17Calculus Partial Derivatives - The Gradient Vector
17Calculus
Gradient Vector
The directional derivative and gradient are very closely related. On this page we give an overview of both but discuss only the gradient in detail. On the directional derivative page you will find a more complete discussion of the directional derivative. However, you need to understand the gradient first.
Topics You Need To Understand For This Page |
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Overview of the Gradient and the Directional Derivative
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.
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).
So you may be asking yourself, what does the gradient actually mean? What does it look like? Well, here is a great video that will give you a more intuitive understanding of the gradient vector.
Trefor Bazett - Geometric Meaning of the Gradient Vector
video by Trefor Bazett |
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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, as you increase the number of variables, you lose the physical interpretation of slope. For three dimensions, the gradient looks like you would expect.
\(\displaystyle{ \nabla h(x,y,z) = \frac{\partial h}{\partial x} \hat{i} + \frac{\partial h}{ \partial y}\hat{j} + \frac{\partial h}{ \partial z}\hat{k} }\) |
Before going on to the directional derivative, let's make sure we understand the gradient by working a few practice problems.
Practice
Unless otherwise instructed, find the gradient vector of these functions.
\( f(x,y) = x\sin y + \cos y \)
Problem Statement |
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Find the gradient of \( f(x,y) = x\sin y + \cos y \)
Final Answer |
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\(\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y - \sin y)\vhat{j}\)
Problem Statement
Find the gradient of \( f(x,y) = x\sin y + \cos y \)
Solution
\(\displaystyle{ \nabla f(x,y) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} }\) |
\( = (\sin y + 0) \vhat{i} + (x\cos y - \sin y)\vhat{j} \) |
\( = (\sin y) \vhat{i} + (x\cos y - \sin y) \vhat{j} \) |
Final Answer
\(\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y - \sin y)\vhat{j}\)
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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
Dr Chris Tisdell - 804 video solution
video by Dr Chris Tisdell |
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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
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
World Wide Center of Mathematics - 2293 video solution
video by World Wide Center of Mathematics |
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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
MIP4U - 2294 video solution
video by MIP4U |
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Given \( f(x,y) = xy \), determine \( \nabla f(-3,2) \).
Problem Statement
Given \( f(x,y) = xy \), determine \( \nabla f(-3,2) \).
Solution
MIP4U - 2295 video solution
video by MIP4U |
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\( f(x,y,z) = z e^{-xy} \)
Problem Statement |
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Find the gradient of \( f(x,y,z) = z e^{-xy} \)
Final Answer |
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\( \nabla f(x,y,z) = -yze^{-xy} \vhat{i} - xz e^{-xy} \vhat{j} + e^{-xy} \vhat{k} \)
Problem Statement
Find the gradient of \( f(x,y,z) = z e^{-xy} \)
Solution
\(\displaystyle{ \nabla f(x,y,z) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} + \frac{\partial f}{\partial z} \vhat{k} }\) |
\(\displaystyle{ z(-y)e^{-xy} \vhat{i} + z (-x) e^{-xy} \vhat{j} + e^{-xy} \vhat{k} }\) |
Final Answer
\( \nabla f(x,y,z) = -yze^{-xy} \vhat{i} - xz e^{-xy} \vhat{j} + e^{-xy} \vhat{k} \)
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\( f(x,y) = 3x + 7y \)
Problem Statement
Find the gradient vector of \( f(x,y) = 3x + 7y \)
Solution
Thomas Wernau - 4340 video solution
video by Thomas Wernau |
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\( f(x,y) = \sin(xy) \)
Problem Statement
Find the gradient vector of \( f(x,y) = \sin(xy) \)
Solution
Thomas Wernau - 4341 video solution
video by Thomas Wernau |
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\( f(x,y) = \ln x + \arctan(xy) \)
Problem Statement
Find the gradient vector of \( f(x,y) = \ln x + \arctan(xy) \)
Solution
Thomas Wernau - 4342 video solution
video by Thomas Wernau |
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\( f(x,y) = xyz^{-1} \)
Problem Statement
Find the gradient vector of \( f(x,y) = xyz^{-1} \)
Solution
Thomas Wernau - 4343 video solution
video by Thomas Wernau |
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Links You May Find Helpful |
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Better Explained: Vector Calculus: Understanding the Gradient |
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