| 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 Dota 2 Counter-Strike Super Smash Bros Heroes of the Storm Other Games |
|
PiGosaur Cup
The PondCast
Kung Fu Cup
IntoTheTV X SOOP
Reynor vs Zoun
Neeb vs Rogue
FEL
Patches Events
AI Arena Tournament
FEL
Sparkling Tuna Cup
Replay Cast
[ Show More ] |
|
|
EPT
