Multi-variable systems?

[Boris Lagutin]

If I understood right in the lecture 9 "Max-Min and Least Squares " of MIT 18.02 course: http://ocw.mit.edu/courses/mathematics/ ... t-squares/ the instructor says that we must find as many partial derivatives as many variables are in a function f(a,b,c....n). Ok, but I'm just curious. If we have 3D-space how we can have more than 2 variables (function is f(a,b) in this case)? Here "a" is for i-axis, "b" for j-axis. Please, notice that I don't talk about parametric equations here, I intuitively understand that values of "a" and "b" may depend on a third variable.

Thanks a lot.

[a.s.h.]

In 3-space our functions will be of two independent variables, i.e. z = f(x, y). For example, z = f(x, y) = x + sin(y). (In 2-space, our functions are functions of one variable, i.e. y = f(x), an example being y = f(x) = tan(x).)

So in the 3-space case, we have two independent variables, so can take partial derivatives with respect to these two (i.e. with respect to x, or y). For example, if we have z = x + sin(y), then there are two (first-order) partial derivatives: ∂z/∂x = 1, and ∂z/∂y = cos(y).

(I haven't seen the lecture, were you referring to a specific part of it?)

[Boris Lagutin]

Thank, a.s.h.

The lecturer doesn't say any specific about this point of my interest. He just mentioned that we should have as many partial derivatives as many variables we have. In this way I thought whether there may be more variables than (N-1) where N is a number of dimensions. That's it.

Have a nice night.

[Boris Lagutin]

Just curious. What kind of stuff may be related to multi-variable function f(x,y,z) (with 3 variables)?

Thanks a lot.

[Boris Lagutin]

Another interesting question. In the lecture 11 "Chain rule" MIT 18.02 course: http://ocw.mit.edu/courses/mathematics/ ... hain-rule/ since 48 min. 00 sec. an example is given. A function f(x,y) where x = r*cos(theta) and y = r*sin(theta). How to make partial differentiation of f(x,y) with respect to theta (angle)?!

Thanks.

[joe]

What kind of stuff may be related to multi-variable function f(x,y,z) (with 3 variables)?

A good example is electric potential, V(x,y,z). You can then get the field components by taking partial derivatives
E_x = - ∂V/∂x etc.
Similarly, for the concentration C(x,y,z), the components of diffusion are proportional to the negative partial derivatives

For the specific MIT example, I'll let you take that up with the staff from MIT.

Joe

[Boris Lagutin]

Thank you very much, Joe. A problem with this MIT course is that it is self-pace. I remember I asked a question and someone of the course classmates answered it only in some weeks. Moreover, this answer was not so satisfactory.

Have a nice weekend.

[a.s.h.]

Another interesting question. In the lecture 11 "Chain rule" MIT 18.02 course: http://ocw.mit.edu/courses/mathematics/ ... hain-rule/ since 48 min. 00 sec. an example is given. A function f(x,y) where x = r*cos(theta) and y = r*sin(theta). How to make partial differentiation of f(x,y) with respect to theta (angle)?!

Thanks.

We have z = f(x, y), where x = r⋅cos(θ), y=r⋅sin(θ).

Then note that ∂x/∂θ = ∂/∂θ (r⋅cos(θ)) = -r⋅sin(θ) = -y.
Similarly, we can see that ∂y/∂θ = r⋅cos(θ) = x.

Therefore,

∂z/∂θ = ∂f/∂x⋅∂x/∂θ + ∂f/∂y⋅∂y/∂θ (multivariable chain rule)
= -y⋅∂f/∂x + x⋅∂f/∂y.

[Boris Lagutin]

Thank you, a.s.h.. Now let's substitute sin(theta) with u and cos(theta) with v. So we have x = r*v and y = r*u. How do we make partial differentiation with respect to v and u? It's clear that u and v change like variables.

thanks.

[a.s.h.]

