# Vector Calculus: Understanding the Gradient

The gradient is a fancy word for derivative, or the rate of change of a function. It’s a vector (a direction to move) that

• Points in the direction of greatest increase of a function (intuition on why)
• Is zero at a local maximum or local minimum (because there is no single direction of increase)

The term “gradient” is typically used for functions with several inputs and a single output (a scalar field). Yes, you can say a line has a gradient (its slope), but using “gradient” for single-variable functions is unnecessarily confusing. Keep it simple.

“Gradient” can refer to gradual changes of color, but we’ll stick to the math definition if that’s ok with you. You’ll see the meanings are related.

Now that we know the gradient is the derivative of a multi-variable function, let’s derive some properties.

The regular, plain-old derivative gives us the rate of change of a single variable, usually x. For example, dF/dx tells us how much the function F changes for a change in x. But if a function takes multiple variables, such as x and y, it will have multiple derivatives: the value of the function will change when we “wiggle” x (dF/dx) and when we wiggle y (dF/dy).

We can represent these multiple rates of change in a vector, with one component for each derivative. Thus, a function that takes 3 variables will have a gradient with 3 components:

• F(x) has one variable and a single derivative: dF/dx
• F(x,y,z) has three variables and three derivatives: (dF/dx, dF/dy, dF/dz)

The gradient of a multi-variable function has a component for each direction.

And just like the regular derivative, the gradient points in the direction of greatest increase (the following article explains why). However, now that we have multiple directions to consider (x, y and z), the direction of greatest increase is no longer simply “forward” or “backward” along the x-axis, like it is with functions of a single variable.

If we have two variables, then our 2-component gradient can specify any direction on a plane. Likewise, with 3 variables, the gradient can specify and direction in 3D space to move to increase our function.

## A Twisted Example

I’m a big fan of examples to help solidify an explanation. Suppose we have a magical oven, with coordinates written on it and a special display screen:

We can type any 3 coordinates (like “3,5,2″) and the display shows us the gradient of the temperature at that point.

The microwave also comes with a convenient clock. Unfortunately, the clock comes at a price — the temperature inside the microwave varies drastically from location to location. But this was well worth it: we really wanted that clock.

With me so far? We type in any coordinate, and the microwave spits out the gradient at that location.

Be careful not to confuse the coordinates and the gradient. The coordinates are the current location, measured on the x-y-z axis. The gradient is a direction to move from our current location, such as move up, down, left or right.

Now suppose we are in need of psychiatric help and put the Pillsbury Dough Boy inside the oven because we think he would taste good. He’s made of cookie dough, right? We place him in a random location inside the oven, and our goal is to cook him as fast as possible. The gradient can help!

The gradient at any location points in the direction of greatest increase of a function. In this case, our function measures temperature. So, the gradient tells us which direction to move the doughboy to get him to a location with a higher temperature, to cook him even faster. Remember that the gradient does not give us the coordinates of where to go; it gives us the direction to move to increase our temperature.

Thus, we would start at a random point like (3,5,2) and check the gradient. In this case, the gradient there is (3,4,5). Now, we wouldn’t actually move an entire 3 units to the right, 4 units back, and 5 units up. The gradient is just a direction, so we’d follow this trajectory for a tiny bit, and then check the gradient again.

We get to a new point, pretty close to our original, which has its own gradient. This new gradient is the new best direction to follow. We’d keep repeating this process: move a bit in the gradient direction, check the gradient, and move a bit in the new gradient direction. Every time we nudged along and follow the gradient, we’d get to a warmer and warmer location.

Eventually, we’d get to the hottest part of the oven and that’s where we’d stay, about to enjoy our fresh cookies.

But before you eat those cookies, let’s make some observations about the gradient. That’s more fun, right?

First, when we reach the hottest point in the oven, what is the gradient there?

Zero. Nada. Zilch. Why? Well, once you are at the maximum location, there is no direction of greatest increase. Any direction you follow will lead to a decrease in temperature. It’s like being at the top of a mountain: any direction you move is downhill. A zero gradient tells you to stay put – you are at the max of the function, and can’t do better.

