| Blogs > Purind |
|
Purind
Canada3562 Posts
d/dx (integral f(x) dx) = f(b) - f(a) Does it work exactly the same way in 2 dimensions? d/dx (integral f(x,t) dx) = f(b,t) - f(a,t) My head is telling me it should be right, but I don't know why. Is this true, and why? I know that if its d/dt (integral of something dx) it follows some other rule, but is what I put above correct?  | ||
|
Nexzico
Sweden93 Posts
When you have f(x,t) you are in 3 dimensions, different "pair number" (x,t) (I have no clue what that's called in english, it's a word for word translation) gives a answer z( which has 3 coordinates; x,y,z) since f(x,t)=z. And if you have a function with 2 or more variables u consider the other variables as (in your case t) constants so.... (1) If f(x,t)= 3t ==> d/dx(f(x,t)) = [3tx] (2) and if f(x,t) = 2x + 3t + 5, then d/dx(f(x,t)) = [x^2 + 3tx +5x] Let's say these functions have the same intervall as your exampl [a,b] then... (1)= 3ta-3tb (2)= a^2 + 3ta + 5a -(b^2 + 3tb + 5b) I'm sorry, I have a hard tim explaining this in Swedish so^^, but I think it works the same way as in 2 dimentions.... the exaples should be correct so, I hope it helps (don't think so but, hey...why not?) You should know that if you want to calculate the volume under the intervall [a,b] you must integrate on the "t" variable too. I have no idea why this works tho.... oh crap, just read your post more thurouly and i think i missunderstood your question.. but no hurt in posting anyway, GL | ||
|
Gyabo
United States329 Posts
On March 29 2008 02:39 Purind wrote: OK so I know that the derivative of the integral just returns the function like so: d/dx (integral f(x) dx) = f(b) - f(a) This should really be: d/dx (integral f(x) dx) = d/dx (f(b) - f(a)) = f(x) The 2nd equation is something I don't believe is possible to exist. You know the derivative of a function is its slope. When you're working with a function that has only 1 variable like f(x), f(x) is normally a curve. The slope of a curve at some point is a line. When you're working with a function that has 2 variables like f(x,t), f(x,t) is normally a surface. The slope of a surface at some point is a plane, not a line. Therefore, you can't attempt to differentiate a function of 2 variables with respect to only 1 variable because the slope of a surface cannot be a line. You can however, take partial derivatives to the 2 variable function. The partial derivative of f(x,t) with respect to x is the slope of the line tangent to the trace of a curve in the plane x=a, where a is part of an arbitrary point (a,b,c) within the surface in 3 dimensional space. To put it in more simple terms, a partial derivative is the slope of part of a function that has more than 1 variable. Your calculus book should probably have more info on this if you want to look into it. I hope I didn't confuse you more with this post. Good luck in your future math studies. | ||
|
Gyabo
United States329 Posts
On March 29 2008 03:49 Nexzico wrote: (1) If f(x,t)= 3t ==> d/dx(f(x,t)) = [3tx] (2) and if f(x,t) = 2x + 3t + 5, then d/dx(f(x,t)) = [x^2 + 3tx + 5x] Be careful with what you're trying to accomplish. The bolded answers are what you should get when you integrate the function, not differentiate. (1) int (3t dx) = 3tx + C (2) int (2x+3t+5 dx) = x^2 + 3tx + 5x + C | ||
|
JacobDaKung
Sweden132 Posts
d/dx (integral f(x,t) dx) = f(x,t) f = X*T X(x) is a function of X only, T(t) is a function of T only. - reasonalbe assumption int( f ) = int(X*T) = T*F, F is primitive function of f. Since T is independent of x you can 'move it' outside the integral. Same goes for the derivation: T*F, dT*F/dx = TdF/dx = Tf = f However I do partially agree with gyabo, need to consider this a partial derivatice. "When you're working with a function that has 2 variables like f(x,t), f(x,t) is normally a surface. The slope of a surface at some point is a plane, not a line. Therefore, you can't attempt to differentiate a function of 2 variables with respect to only 1 variable because the slope of a surface cannot be a line." It can be soo much more then a suraface, consider a particle that is dependent on x-dimension and time. These kind derications are often used when solving PDE's | ||
