Firstly, I'm not even sure how to frame this mathematically, but I'd be curious to know what kind of problem it is and what kind of subjects within maths it requires to be able to solve it.
Here's my problem--
I've set it up in excel to help you visualise it.
There is a different amount of...
I'm working on a green's theorem integral and I'm given a line (0,0) to (1,2) which i parameterise as (t,2t) but the answer is actually (1-t, 2(1-t)) and if I use my parameterisation on the integral I get a different answer!
Yeah, I don't really get why z=1 gives \tan\phi=\pm\sqrt{3} I can get this if I convert to spherical co-ordinates straight away but not by solving the intersection of the rectangular coordinates.
Also while there are no negative in spherical co-ordinate radians I get \phi=\pm\frac{pi}{3}
How is weight not linked to C? to find weight/mass you must take a line integral through that plane, and that line integral is C.
What I was confused about is if I take a 3cm wire and position it from (0,0) to (0,3) and then I move it to some other place in the xy plane suddenly it's weight...
Wasn't sure which section to put this q in.
Just reading now that f(x,y) can represent the density of a semicircular wire and so if you take a line integral of some curve C and f(x,y) you can find the mass of the wire... makes sense.
What I don't get is that if I then move the wire around the...
When I think line integral - I understand when I'm taking a line integral for a function f(x,y) which is in 3D space above a curve that the integral is this curtain type space, just like if you had a 2D function and you find the area under the curve, except now it's turned on its side and it's...
Cool, I see it now.
This.
x^{2}+y^{2}=3z^{2}
\rho^{2}\sin^{2}(\phi)\cos^{2}(\theta)+\rho^{2}\sin^{2}(\phi)\sin^{2}(\theta)=3\rho\cos^{2}(\phi)
\sin^{2}(\phi)=3\cos^{2}(\phi)
\tan^{2}\phi=3
\tan\phi=\pm\sqrt{3}
\phi=\pm\frac{\pi}{3}
Ok. I see you've found the intersection using the...
Homework Statement
Find VR_{z}^{2} = \int \!\!\! \int \!\!\! \int_{E} (x^{2} + y^{2})dV given a constant density lying above upper half of x^{2}+y^{2} = 3z^{2} and below x^{2}+y^{2}+z^{2} = 4z.
Homework Equations
The Attempt at a Solution
Why does it say upper half of x^{2}+y^{2} =...
I've often been plagued by the thought of being unintelligent and slow compared to others. I seem to take much longer to understand concepts and work out problems. With that said, I do really well at school (but also put in much more time than others). I'm also quite obsessive when it comes to...
I'm trying to understand a function which draws a koch curve using a library which draws (lt = left turn, fd = move forward, t represents an object class).
def koch(t, n):
if n<3:
fd(t, n)
return
m = n/3.0
koch(t, m)
lt(t, 60)
koch(t, m)
rt(t, 120)...
Hi,
I'm working through thinkpython and there is an exercise which requires drawing flowers and arcs. I'm having some trouble understanding the arc function.
def arc(t, r, angle):
"""Draws an arc with the given radius and angle.
t: Turtle
r: radius
angle: angle subtended by the...
Homework Statement
The boundary of a lamina consists of the semicircles y=\sqrt{1-x^{2}} and y=\sqrt{4-x^{2}} together with the portions of the x-axis that join them. Find the centre of mass of the lamina if the density at any point is proportion to its distance from the origin.
Homework...
I know what I did wrong and I finally understand the question. I was taking the second derivative of dz/dr and used the product rule but r is independent of \theta so when I treat \theta as a constant I get the same answer as you.
Thanks again for helping me out guys! much appreciated.
That left window with document structure is great - exactly what I'm looking for. It didn't work with my custom homework templates though as I'm using some other structure ( "problem" instead of chapter/section).
Also that jump to tex isn't working very well.
I think it's a software issue and...
But this was actually derived when finding dz/dr (z w.r.t to r)... How else would you come up with this operator? How then do I know I can remove the z and plug anything else in?
I have the same problem with \cos\theta\frac{\partial}{\partial r}\left[\frac{\partial z}{\partial x}\right]. How...
I know that when expanded that becomes \frac{\partial^{2}z}{\partial r^{2}}=\cos^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}+2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\sin^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}
(the answer) but I don't get how we go from...
:cry:
I will look this up again.
I just stumbled on that answer by chance because I was actually doing a product for some reason.
Now I get this:
\frac{\partial^{2}z}{\partial r^{2}}=\frac{\partial}{\partial r}(\cos\theta\frac{\partial z}{\partial x})+\frac{\partial}{\partial...
I did this and it's wrong. I've just taken the product, not the 2nd derivative (which I still haven't figured out) but for some reason my answer was correct.
hmmm maybe I should be splitting up my assignment question into different files then...
Is there an edit environment where I can have a main document and then click on the titled links to separate files?
Latex produces some really good looking documents and because of this I tried to pick it up. I was hoping it would make typing up my maths assignment a much quicker process.
What I find insanely frustrating is that because of all the marks up in a .tex file it becomes impossible to navigate...
Homework Statement
Use \frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}
and \frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y} to show that...
Ok, I see what I did. I was trying to take a derivative of a constant....
\frac{\partial z}{\partial r}(\frac{\partial z}{\partial r})=(\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y})(\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial...
Homework Statement
Use a double integral to find the area of the region enclosed within both circles of r=cosθ and r=sinθ
Homework Equations
The Attempt at a Solution
I begin by finding the region in polar co-ordinates.
For r=\cos\theta
0\leq r \leq\cos\theta...
Were not supposed to do it using a Taylor expansion. Last time I used polar co-ordinates to find a limit where no method was specified and lost marks. Besides, I forgot Taylor expansions.
Hey Tim. So I can't apply product rule to find second derivatives of that? If that was an equation without...