Question 1

How would I be able to type in the symbol "nabla" on a mac keyboard?

Accepted Answer

I know this is a bit late, but hopefully still helpful for someone. On recent versions of OS X, you can get to the "Character Viewer" from any application by going to Edit > Emoji and Symbols. If you don't see the symbol you are looking for in the list (many math symbols aren't there), click the button on the upper right corner to expand the list, giving you the full character viewer. From there, the "Math Symbols" section includes many helpful symbols, including ∇!

Question 2

Does anyone know of a good online resource that has Multivariable practice questions?

Accepted Answer

I just saw http://www.leadinglesson.com/, not sure how good they are but the professor seems to have some chops in this area.

Question 3

I have a question about the gradient: after watching linear algebra videos on linear transformations, we've learned that a transformation T, which takes place from IR^n ---» IR^m represents a matrix with m rows and n columns. Now being aware of this fact, let's assume a function f(x,y) = x^2 - xy, where f: IR^2 ---» IR. Why is the gradient represented as a 2x1 matrix (2 rows and 1 column) and not as a 1x2 matrix (1 row and 2 columns)? The gradient here is represented as a vector field, not as a scalar field, but why?

Accepted Answer

Whether you represent the gradient as a `2x1` or as a `1x2` matrix (column vector vs. row vector) does not really matter, as they can be transformed to each other by matrix transposition. If `*a*` is a point in `*R*²`, we have, by definition, that the gradient of `ƒ` at `*a*` is given by the vector

`∇ƒ(*a*) = (∂ƒ/∂x(*a*), ∂ƒ/∂y(*a*)),`

provided the partial derivatives `∂ƒ/∂x` and `∂ƒ/∂y` of `ƒ` exist at `*a*`. Note that `∇ƒ(*a*)` is a _vector_. Thus `∇ƒ` maps a vector `*a*` in `*R*²` to the vector `∇ƒ(*a*)` in `*R*²`, so that `∇ƒ: *R*² ➝ *R*²` is a vector field (and not a scalar field).

*Edit*
Going slightly on a tangent here: the gradient `∇ƒ` is closely related to the _(total) derivative_ of `ƒ`. The total derivative of `ƒ` at `*a*` (if it exists) is the unique linear transformation `ƒ'(*a*): *R*² ➝ *R*` such that

`|ƒ(*x*) - ƒ(*a*) - ƒ'(*a*)(*x* - *a*)| / ‖*x* - *a*‖ ➝ 0`

as `*x* ➝ *a*`. In this case, the matrix of `ƒ'(*a*)` (that is, the matrix representation of the linear transformation `ƒ'(*a*)`) is given by the `1x2` matrix

`Dƒ(*a*) = [∂ƒ/∂x(*a*)  ∂ƒ/∂y(*a*)].`

Question 4

Where does the gradient point if there are 2 equally steep directions to go in. For that sake, there could be any n number of directions. How does the gradient decide which one to point in if they are equal?
An example I can think of is the the origin in the graph z = x^2 - y^2. If you go along either x axis, the curve will increase exponentially (but equally) on both sides. What does the gradient vector do in such cases? (In the case of the origin of x^2 - y^2, I believe it gives the 0 vector, as if we're at a local maxima -- which makes sense along the y direction but not along the x direction...)

Accepted Answer

If you actually take the gradient,  it becomes [2x, -2y]. so at x-axis, put y = 0, and the gradient becomes [2x, 0]. Now If you are at x = 0, then gradient is [0,0] which does not tell you to go anywhere i.e. does not point in any direction. but as you deviate slightly in any direction, [h,0] or [-h,0], gradient start pointing in a specific direction, which is the direction of steepest ascent.

Question 5

I'm missing exercises on multivarible calculus. 
This was a great article thought.

Accepted Answer

To find the gradient you find the partial derivatives of the function with respect to each input variable. then you make a vector with del f/del x as the x-component, del f/del y as the y-component and so on...