But what if there are two nearby maximums, like two mountains next to each other? You could be at the top of one mountain, but have a bigger peak next to you. In order to get to the highest point, you have to go downhill first.

Ah, now we are venturing into the not-so-pretty underbelly of the gradient. Finding the maximum in regular (single variable) functions means we find all the places where the derivative is zero: there is no direction of greatest increase. If you recall, the regular derivative will point to local minimums and maximums, and the absolute max/min must be tested from these candidate locations.

The same principle applies to the gradient, a generalization of the derivative. You must find multiple locations where the gradient is zero — you’ll have to test these points to see which one is the global maximum. Again, the top of each hill has a zero gradient — you need to compare the height at each to see which one is higher. Now that we have cleared that up, go enjoy your cookie.

## Mathematics

We know the definition of the gradient: a derivative for each variable of a function. The gradient symbol is usually an upside-down delta, and called “del” (this makes a bit of sense – delta indicates change in one variable, and the gradient is the change in for all variables). Taking our group of 3 derivatives above

$\displaystyle{grad F(x,y,z) = \nabla F(x,y,z) = (\frac{dF}{dx},\frac{dF}{dy},\frac{dF}{dz})}$

Notice how the x-component of the gradient is the partial derivative with respect to x (similar for y and z). For a one variable function, there is no y-component at all, so the gradient reduces to the derivative.

Also, notice how the gradient can itself be a function!

$\displaystyle{F(x,y,z) = x + y^2 + z^3 }$

$\displaystyle{\nabla F(x,y,z) = (1, 2y, 3z^2)}$

If we want to find the direction to move to increase our function the fastest, we plug in our current coordinates (such as 3,4,5) into the equation and get:

$\displaystyle{direction = (1, 2(4), 3(5)^2) = (1, 8, 75)}$

So, this new vector (1, 8, 75) would be the direction we’d move in to increase the value of our function. In this case, our x-component doesn’t add much to the value of the function: the partial derivative is always 1.

Obvious applications of the gradient are finding the max/min of multivariable functions. Another less obvious but related application is finding the maximum of a constrained function: a function whose x and y values have to lie in a certain domain, i.e. find the maximum of all points constrained to lie along a circle. Solving this calls for my boy Lagrange, but all in due time, all in due time: enjoy the gradient for now.

The key insight is to recognize the gradient as the generalization of the derivative. The gradient points to the direction of greatest increase; keeping following the gradient, and you will reach the local maximum.

## Questions

Why is the gradient perpendicular to lines of equal potential?

Lines of equal potential (“equipotential”) are the points with the same energy (or value for F(x,y,z)). In the simplest case, a circle represents all items the same distance from the center.

The gradient represents the direction of greatest change. If it had any component along the line of equipotential, then that energy would be wasted (as it’s moving closer to a point at the same energy). When the gradient is perpendicular to the equipotential points, it is moving as far from them as possible (this article explains why the gradient is the direction of greatest increase — it’s the direction that maximizes the varying tradeoffs inside a circle).

## Questions & Contributions

1. me says:

i like it… well explained.

2. Jane says:

Super!!!

3. Chris says:

You are the man! Nice work!

4. Kalid says:

5. gaurav says:

i was always looking for conceptual and practical examples and yes i finally got.

6. Harry says:

Awsome!

7. Palo says:

well you made a good explanation, that even a not-so-smart guy gets it, but i think you missed the obvious -> WHY does gradient show the direction of the greatest increase.

I think that the principle of the gradient is quite easy, but understanding why does it work the way it does is a bit tricky and you should have focued on it more.

It would be interesting if you would somehow add it to this good article. Inspiration http://mathforum.org/library/drmath/view/68326.html

good luck !

8. Hi Palo, that’s a great point! I’ve been feeling a bit guilty, if you can imagine it, because I’ve lacked that explanation

I’m probably going to do a separate article on the reason *why* the gradient points in the direction of greatest increase — I have another explanation that it works well with. Thanks for the link and feedback!

9. Your introduction is not quite correct:

You claim: “Points in the direction of greatest increase of a function”.