Thank you, a.s.h.. Now let's substitute sin(theta) with u and cos(theta) with v. So we have x = r*v and y = r*u. How do we make partial differentiation with respect to v and u? It's clear that u and v change like variables.

thanks.

Let's consider z = f(x, y), where x = x(r, u, v) = rv, and y = y(r, u, v) = ru.

∴ ∂z/∂v = ∂f/∂x⋅∂x/∂v + ∂f/∂y⋅∂y/∂v (multivariable chain rule)

= ∂f/∂x⋅r + ∂f/∂y⋅0 (since ∂x/∂v = ∂/∂v(rv) = r, and ∂y/∂v = ∂/∂v(ru) = 0)
= r⋅∂f/∂x.

You can use a similar process to find ∂z/∂u.
(cont.)

[a.s.h.]

We do not worry about whether u and v are related here, because the notation ∂z/∂v means basically the rate of change of z if we could hold r and u constant while only changing v.

e.g. Say z = f(x, y) = x² + y, then z = r²⋅v² + ru, since x = rv, and y = ru. But since
z = r²⋅v² + ru,

∂z/∂v = ∂/∂v(r²⋅v² + ru)
= 2r²⋅v.

You can check that this agrees with the formula derived above (∂z/∂v = r⋅∂f/∂x), because

r⋅∂f/∂x = r⋅∂/∂x(x² + y)
= r⋅(2x)
= r⋅(2rv)
= 2r²⋅v.

[Boris Lagutin]

Great Thanks, a.s.h.. I have something to say about your last posts here but now I would like to ask a question about Gradient. In the lecture 12 "Gradient" MIT 18.02 course: http://ocw.mit.edu/courses/mathematics/ ... -gradient/ since 5 min. 30 sec. the following example is given. The plane equation: w = a_1*x + a_2*y + a_3*z where Gradient is <a_1,a_2,a_3>. In other words, the gradient is perpendicular to the plane, however, I know that Gradient is perpendicular to the level curve (not level surface). Could you clarify this point? If the gradient may be perpendicular to the plane then may the gradient be perpendicular to some level surface (please, see my drawing #2)? Thanks.

[Boris Lagutin]

Another obscure point is here: https://www.khanacademy.org/math/multiv ... gradient-1 In this video of Khanacademy all Gradient vectors are parallel to x,y-plane (bottom) despite the surface pictured is curved. If I understand right any gradient vector must be perpendicular to a tangent plane (surface) at each point on the surface? If so then why do all the gradient vectors are parallel to x,y-plane in the video?

Thanks.

[joe]

Gday Boris

I think that you might be confusing some related but different issues here.

z = a_1*x + a_2*y
is a plane in 3 dimensions. The slope of this plane in the x direction is a_1.

To make an analogy with
w = a_1*x + a_2*y + a_3*z
you need 4 dimensions, which is impossible to visualise, and hard to imagine.

The del (∇w) of a scalar field like
w = a_1*x + a_2*y + a_3*z
is at right angles to an equipotential; an equipotential is a surface having equal w. e.g. the plane
z = (1/a_3)(a_1*x + a_2*y -w) where w is constant

For a practical example, consider electric potential due to a point charge:
V = kQ/r
- ∇V gives the electric field
E = -kQ/r^2 in the r direction.

================
It's a bit surprising that MIT is using the UNSW discussion forum for its courses. MIT is a fairly rich institution with high investments and high fees. Don't they have their own forum?

[Boris Lagutin]

Good morning Joe,

Thank you. I am a learner and not a MIT student. Truly I tried to find any free UNSW Calculus forums (like this physics forum) but failed. It's pity. In terms of MIT 18.02 course I can say that it is a self-pace course and waiting for answers takes long in my opinion. I need to get answers as soon as possible and I use different forums. Another advantage of it is that I may see different answers which give different perspectives to understand a same point.

P.S. By the way, it's a problem to find an online multi-variable calculus course (not self-pace) with real-time instructors and scheduled quizzes in English language. Ok, those were my last questions about calculus here. Sorry.