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Integrating/differentiating several variables

  1. Jul 7, 2011 #1
    Hi all! i am just a high school student and have only taken AP calculus AB, but i have a question which hopefully some of you will be able to answer. how do you take the derivative/integral of a multivariable equation? how is it decided which variable you are taking derivatives/integrals with respect to, and what is this doing conceptually?

    im sorry is this is a dumb question, but all the other websites i've tried on the web use terminology and symbols im not familiar with, and are very confusing
     
  2. jcsd
  3. Jul 7, 2011 #2
    Hi mexijo!! :smile:
    Firstly sorry, i don't understand what you mean by "AP" and "AB".

    You said that you are in a high school at present. Trying to go into Multi-Variable Calculus at this stage is dangerous (not exactly :biggrin:). You should first start with the single-variable calculus and that is also fun. Why you want to go to higher level right now?

    To know with what respect to we are taking the integral, we take a example:-
    [tex]\int x^2 dx[/tex]
    Do you understand what this dx? This "dx" conveys that we are integrating with respect to x.
     
  4. Jul 7, 2011 #3
    Like the poster above said, the [itex]dx[/itex] tells you that you're working with respect to [itex]x[/itex].

    @Pranav-Arora: AP Calculus AB is an Advanced Placement Calculus class, which is basically the Calculus 1 and 2 you find in colleges. Actually, a high enough score in the Advanced Placement test gives you the college credits for Calculus 1 and 2 when you enter, meaning you get to skip them and go straight to multivariable as a freshman.


    Now, back to the topic at hand, let's say you have a function of several variables
    [tex]f(x,y) = 3e^xsin\; y + x[/tex]
    When you differentiate, you hold the variable that you aren't differentiating with respect to constant.
    Here's differentiating with respect to x:
    [tex]\frac{\partial}{\partial x}f(x,y) = \frac{\partial}{\partial x}(3e^xsin\; y + x) = 3e^xsin\; y + 1[/tex]
    The squiggly d just tells you that this is a partial derivative, because you're only taking the derivative of one part (the [itex]x[/itex] part, in this case). Since you're only differentiating the [itex]x[/itex] part, just pretend that [itex]y[/itex] is a constant, like [itex]\pi[/itex] (I like to think in terms of letter constants since that makes it clear that it is a letter, but a constant one).
    Now, when differentiating with respect to y:
    [tex]\frac{\partial}{\partial y}f(x,y) = \frac{\partial}{\partial y}(3e^xsin\; y + x) = 3e^xcos\; y[/tex]
    Since the [itex]x[/itex] is constant when you differentiate with respect to [itex]y[/itex], it gets dropped out just like any other constant.
    Another notation for a partial derivative is subscript notation, where the variable you're differentiating with respect to is the subscript of the function.
    [tex]\frac{\partial}{\partial x}f(x,y) = f_x(x,y)[/tex]

    Conceptually, a derivative is how sensitive a function is to change, or, more physically speaking, the "speed" at which it changes. Another good concept is slope, you're finding the slope of a line tangent to that particular point on the curve.
    For partial derivatives, remember that you're differentiating with respect to one variable, and holding all others constant. That means you're finding the slope or the rate of change in that particular direction, regardless of what the others are doing.

    For integration, you have multiple integration:

    2 integrals (Double Integral)
    [tex]\int \int_R f(x,y)dA[/tex]

    3 integrals (Triple Integral)
    [tex]\int \int \int_R f(x,y, z)dV[/tex]

    Don't get too confused over what [itex]R[/itex] is, it's just a placeholder for all of the limits of integration. It is the region you're integrating over.
    Just like when you do a definite integral in single variable calculus, in multivariable you have to go from "a to b" on one and "c to d" on the next, etc.
    Also, as for what [itex]dA[/itex] and [itex]dV[/itex], it's the product of [itex]dx[/itex] and [itex]dy[/itex] (for double) and [itex]dz[/itex] (all three for triple). [itex]dA[/itex] just stands for [itex]dxdy[/itex] or [itex]dydx[/itex] (depending on which order you integrate in), and, from an intuitive standpoint, is a differential of area (an extremely small area), that you multiply by a height ([itex]f(x, y)[/itex]) and sum up ([itex]\int[/itex]). In single variable calculus, you used [itex]dx[/itex] when integrating (or [itex]dy[/itex], sometimes). Remember that when you do a definite integral, you're actually summing up an infinite amount of infinitely thin rectangles to find area. In multivariable, now you're summing up cuboids to find volume (for double integrals).

    So for a double integral, let's just say you have the following:
    [tex]\int_0^3 \int_2^3 (2y+x)dxdy[/tex]
    So, just like partial derivatives, whatever you're integrating with respect to, hold everything else constant. In these integrals, you work from the inside out.
    So, after the first round of integrating, you have:
    [tex]\int_0^1 \left[2xy + \frac{x^2}{2} \right]_{x=2}^{x=3}dy = [/tex]
    Now just plug in the limits of integration, and integrate again:
    [tex]\int_0^1 \left(2y + \frac{5}{2}\right)dy = \left[y^2 + \frac{5}{2}y \right]_0^1 = 1+\frac{5}{2}=\frac{7}{2}[/tex]
    So we have found the volume of this region, it is [itex]\frac{7}{2}[/itex].

    You might also run into an indefinite integral, although I first saw a multivariable indefinite integral in differential equations.
    Remember in an indefinite integral, you always have to add a constant. The reason is because, since this is also called the antiderivative, the added constant could have been in the function that you integrate into. Let's just say [itex]F\,'(x) = f(x)[/itex], if [itex]F(x)[/itex] has any constants, when you differentiate into [itex]f(x)[/itex], they're gone. So when you integrate, you have to account for any abitrary constant that may have been in [itex]F(x)[/itex]
    For multivariable, however, these arbitrary constants are now functions of the variable you held constant. So, if [itex]f(x, y) = F_x(x,y)[/itex], let's say for [itex]f(x,y) = 3xy^2[/itex], we find the integral with respect to x.
    [tex]\int f(x,y)dx = \int 3xy^2dx = \frac{3}{2}x^2y^2 + c(y)[/tex]
    Basically, if you had any purely [itex]y[/itex] terms in the function [itex]F(x,y)[/itex], they would have been dropped out when differentiating with respect to [itex]x[/itex], since they're held constant. So, not only do with have to account for any constant, we have to account for any function of only [itex]y[/itex] when integrating with respect to [itex]x[/itex]. The same is true for integrating a three variable expression:
    [tex]\int f(x,y,z)dx = F(x, y, z) + c(y, z)[/tex]


    Hopefully this has helped you. If you want to go further into the subject, other than this very simple overview of only 2 topics in multivariable calculus. If you want to study further, I highly recommend going over to MIT's OpenCourseWare at http://ocw.mit.edu" [Broken].
     
    Last edited by a moderator: May 5, 2017
  5. Jul 7, 2011 #4
    I too use those resources and that Calculus book too. They are awesome.
     
    Last edited by a moderator: May 5, 2017
  6. Jul 8, 2011 #5
    thanks so much Brianmath!! i got a 5 on the ap exam btw, and im interested in what multivariable calculus is really all about and wanted to teach myself, but i couldnt find anything that seemed geared towards people who didnt already know the subject. i checked out the MIT opencourseware and it looks amazing. awesome. thanks.
     
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