Why? It can also point in the direction of greatest decrease of a function.

A gradient is one or more directional derivatives. These derivatives are considered in a particular direction. In the case of single variable calculus, we generally talk about a directional derivative when we consider multiples of the x unit vector, i.e. k*(1,0). To consider the y unit vector, we deal with the partial derivatives with respect to y in a given direction. In three dimensions, the 3 partial derivatives form what we now call a ‘gradient’.

So in fact it is incorrect to call this a slope or anything else except to say that it describes the partial derivatives of a point in the direction of a given vector in space.

Does this make sense? Please visit my blog for some more interesting reading.

http://mathphile.blogspot.com/

10. Hi John, thanks for writing. You’re right, the formal definition of a gradient is a set of directional derivatives.

But when thinking about the intuitive meaning, I think it’s ok to consider the gradient as a vector that “points” in the direction of greatest increase (i.e. if you follow that direction your function will tend towards a local maximum).

Unless I’m mistaken, the gradient vector always points in the direction of greatest increase (greatest decrease would be in the opposite direction).

11. What I was saying is that it points either one way or the other, it is not restricted to the direction of greatest increase. As a simple example, consider what happens when you differentiate a parabola: You set the derivative equal to 0 and then you determine that it has either a maximum or a minimum at its turning point. It is not always a maximum just as it is not always a minimum. Think I have explained this correctly now.

12. sqib says:

good john you have done a great job.

13. Kalid says:

Hi John, thanks for the clarification. I’d still politely disagree and say that in general, the gradient points in the direction of greatest increase :).

In the case of 2 dimensions, the gradient/slope only gives a forward or backward direction. A positive slope means travel “forward” and a negative slope means travel “backwards”.

Consider f(x) = x^2, a regular parobola. The gradient is zero at the minimum (x=0), and there is no *single* direction to go. At x = -1, the slope is negative, which means travel “backwards” (to x = -2) to increase your value. Similarly, at x = 1, you travel forward (to x = 2) to increase your value.

But, as you mention, strange things can happen when the derivative = 0. It can mean you are at a local maximum (no way to improve), or at a local minimum (no single direction to improve your position — forward or back will help). I consider the corner case of zero an exception to the general rule / intuition that the gradient is “the direction to follow” if you want to improve your function.

14. Vidhya says:

Wonderful explanation!

15. Thanks Vidhya, glad you liked it.

16. bihazo says:

hi john keep it up you done a great job

17. Travis says:

Thanks a bunch! I didn’t think it could be this simple to find the maximum increase at a point, so I thought I’d look it up. Thanks to your great explaination, it turn out it was as easy as it seemed it should be. Great job! Thanks!

Travis

18. Awesome, glad it worked for you

19. caitlyn says:

thanks!!!!

20. Kalid says:

Hi Caitlyn, you’re welcome.

21. Thanks! The sadistic microwave example helped a lot.

22. Kalid says:

Awesome, glad it was useful :).

23. Hello Kalid,
time. Am sorry you do not agree.

Let me give you an example:
Suppose we are dealing with pressure
and height in a certain ‘cubic’
area. Suppose that the middle of the
cube height is 0 meters. Also suppose
that we have a whirlpool generated in the
cube such that the pressure rate increases
as we go below the middle of the cube.
Anything below is negative height and anything above
is positive height. Now, as one rises
higher in the cube, the pressure decreases.
If we find the gradient, then according to
your definition (and many others’), then
the gradient vector for the rate of greatest
increase will point below the middle of the
cube, not above. But above the middle we
find the greatest ‘decrease’ in rate of pressure.

In this example, greatest increase points
downwards and greatest decrease upwards.

It would probably be better to define
gradient as a vector that points in a
direction of greatest increase or decrease.
It’s additive inverse will point in the
diretion of greatest decrease or increase
respectively. For most physical phenomena,
your definition would generally be true.
But what happens when you have an anomaly?
Make sense?

24. I do not believe I have the best answer to this question but like yourself, I am a believer in trying to find the best possible explanation. Once again, I like your website. Keep up the good work Kalid!