|
Nexzico
Sweden93 Posts
On March 29 2008 04:48 Gyabo wrote: Be careful with what you're trying to accomplish. The bolded answers are what you should get when you integrate the function, not differentiate. Yea that was the thing i realized after reading through it again, I was fooled by the F(a)-F(b) and just assumed that he was asking about interigation, by bad -_- (just took a second look at my post, and I mixed the d/dx term with the "dx" term that usually is behind the function which you are integrating... soz again  | ||
|
Purind
Canada3562 Posts
On March 29 2008 05:57 JacobDaKung wrote: d/dx (integral f(x) dx) = f(x) d/dx (integral f(x,t) dx) = f(x,t) f = X*T X(x) is a function of X only, T(t) is a function of T only. - reasonalbe assumption int( f ) = int(X*T) = T*F, F is primitive function of f. Since T is independent of x you can 'move it' outside the integral. Same goes for the derivation: T*F, dT*F/dx = TdF/dx = Tf = f However I do partially agree with gyabo, need to consider this a partial derivatice. "When you're working with a function that has 2 variables like f(x,t), f(x,t) is normally a surface. The slope of a surface at some point is a plane, not a line. Therefore, you can't attempt to differentiate a function of 2 variables with respect to only 1 variable because the slope of a surface cannot be a line." It can be soo much more then a suraface, consider a particle that is dependent on x-dimension and time. These kind derications are often used when solving PDE's I'm not quite sure I understand this bolded part. You are taking a function of x and t (f(x,t)) and expressing it as a product of two functions, X(x) and T(t). I don't think you can necessarily do that. f(x,t) = sin (sqrt xt) How do you split that into products like you did there? On March 29 2008 04:41 Gyabo wrote: You can however, take partial derivatives to the 2 variable function. The partial derivative of f(x,t) with respect to x is the slope of the line tangent to the trace of a curve in the plane x=a, where a is part of an arbitrary point (a,b,c) within the surface in 3 dimensional space. To put it in more simple terms, a partial derivative is the slope of part of a function that has more than 1 variable. Your calculus book should probably have more info on this if you want to look into it. . I meant to put partials in the 2-D (or 3-D or however you wanna look at it, I see it as 2-D) but i didn't wanna bog down the post with partial f partial x and all that, so I kept it compact with my improper use of d. So is the same thing true when I do the partials instead of an ordinary d/dx? | ||
|
Gyabo
United States329 Posts
On March 29 2008 05:57 JacobDaKung wrote: It can be soo much more then a suraface, consider a particle that is dependent on x-dimension and time. These kind derications are often used when solving PDE's Ya this is true. Perhaps I should have clarified by saying "a possible geometric representation of a 2 variable function is...etc." | ||
|
JacobDaKung
Sweden132 Posts
since t is concidered a constant in respect to x. | ||
|
Purind
Canada3562 Posts
I just pointed out a function that disproves your sketchy assumption, but the explanation you just made is a reason why your earlier proof isn't even necessary. | ||
| ||
StarCraft 2 StarCraft: Brood War Rain Dota 2 4074Horang2 1983Shuttle 793firebathero  197scan(afreeca)  51Mong 40Rock 40sSak 32Aegong 27JulyZerg  19[ Show more ] Counter-Strike Other Games ceh9574 FrodaN503 KnowMe340 Lowko294 Fuzer  263Hui .254 Sick196 Liquid`VortiX189 Mew2King137 ArmadaUGS61 QueenE44 Trikslyr44 Beastyqt16 |
|
Replay Cast
WardiTV Korean Royale
OSC
Replay Cast
Replay Cast
Kung Fu Cup
Classic vs Solar
herO vs Cure
Reynor vs GuMiho
ByuN vs ShoWTimE
Tenacious Turtle Tussle
The PondCast
RSL Revival
Solar vs Zoun
MaxPax vs Bunny
Kung Fu Cup
[ Show More ] |
|
|
EPT
