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)

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

March 28 2008 18:49 GMT

I think it's true..(except when you have f(x) you are in 2 dimentions,the x-axis and y-axis (i think it's called axis) tells you this.

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

March 28 2008 19:41 GMT

I don't think that's quite right. First of all, your 1st equation is incorrect, despite you explaining it correctly.

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

March 28 2008 19:48 GMT

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

March 28 2008 20:57 GMT

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

Nexzico

Sweden93 Posts

March 28 2008 20:58 GMT

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

March 28 2008 21:09 GMT

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

March 28 2008 21:20 GMT

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

March 28 2008 21:53 GMT

for f(x,t) = sin (sqrt xt), I m not quite sure but it is probably doable with eulers identeti's or complex exponetials. However it isnt needed since, int(sin(sqrt(xt)dx) <=> int(sin(C*sqrt(x))dx
since t is concidered a constant in respect to x.

Purind

Canada3562 Posts

March 29 2008 16:12 GMT

#10

Yeah it just occured to me, d/dx (partial, not regular d) is the derivative wrt x while keeping t constant, so we can replace d/dx (partial) f(x,t) with d/dx (ordinary) f(x) (at to) where to is some random time that you are keeping constant, effectively changing the 2-D problem into a 1-D one.

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.