25. Okay, I think I have the best answer. If f is a real-valued function, then del(f) or gradient of f points to the greatest increase, whereas -del(f) points t0 the greatest decrease.

For once planet math has some decent information on this since I last checked:

I do not endorse everything Planet Math publishes but this particular information appears to be correct. In any event, it clears up the previous confusion I think.

26. Hi John, thanks for the comment! Yes, that’s an important distinction to make: the positive gradient is the greatest increase, and the negative gradient is the greatest decrease. Thanks for helping clarify :).

27. Jared says:

Thank you!

28. Bigmouth says:

This actually makes sense to me. Thanks!

29. Kalid says:

30. Anonymous says:

did not grasp the idea

31. Kalid says:

Be more specific. The gradient is the direction to move that gives you the biggest increase.

32. Shaheen says:

It helps me a lot. But I have some doubt still now.Is it the same concept for gradient of each vertex in a triangle mesh?

Thanks so much.

33. JohnnyT says:

Kalid

Thanks for the great explanations! I thought I was math-retarded for some time; however your writings actually make sense to me!

Take care!

Johnny T

34. Kalid says:

@Shaheen: Thanks, glad you enjoyed it. I’m not sure I understand the question: in a triangle mesh, you could measure the gradient at each vertex to find the “best” direction to move. Again, not sure if this is your question.

@Johnny T: Thank you for the comment! Yes, when a subject seems difficult (as vector calculus was for me) sometimes it’s just because the explanation wasn’t clicking properly. Thanks for dropping by.

35. wali khan says:

well done,excellent explaination with solid examples

36. Kalid says:

Thanks Wali, glad you enjoyed it.

37. j.sathish kumar says:

thanks
but i have some doubts.how the differentaion gives the maximum space rate of change. as per my understandings differentiation only is difference between two point in the region say p1 and p2.can u clarify

38. leon says:

Thanks a lot for explaining the concept.

39. sophie says:

i was having so much trouble understanding this and now its all clear thank you so much!

40. @lon, sophie: Thanks, glad you enjoyed it!

41. Ryan Johnson says:

Jesus. This was a lot better explained than in my text book and by my professor. I thought we were using the gradient as the normal vector but I really doubted that it could be that.

42. @Ryan: Thanks! I struggled with this concept for a while also.

43. Ranjeet Kumar says:

thanks ! this explanation made me clear how to find the direction of smallest change.It is just the 90 degree rotation of gradiant(the direction of largest change).

44. Shakeel Ahmed says:

Thanks very much for your effort

45. Bill says:

Um — in your microwave example, aren’t you pushing the doughboy out the back of the microwave? (Just wanted to understand the concept). I love these essays, btw, keep them coming!

46. Hehehe says:

I loved the microwave analogy.also thanks for clarifying the upsidedown delta now everything makes more sense

47. RAHUL says:

stil im confused between scalar field and vector field….

how can such a mathematical expression denote the max change? pls i didnt understand the relation of this with mathematics. pls reply sir.

49. nat2_bam2 says:

thank you soo much!!

its a big help for our project…

50. Kalid says:

@Rahul: A scalar field returns a single value (x), but a vector field returns multiple values (x,y,z). Usually the multiple values (x,y,z) are taken as a “direction” to follow.

@aradhita: Hi, that’s a question I need to get into in a later post.

@nat2_bam2: Thanks!

51. Migs says:

Hi kalid! i read your explanation. oh this is very helpful! by the way can you give an example on how to apply this on a situation of the classic “mountain and mountain climber” problem? hope you will reply. thanks again your explanations were clear

52. @Migs: Great question. The classic “mountain climber” problem is when the vector field gives the height of the mountain (z) at a certain position (x,y), so z = f(x,y).

The gradient at any position x,y will give you the direction of the _greatest increase_ in z. That is, the gradient will point in the “most uphill”. Following the gradient will give you the shortest path the the top of the mountain (technically, the top of the nearest local maximum). How this helps!

53. vignesh says:

beautiful…well said

54. akansha says:

thanks a lot for the wonderful explanation!!!

