction we are interested in. The equation looks like this.

Here is a video going through a proof showing that the directional derivative is the dot product of the gradient with a direction vector. This video is not required for understanding how to use the dot product version of the directional derivative but it will help you understand it. So it is a good video to watch.

Dr Chris Tisdell - Proof: Directional Derivative is a Dot Product [8mins-6secs]

video by Dr Chris Tisdell

Here is a video clip discussing the properties of the directional derivative based on the use of the dot product.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 2) [13mins-23secs]

video by Dr Chris Tisdell

From the last video clip, you now know that the direction of maximum increase of a function f is the gradient \(\nabla f\) and the direction of maximum decrease is \(-\nabla f\). ( Remember that these are vectors. )

Since the gradient is just a partial derivative, the basic rules of algebra are followed as well as the chain rule. If you would like to confirm these rules, here is a quick video clip to watch.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 3) [2mins-25secs]

video by Dr Chris Tisdell

Calculate the derivative of \( f(x,y) = xe^y + \cos(xy) \) at the point \( (2,0) \) in the direction of \( \vec{v} = 3\hat{i} - 4\hat{j} \).

Problem Statement

Calculate the derivative of \( f(x,y) = xe^y + \cos(xy) \) at the point \( (2,0) \) in the direction of \( \vec{v} = 3\hat{i} - 4\hat{j} \).

Solution

799 solution video

video by Dr Chris Tisdell

close solution

At the point \( (1,1) \), determine the directions in which \( f(x,y) = x^2/2 + y^2/2 \) (a) increases most rapidly; (b) decreases most rapidly; (c) has zero change

Problem Statement

At the point \( (1,1) \), determine the directions in which \( f(x,y) = x^2/2 + y^2/2 \) (a) increases most rapidly; (b) decreases most rapidly; (c) has zero change

Solution

800 solution video

video by Dr Chris Tisdell

close solution

Calculate the derivative of \( f(x,y,z) = x^3 - xy^2-z \) at the point \( (1,1,0) \) in the direction \( \vec{v} = 2\hat{i} - 3\hat{j} + 6\hat{k}\).

Problem Statement

Calculate the derivative of \( f(x,y,z) = x^3 - xy^2-z \) at the point \( (1,1,0) \) in the direction \( \vec{v} = 2\hat{i} - 3\hat{j} + 6\hat{k}\).

Solution

801 solution video

video by Dr Chris Tisdell

close solution

Calculate the directional derivative of \( f(x,y) = 1 - x^2/2 - y^4/4 \) in the direction \(\hat{i} + \hat{j}\) at the point \( (1,2) \).

Problem Statement

Calculate the directional derivative of \( f(x,y) = 1 - x^2/2 - y^4/4 \) in the direction \(\hat{i} + \hat{j}\) at the point \( (1,2) \).

Solution

803 solution video

video by Dr Chris Tisdell

close solution

Calculate the directional derivative of \( f(x,y) = x^2 + 2xy + 3y^2 \) at the point \( (2,1) \) in the direction \(\hat{i}+\hat{j}\).

Problem Statement

Calculate the directional derivative of \( f(x,y) = x^2 + 2xy + 3y^2 \) at the point \( (2,1) \) in the direction \(\hat{i}+\hat{j}\).

Solution

807 solution video

video by Krista King Math

close solution

Calculate the maximum directional derivative and its direction for \( f(x,y) = 2x^2 + 3xy + 4y^2 \) at \( (1,1) \).

Problem Statement

Calculate the maximum directional derivative and its direction for \( f(x,y) = 2x^2 + 3xy + 4y^2 \) at \( (1,1) \).

Solution

808 solution video

video by Krista King Math

close solution

Calculate the directional derivative of \( f(x,y) = x^2y^3 - y^4 \) at \( (2,1) \) in the direction given by the angle \( \theta = \pi/4 \).

Problem Statement

Calculate the directional derivative of \( f(x,y) = x^2y^3 - y^4 \) at \( (2,1) \) in the direction given by the angle \( \theta = \pi/4 \).

Solution

810 solution video

video by PatrickJMT

close solution

Calculate the value of the maximum rate of change of the function \( f(x,y) = \sin(xy) \) at the point \( (1,0) \).

Problem Statement

Calculate the value of the maximum rate of change of the function \( f(x,y) = \sin(xy) \) at the point \( (1,0) \).

