Since there are a lot of stupid threads now, let this be the one all-purpose thread where you can ask a person who knows calculus anything.
>>7927959
I haven't done calculus in a while, but I never quite understood the equation of work. Please, explain as if you're talking to someone with an IQ of 70.
>>7927993
It is force times how far out you are in the line of work. Meaning, if you are doing work on a pole extending outwards, the force of gravity at each point times your distance is the ammiunt of work at that point. It is defined to be force times distance.
>>7928001
Thank you.
>>7928009
Your welcome Mate.
What is calculus?
Why is d/dx treated as if it were a fraction? moreover, why does separation of variables work if they aren't supposed to be treated as fractions? Why do things like total differential even exist?
>>7928068
[math]\frac{d}{dx}[/math] is Leibniz notation for "derivative of x", never treat it as a fraction. You can think of it as a fraction when doing u-sub or something but do not actually computate it as a fraction.
>>7928068
Just to let you onow the other guy was not Op, i am. Treat it as a fractions and thinks work out, treat the "dx" in the integrals as multiplication and it works out.
>>7928077
false
you can cancel off the d's to get 1/x
>>7928086
Op here, please, go back to /b/.
>>7928068
Now, not to say they exactly represent those things, but you are allowed to manipulate them, and they are given definitions that make them act like fractions. (dy/dx)(dx) is the definition of dy, and dx is defined to be ewual to delta x.
>>7928077
how is it different from dy/dx? Can you not manipulate it such that d(f(x))=kdx (where k is any term you want)? Or are you not saying you can't and i'm being stupid?
>>7928108
He's saying that they aren't exactly fractions, but playing pretend is allowed.
>>7928115
Ah, that cleared it up. Thanks
>>7928129
You're welcome matey, - Op
For this question......
>>7928174
Why do we define dS as what I boxed in red?
>>7928177
Uh..... This is the sort of thing in an analysys class. I would assume there is a proof of it, but I wouldn't know it.
>>7928090
just take d (x^1/2)/dx
you cancel off the d's to get x^1/2/x
which of course simplifies to 1/x^1/2
Q.E.D.
>>7928187
The class is Multivariable calculus (I haven't taken a proofs class yet), I just never saw in the book where it said: "dS = this..."
Wasn't sure if it was something intuitive that I missed or something. I guess more of my question was, if it's true for all surface integrals?
>>7928195
Dunno. - useless Op
>>7928212
>useless
Nah, it's fine.
>>7928214
K man. - Op
>>7928177
In differential geometry, n-dimensional "surfaces" (manifolds) can characterized entirely by their intrinsic geometry.
This information is contained in a object called the metric tensor of the "surface", [math]{{g_{ij}}}[/math].
So let [math]g = \det \left( {{g_{ij}}} \right)[/math].
Then an n-dim volume form on the "surface" (manifold) is written as [math] dV = \sqrt g d{x_1} \wedge ... \wedge d{x_n} [/math]
The typical thing you think of as a surface is a 2D object embedded into a 3D space.
So the "volume form" for a typical 2D surface, would actually be a surface area form [math] dA = \sqrt g d{x_1} \wedge d{x_2}[/math].
At the level of Multivariable Calculus, the wedge product is typically ignored. So I will do that from now on.
Define a Surface by a map, [math] f:U \to {\mathbb{R}^3} [/math], where U is a subset of R^2.
Take a general coordinate, [math] \left( {{x_1},{x_2}} \right) \in U [/math].
Then the geometric factor sqrt(g) can take the form: [math] \sqrt g = \left\| {\frac{{\partial f}}{{\partial {x_1}}} \times \frac{{\partial f}}{{\partial {x_2}}}} \right\| [/math].
So then our Surface Area element is: [math] dA = \left\| {\frac{{\partial f}}{{\partial {x_1}}} \times \frac{{\partial f}}{{\partial {x_2}}}} \right\|d{x_1}d{x_2} [/math]
Now the Vector Surface Area element is just: [math] d\vec S = \hat \nu dA [/math].
Where nu is the unit normal to the surface. Defined as: [math] \hat \nu = \pm \frac{{\frac{{\partial f}}{{\partial {x_1}}} \times \frac{{\partial f}}{{\partial {x_2}}}}}{{\left\| {\frac{{\partial f}}{{\partial {x_1}}} \times \frac{{\partial f}}{{\partial {x_2}}}} \right\|}} [/math]
So finally our general Vector Surface Area element can be written: [math] d\vec S = \pm \left( {\frac{{\partial f}}{{\partial {x_1}}} \times \frac{{\partial f}}{{\partial {x_2}}}} \right)d{x_1}d{x_2} [/math]
In your specific question: [math] f\left( {{x_1},{x_2}} \right) = \left( {{x_1},{x_2},\sqrt {{x_1}^2 + {x_2}^2} } \right) [/math]
>>7928223
Trying 2 beat Op now are we?
>>7928225
Well OP is clearly useless.
>>7928231
op is probs first year that took calc 1 and is currently doing calc 2
that's all that's needed for knowing calculus in his mind
http://www.allaboutcircuits.com/worksheets/calculus-for-electric-circuits/
Scroll to question 2 and look at the first and second graphs. Is the first supposed to be the derivative of the second? Because I really don't think it is. It would have to be shifted to the left by 2.
Right?
>>7928233
Once you learn the power rule, thats all the calculus you really need.
Can someone explain vector and parametric functions to me like I'm a retard?
>>7927959
How to show that the Cantor set is perfect but with measure zero?