55. Kalid says:

@akansha: You’re welcome!

56. Very nice! Keep up. Thanks a lot

57. Florencia says:

Very nice article!!
Hope to see how to find the maximum of a constrained function soon!!
Thanks a lot!!

58. Kalid says:

@Florencia: Glad you liked it! Thanks for the suggestion.

59. ab says:

Very good explanation by the way. So if you are on a landscape given by z=cosy-cosx and u want to get from (0,0,0) to (4pi,0,0) by moving in the direction of the gradient in the positive x-direction how would u explain that? What would that path look like?

60. P-F says:

Thanks for the great explanation. Another topic that would be very interesting for you to cover is the Jacobian, which causes pain for many, many students (including myself).

61. Kalid says:

@P-F: Thanks for the note — I think the Jacobian, and linear algebra in general, would be great to cover. I’ve forgotten a lot of it and am looking to relearning :).

62. Mark Soric says:

Just wondering something. In that case of f(x,y) = X^2 + y^2, a paraboloid – how can the gradient by perpendicular to the tangent plane at all point and only have components in x and y…

How can it point in any other direction other than parallel to the xy plane?

I’m lost here.

63. prabu says:

thank you kalil. wonderful explanation.

64. Kalid says:

65. Ashraful says:

It was a great explanation! But I have a specific problem with gradients. Is there any functions that cant be expressed as gradient of any parameter? What could be the properties of that function?

66. Ashraful says:

May I could be more specific about my previous problem. If a function is constant in all direction, is it possible to express the function as gradient?

67. Kalid says:

I’m not sure if I understand the question — the gradient of a constant function would be a 0 vector [perhaps technically (0,0)], that is, there is no direction of greatest increase. If it helps, think of the gradient in terms of a derivative (the derivative of a constant function is 0).

68. Kinar says:

Math professional!

69. Anonymous says:

Thank you for getting to the heart of why del is required and how to intuitively understand it. Its the first time I understand it so well despite reading so much about it before!

70. Anonymous says:

damn! i got it now

71. Anonymous says:

math is so beautiful

72. bob clear says:

WOW! great explanation…. thanks dude.:D

73. Kalid says:

@bob: Thanks!

74. Kalid says:

@Anonymous: Agreed :).

75. Jose says:

Great explanation helped me explain my brother! Nice job! Gonna bookmark it for further needs I might have with it.

76. js says:

great explanation and example

77. Kalid says:

@js: Thanks!

78. shreedhar says:

hey, explained really well. But still you didn’t provide any sign of why the gradient would always point in the direction of maximum increase…

79. Nick Pellatz says:

I don’t usually comment on blogs, but this is a great explanation. Way better than my text book. A+++++++++

80. @shreedhar: Thanks — I’d like to cover that in a follow-up article. I need to get a nice, intuitive explanation for it first ;).

81. Al Paquette says:

Man! I just love this kind of explanation. It’s so clear and concise, and it shows me that the author really understands the concept himself.

All mathematics should be taught this way. Go from the specific to the general (abstract). Not the other way around, which is the path usually followed by the type who wants to show off his prowess with math symbols and equations.

82. Anonymous says:

nice explanation

83. Anonymous says:

84. @Al: Thank you! I think one of the big problems in math teaching (especially) is just trying to get things explained without the professor’s “prowess” getting in the way, as you say.

@Anonymous: How could you eat cookies when there’s gradients to be studied?

85. Pandia says:

Nice work!! Thanks man:)

86. Kalid says:

@Pandia: No prob!

87. Gaton says:

Thanks!

88. Marwa says:

Keep it up.

89. Kalid says:

90. Vitor P.F. says:

Great !! Congrats

91. Alex says:

Awesome! I was cracking my head trying to figure out HW, only to realize how basic it was after reading through ur page. Thanks!

92. dianne says:

goodluck with my exam tom. ^^

93. Anonymous says:

It wasn’t a bad explanation but I wish you had explained ‘why’ the gradient is the perpendicular vector of the function its derivatives were derived from. This still bothers me a little.

