I don't like the term dependent variable vs independent variable. I prefer to say that "y is a function of x" or vice versa; meaning that if y is a function of x then it can be put into the form y=f(x).
Remember that an ellipse is not a function; it doesn't pass the vertical line rule (in my...
You may be :P
Totally agree with this, it turns out that if you look at the average velocity over two points and you keep making the points closer to each other, the average velocity keeps getting closer and closer to a specific number. If you make the points infinitely close, the average...
Velocity is defined as ##\frac{\Delta Displacement}{\Delta Time}## or ##\frac{dx}{dt}## or any one of probably dozens of equivalent definitions. What are you looking for?
Because when you take a line integral from (0,0) to (0,3) in a density field z = f(x,y) you're not just finding out what the mass of a 3 cm wire is you're finding out what the mass of a 3 cm wire is when density is defined by z = f(x,y).
In your mind you're thinking that the 3 cm wire already...
f(x,y) gives you a value of density for all points in the x-y plane. Taking a line integral through that plane is like 'cutting out' a wire from that density plane.
Unless f(x,y) is constant, there is no reason to assume that one wire cut out would be the same density as another.
I think what...
Well, you're learn all this in your next course. But the key here (to be brief) is the difference between definite integrals and indefinite integrals.
Indefinite integrals are obviously related to derivatives because its just the opposite of differentiation (no magic involved here, that's...
And to show yourself it, try solving the differential equation ##\frac{dy}{dx} = y##
It doesn't really prove much, but its a nice way to convince yourself that what slider142 said is true!
You can do surface areas in single variable calculus, but they are just as ugly as they are in multivariable. The reason is that instead of using dx you have to use ds where ds = arc length = ##\sqrt{1+ (\frac{dy}{dx})^2} dx##
The general formula for surface area is ##SA = \int 2 \pi y...
And just to add on to this, ##\frac{d}{dx}## is just notation for a 'derivative operator', i.e. ##\frac{d}{dx}## is saying "take the derivative of this:".
##\frac{dy}{dy}##, which is not a derivative operator but in fact a derivative itself can't even be considered a fraction rigorously...
No. The key here was in your original post: "goes to zero." The key thing about infinitesimals is that they may be infinitely close to zero, but they are not zero. If they were, then you couldn't divide by them.
The way you should look at it is that you are considering the limit as the...
It's completely up to you. Of course sometimes you might be asked to re-parametrize or not depending on the problem, but in general it doesn't make a difference. The whole point of parameterizing is to re-write functions in a different way, with the goal of making your parametrization simpler...
Here's my two cents on this, know that that's all this is:
Finding Area and Finding Slope really aren't inverses of each other. The confusion is in the difference between definite integrals and indefinite integrals. Indefinite integrals are defined as simply the reverse of differentiation, so...
I'm assuming your familiar with multivariable calculus up to this.
Realize that when you are taking a double or triple integral what you are doing is summing up all the values of a function in a region of space or a region of the plane.
A line integral is what you get if instead of wanting to...
Hey all,
I was working through some problems in my spare time when I realized that I wasn't so satisfied with my understanding of how to use Greens theorem with holes. Can someone refresh my memory?
More specifically:
Lets say I want to take the line integral in some vector field of a curve C...
Like mfb said, 99% of the time when you're asked for a line integral of a scalar field you'll want it with respect to arc length, and then you'll want the integral with ds in it. As you showed in your post, ##\int_C{F(x,y) dx}## + ##\int_C{F(x,y) dy}## ≠ ##\int_C{F(x,y) ds}##, so the line...
The idea of the chain rule is that if you have a function f(p) where p is a function of x: p(x), and you want to find the rate at which f changes with respect to a change in x, you take the rate at which f changes with respect to p and multiply that by the rate at which p changes with regards to...
Well, you can look at the derivatives like fractions. Look at: ##\frac{dv}{dr} \cdot \frac{dr}{dt}##. Notice that this quantity has a ##dr## in both the numerator and the denominator so they cancel out, meaning ##\frac{dv}{dr} \cdot \frac{dr}{dt} = \frac{dv}{1} \cdot \frac{1}{dt} =...
\frac{\partial^2 p(x)}{\partial p(x) \partial x} = \frac{\partial^2 p(x)}{\partial x \partial p(x)}
I'm not really equipped to answer your question with a rigorous proof, but to expedite other people helping you I fixed your latex.
Well, first before we help you, you should say what you've gotten so far. Though based off the information given, I'll guess you know about Green's Theorem, in which case this is a fairly easy problem.
First of all, df is not the derivative of f, it is the differential of f, ##\frac{df}{dx}## is the derivative of x (assuming f is a function of x).
Second of all, why would you think that it's derivative would have a different domain?
There isn't really a proof for tabular integration, it's just shorthand for integration by parts. If you did a integration by parts using placeholder functions for u and v (meaning the actual functions could be whatever you want) you'll see that it expands to the same thing that tabular...
As you may or may not see, evaluating it with regards to x first will let you solve the whole thing. The trick is just the figure out the new bounds on the integrals - it's not as simple as just switching the order.
What I would do is the draw a graph and draw the lines x=4, x=0, y=2 and y=x/2...
Well, I have an algebra-heavy way. There might be an easier way but I can't think of it right now.Lets say that the shape in figure A is a rectangle of width ##W## and height ##H## that has a triangle of width ##w## and height ##h## cut out of it. This would mean that the total area of the shape...