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external links you may find helpful |
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Better Explained: Vector Calculus: Understanding the Gradient |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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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.
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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.
video by Dr Chris Tisdell
Gradient |
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\(\displaystyle{ \nabla g(x,y) = \frac{\partial g}{\partial x} \hat{i} + \frac{\partial g}{ \partial y}\hat{j} }\)
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 |
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If \( f(x,y,z) = x^3+y+z^2 \), calculate \( \nabla f \) and \( \nabla f(1,2,3) \).
Solution |
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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 |
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Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)
Hint |
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This problem uses unusual variables \( \tau \) and \( \kappa \). However, they work just as if you had t and x.
Problem Statement |
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Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)
Hint |
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This problem uses unusual variables \( \tau \) and \( \kappa \). However, they work just as if you had t and x.
Solution |
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video by World Wide Center of Mathematics
close solution |
Calculate the gradient vector field of \( f(x,y) = \ln(2x+5y) \).
Problem Statement |
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Calculate the gradient vector field of \( f(x,y) = \ln(2x+5y) \).
Solution |
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video by MIP4U
close solution |
Given \( f(x,y) = xy \), determine \( \nabla f(-3,2) \).
Problem Statement |
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Given \( f(x,y) = xy \), determine \( \nabla f(-3,2) \).
Solution |
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video by MIP4U
close solution |
Directional Derivative |
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\(\displaystyle{ D_{\vec{u}}f = \nabla f \cdot \vec{u} }\)
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.
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.
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.
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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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Calculate the maximum directional derivative and its direction for \( f(x,y) = 2x^2 + 3xy + 4y^2 \) at \( (1,1) \).
Solution |
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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 |
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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 |
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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 |
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Calculate the value of the maximum rate of change of the function \( f(x,y) = \sin(xy) \) at the point \( (1,0) \).
Solution |
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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 |
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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 |
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video by Dr Chris Tisdell
close solution |
Applications |
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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 |
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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 |
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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.
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 |
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Calculate the tangent plane and the normal line to the surface \( x^2 + y^2 + z = 9 \) at the point \( (1,2,4) \).
Solution |
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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 |
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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 |
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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 |
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Calculate the equation of the tangent plane to \( x^4 + xy + y^2 = 19 \) at the point \( (2,-3) \).
Solution |
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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 |
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Calculate the equation of the tangent plane to the surface \(f(x,y)=2y\cos(5x-3y)\) at the point \((3,5,10)\).
Solution |
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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.
video by Dr Chris Tisdell
Calculate the potential function for \( \nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k} \).
Problem Statement |
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Calculate the potential function for \( \nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k} \).
Solution |
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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.
\(\displaystyle{ \nabla h(x,y,z) = \frac{\partial h}{\partial x} \hat{i} + \frac{\partial h}{ \partial y}\hat{j} + \frac{\partial h}{ \partial z}\hat{k} }\) |