Also, if we have a function with three variables, shouldn’t the independent variable be considered? By considered I mean, if I have a function F(x, y z), then I am saying that w = F(x, y, z), and this function can not be graphed since it has 4 dimentions. A normal F(x, y) can be graphed since you considered the Z, X, and Y of the graph.

From the book I read, I interpreted that the original function has a constant value for ‘w’, hence producing a graph with a new function F2(x, y). However I still didn’t see the math that proves that the gradient of the function F(x, y, z) is actually the vector that is perpendicular to the surface of the graph from which its derivatives were derived from. If you could prove this, it would be really helpful.

94. thx man very much
I understood it totally from u
my regards

95. Burton says:

Thank you very much! This made perfect sense and it really helped me out.

96. Kalid says:

@Burton: Thanks!

97. panchito says:

Cool! Thanks

98. Donglin says:

good explain, it solved my problem

99. Anonymous says:

kalid , are u professor

100. shrikant says:

such wonderful explanation……..wow

101. Kalid says:

@shrikant: Thanks!

102. Mrigeh says:

I love you!

103. Kalid says:

@Mrigeh:

104. hyaa says:

i want to ask, once knowing the maximum rate of change of temperature in your’s microwave example, how we can attain that particular place without moving our coordinates positions as mentioned by microwave for example when we choose coordinates (3,5,2) we obtain gradient as (3,4,5). now from where we get the information that which coordinates should be selected next time that gives us maximum gradient? should we choose (3,4,5) coordinates?

105. Kalid says:

@hyaa: I’m sorry, I don’t think I understand the question. The gradient gives you the direction (not coordinates) of the greatest increase in your current value. You have to follow the gradient for a bit, get to a new point, get the gradient there, follow it for a bit… and so on to maximize your value.

Think of the gradient as a compass which points towards your greatest increase. A compass doesn’t give you the coordinates of North, but tells you how to get there from your current position. Hope that helps.

106. Zita says:

Hi, I still have a question. If there is a function h(x,y)to denote the height of a mountain at position(x,y). Can I use the knowledge of gradient to locate the top of the mountain and how?

107. kalid says:

@Zita: Yep — you start at any point, and keep following the gradient of h to find the top.

108. max says:

great, really good thank you,it would be comprehensive if you explain that ‘why’ the gradient is the perpendicular vector of the function its derivatives were derived from

109. kalid says:

@max: Great question. Going to add it as a Q & A at the end of the article.

110. Patel Ankit says:

Excellent explanation, I think if you provide your ebook for free of cost it would really be helpful for the poorer students to strengthen thier grass-roots.
GOOD JOB, KEEP IT UP………….

111. Chico says:

Consider the directional derivative, f_u.
f_u = f_x u_1 + f_y u_2 (it takes some effort to see this definition of f_u)
=grad(f) dot u (u is a unit vector)
Thus, it is clear that the directional derivative, f_u, is maxed when cos@=1.
It follows that @=0 and the directional derivative, f_u, is attained when u is in the direction of the gradient. Therefore, the gradient does indeed give the direction of greatest increase.

Note that f_u is minimized when cos@=-1. Thus, @=pi, and u is in the opposite direction of the gradient. QED

ps
I am a nerdy math professor who likes demonstrating mathematical prowess. Thanks for the microwave intuition builder. My students are going to like that.

112. Deniz says:

I already knew this but you gave me a better intuition of it and I like your style of writing! Thank you!

113. kalid says:

@Chico: Awesome, thanks for sharing! I like that a lot — lining up with the gradient (out of all possible directional derivatives) will give you the best return (cosine = 1). That clicks for me.

Glad you enjoyed the microwave intuition, I love searching for little analogies.

114. kalid says:

@Deniz: Thanks! And you’re welcome :).

115. beant singh says:

ah highly informative and excellently defined

116. kalid says:

could u explain why does gradient is zero for local minimum?

119. kalid says:

Madhu: Those cookies are delicious :). The gradient behaves like the first derivative (rate of change). In regular calculus, d/dx = 0 means your function is not changing [therefore you are at a max or min].