Question 6

This is regarding the question "Why are the vectors in this vector field so small along the upward diagonal stripe?"

I understood why the vector along the line y=2x is going to be 0 but how can you deduce that the vectors close to that line have small horizontal component? Cant they just jump around? Is it because the functions are linear? What is the intuition behind this? Thanks.

Accepted Answer

I’m not sure why I am doing this: answering a 2-year old question by someone who probably already got on with his life. Kunjaan unfortunately does not  refer explicitly to the “reflection question” under example 1 of the gradient article.
In the answer that you can find by clicking “show answer”, the author explains (quite lucidly in my opinion) that de gradient (of which the x-component=2x-y and y-component=x) is smallest in the vicinity of the line y=2x and close to the y-axis. In that region the gradient approaches (0,0) wich gives us small vectors. Just read the text closely.
If you formulate your questions precisely, you are more likely to get a timely answer.

Question 7

My question is suppose we are standing in the xy-plane now-

1.the gradient of the function shows us the direction of the steepest accent?

OR

2.the gradient of the function shows us its value or length by which we can see that in which way its length is minimum and by that we can get the steepest accent of the function?

I am actually confused about the direction of the gradient weather its parallel to the xy-plane or in the direction of the steepest accent?

Accepted Answer

The gradient gives us a vector, specifically a 2D vector. This vector is going to be parallel to the xy axis.

Now, you will have a point (x,y,z) on a graph of f(x,y) the gradient says at point (x,y,z) if you rotate yourself to face the vector you get from the gradient at that point, if you proceed forward the rate of change relative to the z axis will be the greatest in that direction. I will do an example.

let's just use f(x,y) = x^2 + y^2
gradient = <2x, 2y>

let's use the point (2, 3, 13) here the gradient is <2*2, 2*3> = <4, 6>. If you need find the angle on the xy plane .

Now, on point (2,3,13) if you imagine yourself standing there holding a compss, you would use the compass to you are facing in the direction of the gradient. Now, once you are doing that, walking forward gives you the fastest path up the "hill" you are standing on.

Worth saying the negative gradient is the steepest path down the hill.

So the vector that would point up the hill, so with some measure not parallel with the xy plane is different.

Let me know if that didn't help

Question 8

"If you imagine standing at a point (x0,y0,…) in the input space of f, the vector ∇f(x0,y0,…) tells you which direction you should travel to increase the value of f most rapidly."

So this means that the gradient does not point towards the top of a mountain, but to the steepest point, correct?

If I have a tilted plane, what would the gradient be? Zero?

Accepted Answer

Might be a little late, but I'll answer in case someone finds it useful...

the gradient points to the steepest path, like the text you quoted says. I does not point to the steepest point. If you were trying to climb a mountain as quickly as posible, you could use the gradient as a "compass" that would always tell you the fastest way to get to the top (without considering physical restrictions, of course).

It's easy to see a plane example, let's say: f(x,y) = x + y 
what's the gradient? [1,1] 
what does it mean? it means that, no matter where you are, the steepest point is always in that direction ([1,1]).

To get a gradient that is always zero you function would have to be constant.


Name	Symbol	Example
Derivative	$\frac{d}{d x}$ ‍	$\frac{d}{d x} (x^{2}) = 2 x$ ‍
Partial derivative	$\frac{\partial}{\partial x}$ ‍	$\frac{\partial}{\partial x} (x^{2} - x y) = 2 x - y$ ‍
Gradient	$\nabla$ ‍	$\nabla (x^{2} - x y) = [\begin{matrix} 2 x - y \\ - x \end{matrix}]$ ‍

Multivariable calculus

Course: Multivariable calculus > Unit 2

The gradient

What you need to be familiar with before starting this lesson:

What we're building toward

Definition

Example 1: Two dimensions

Example 2: Three dimensions

Interpreting the gradient

Example 3: What local maxima look like

The gradient is perpendicular to contour lines

The del operator

Summary

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