Solution

1522 solution video

video by Krista King Math

close solution

Calculate the direction of the maximum rate of change of the scalar field \( \phi(x,y,z) = -2xy + x\ln(y+z) \) at \( P(1,3,-2) \) and the maximum value of the directional derivative at P.

Problem Statement

Calculate the direction of the maximum rate of change of the scalar field \( \phi(x,y,z) = -2xy + x\ln(y+z) \) at \( P(1,3,-2) \) and the maximum value of the directional derivative at P.

Solution

1776 solution video

video by Dr Chris Tisdell

close solution

Applications

There are several applications where the gradient is used and has additional meanings. Here are three of them.

temperature application

When the function f is a temperature, the gradient can be interpreted as the change in temperature.

If we have a material where the heat flow (from hot to cold) is given by \( T = x^3 - 3xy^2 \), determine the direction of maximum decrease of temperature at the point \( (1,2) \).

Problem Statement

If we have a material where the heat flow (from hot to cold) is given by \( T = x^3 - 3xy^2 \), determine the direction of maximum decrease of temperature at the point \( (1,2) \).

Solution

802 solution video

video by Dr Chris Tisdell

close solution

calculating a tangent plane

To determine the equation of a plane, we need one point and a normal vector. Since the gradient is normal to the level curves (and also to the tangent plane), we have a normal vector and we are usually given a point in the problem statement. This is enough information to determine the equation of the tangent plane.

Here is a video clip with a great explanation of this idea. He also explains how using the gradient notation simplifies the notation significantly.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 4) [5mins-39secs]

video by Dr Chris Tisdell

Calculate the tangent plane and the normal line to the surface \( x^2 + y^2 + z = 9 \) at the point \( (1,2,4) \).

Problem Statement

Calculate the tangent plane and the normal line to the surface \( x^2 + y^2 + z = 9 \) at the point \( (1,2,4) \).

Solution

805 solution video

video by Dr Chris Tisdell

close solution

Calculate the gradient, directional derivative and equation of the tangent plane of \( g(x,y,z) = \sin(xyz) \) at the point \( (\pi, 1/2, 1/2) \) in the direction \( \vec{v} = \langle \sqrt{3},-2,3 \rangle \).

Problem Statement

Calculate the gradient, directional derivative and equation of the tangent plane of \( g(x,y,z) = \sin(xyz) \) at the point \( (\pi, 1/2, 1/2) \) in the direction \( \vec{v} = \langle \sqrt{3},-2,3 \rangle \).

Solution

806 solution video

video by Dr Chris Tisdell

close solution

Calculate the equation of the tangent plane to \( x^4 + xy + y^2 = 19 \) at the point \( (2,-3) \).

Problem Statement

Calculate the equation of the tangent plane to \( x^4 + xy + y^2 = 19 \) at the point \( (2,-3) \).

Solution

809 solution video

video by Krista King Math

close solution

Calculate the equation of the tangent plane to the surface \(f(x,y)=2y\cos(5x-3y)\) at the point \((3,5,10)\).

Problem Statement

Calculate the equation of the tangent plane to the surface \(f(x,y)=2y\cos(5x-3y)\) at the point \((3,5,10)\).

Solution

2201 solution video

video by MIP4U

close solution

gradient and potential function

These types of problems will often be worked in the opposite direction than what we have discussed so far, i.e. you may be given a gradient and then expected to come up with the potential function. These are not hard but require some changes in your thinking to solve. The idea is that you are given a gradient and you need to come up with the original function. Here is a great video, with examples, showing how this works.

Dr Chris Tisdell - Potential Function Example (Gradient) [5mins-19secs]

video by Dr Chris Tisdell

Calculate the potential function for \( \nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k} \).

Problem Statement

Calculate the potential function for \( \nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k} \).

Solution

811 solution video

video by PatrickJMT

close solution

Functions of More Than Two Variables

You can easily extend the definition of the gradient and directional derivative to functions of 3 or more variables. However, you lose the physical interpretation of slope. For three dimensions, the gradient looks like you would expect.

\(\displaystyle{ \nabla h(x,y,z) = \frac{\partial h}{\partial x} \hat{i} + \frac{\partial h}{ \partial y}\hat{j} + \frac{\partial h}{ \partial z}\hat{k} }\)

Real Time Web Analytics