Similarly, when the gradient is zero, it means your function is not changing when you move.

120. Anonymous says:

wow gr8

121. Yogesh says:

Hey, could you explain WHY gradient points in the direction of maximum increase?
Also, could you explain a vector field? In case of scalar field, what I imagine is as follows: Consider the 3D space, and for each point my scalar function returns a value.. And with a intensity proportional to that value, a black point appears(Greyscale).. I cant picture anything about a vector field though.. Please consider..
And again, thanks for all the Vector calculus explanations..

122. kalid says:

Hi Yogesh, thanks for the note.

The essence is “you have a circle of possible directions, the individual derivatives (df/dx and df/dy) give you the tradeoff as you change directions, so find the direction that makes the best use of that tradeoff”.

A vector field is tricky. Imagine your same 3d space, but instead of a point (a single value) imagine that there is wind blowing through it. Each position in your space could feel a different “push” (strength and angle) from the wind. For example, in a cyclone, the push might be in a circle (each individual point is pushed in a different instantaneous direction), with a dead spot in the middle. With a steady wind, every point might feel the same “push”.

123. someone says:

You are a gift to mankind! Well to me you are.

124. Jose says:

People confused by gradient pointing towards maximally increasing direction, or perpendicularity of gradient to equipotential.

125. Baljeet Kaur says:

its awsome…it clearly explains the actual physical significance of a gradient..thank you..:)

126. kalid says:

127. manju says:

thks for complete explanation

128. Kevin @ http://kldavenport.com says:

I stumbled across your site looking for one specific aspect of gradients and ended up reading the whole post. You did a great job of distilling these concepts Kalid.

129. Kalid says:

Thanks Kevin, I appreciate it!

130. Rick says:

I went over to Wikipedia and read an article on a similar topic. It was so much more difficult to understand, and Wikipedia is easier than most math texts. It makes me so angry that most math books seem to go out of their way to make mathematics unnecessarily difficult. Maybe, with more people beginning to write internet articles like this, the math obfuscators won’t be able to get away with it much longer. I look forward to the day when students realize that math can be the easiest class in school.

Thanks and keep up the good work Kalid!

131. Kalid says:

Thanks Rick! I can relate to the feeling of frustration, it’s what drove me to write (after having an aha momne, it quickly turned to: “Why couldn’t they explain it like that in the first place?”).

I do think math has the potential to become the easiest subject. Its objectivity, which could be seen as offputting, is a great indicator of when something has truly clicked. As a result, we can quickly determine whether an analogy is helping solve the problem before us.

Appreciate the support :).

132. Marco P says:

Lovely!

133. Khasud says:

Jazak Allah khair

134. An easy and interesting approach. Hats off!

135. ATHUL P ANAND says:

awesme dude!!gud work!!!well explained…ws really helpful!

136. lashiwe says:

Thank u so much,your explanations really helped…- must confess am not all that good in mathematics I could really use some more help.but thank you

137. Kalid says:

Another explanation that I posted on reddit:

I imagine a vector field like a grid. Off in the distance is a billboard showing the amount of money we’ll get for being at a certain position on the grid (value of the function).

From where we’re standing, we can take a step in any direction. The “work” is the same (i.e., I moved 1 unit) but the payoff (net amount of cash we gain) can differ drastically depending on if we go North, West, South, NorthEast, etc.

The gradient is the “payoff” in the sense it points to the direction of greatest reward (if it’s zero, it means you are already at the max reward, i.e. any step you take will diminish your earnings).

A surface like “z = 3″ is basically saying “show me all the positions in this grid where the reward is $3″. On the grid a path is drawn, highlighting all the positions of this equal payoff. If you started following this path, your payoff would never change. But… what would the gradient be? The gradient is pointing in the direction of greatest increase, so should have nothing in common with the path of 0 increase. In other words, the projection of the gradient onto the surface should be 0, i.e. it’s normal. The gradient has zero inclination for you to go anywhere near the path of zero gains. Ok, fine, that’s what the gradient should do. How do we show it actually maximizes the payoff? Here’s how I figure it: on a circle (showing all possible paths), we can basically make any tradeoff of x and y that we want. At 45-degrees we can trade them 1-for-1, at higher values we can get 2 units x-distance for 1 unit of y distance, or 10 to 1, or a million to 1 (at angles close to 0 or 90). The direction of the gradient is calculated to maximize the tradeoff based on dz/dx and dz/dy, i.e. it figures out how much reward we get for moving in each direction and allocates “effort” appropriately. If we get$20 for moving the x direction and \$10 for moving in the y direction, then our direction should favor x, but only at a 1:2 tradeoff. I.e., if we can trade 1 x for 3 y’s then we should keep trading (adjusting the angle) until we get 1x for 2y’s.
In other words, you can prove that the gradient direction is the direction which maximizes z assuming you are moving 1 infinitesimal unit and are getting rewarded by dz/dx and dz/dy. And by definition, this profit-maximizing direction would not waste any energy along the profit-maintaining path that must have both dz/dx and dz/dy of 0 (the equal-valued path must not change the amount of z).

138. Really great explanation! The only thing is that in your definition of the ‘del’ the partial derivatives should have a lower-case delta ($\displaystyle{\delta}$ instead of ‘d’

139. Ajinkya says:

I would be really glad if you could tell me the derivation for the formula for gradient of a scalar in terms of the nabla/del operator…
Or if you could tell a link where I can find it…

140. pari says:

Hi kalid, it is a great explanation at lest for people like me.
I think one should get this overview before getting into the actual concept.
thanks a lot!

141. Naveen says:

Thanks a bunch !

142. Rajmohan says:

very good article

143. Aymun says:

MashaAllah you posted this in 2007and to this date people are getting benefit from it. Loved it. Thank you

144. Anonymous says:

what is green’s theorem? can anyone explain.

145. sanu says:

but the world of mathematics is very very very interesting when people like you teaching us.
thanks a lot….

146. Osama says:

Awesome work But if you can help me with this problem,? My professor assigned me “physical significance of gradient” that would be my first assignment confused what to submitt? ?? (application/physical significance)

147. Amir sultan says:

Good explanation

148. Taps patel says:

Thank you so much

149. J Scrib says:

Thanks a lot. this really helped me understand it better!

150. Manikanta says:

A gradient is Vector differentiation operator applied on a scalar function. Not strictly the derivative of vector functions as you said in opening paragraph, both are different. However, gradient can be treated as a derivative of a special vector for which all the vector components have same function, whose magnitude is the scalar function and directed in the direction of f_x=f_y=f_z from notation f(vector) = (f_x,f_y,f_z).

151. Nice woke it realy open but wish ther was a summarie of it

152. Aqeel says:

how to find max rate of change of scalar field

153. VU Khan says:

Pick an area that has small hill. Starting at the bottom of the hill

2. In particular, describe how to determine the direction in which the gradient points at a given point on the hill.

3. What should you do if you encounter a wall ?

4. Discuss whether the gradient path is the shortest path, the quickest path or the easiest path.

154. Tim McIntyre says:

Has your page been changed? I used it last year as a reference for a course I am teaching on fields. This year a student immediately picked out an issue with the statement “The term gradient (grad) typically refers to the derivative of vector functions”. It refers to the derivative of scalar functions. All your examples are scalars – temperature, height and “x+y^2+z^3″. Could this be corrected please?

155. Tim says:

I’d suggest it would be better to write “The term gradient (grad) typically refers to the derivative of a scalar field”. The term “vector function” is somewhat confusing.

156. kalid says:

Hi Tim, great feedback — I just updated the article. (I’d incorrectly used “vector function” as a function that took a vector as input, vs. a vector-valued one that returned a vector.) Thanks!

157. Tim says:

Hi Kalid,

thanks for the quick correction! Just in time for Thursday’s lecture.

Tim

158. Minie says:

I have been trying to ‘reload’ these ‘memories’ because I am planning to study again. I really needed this website. Thank you

159. Dhar says:

Thank you so much for this example! I’ve been struggling the way my professor explains it, but this just clicked@