Homework Statement
Let B be an m×n matrix with complex entries. Then by B* we denote the n×m matrix that is obtained by forming the transpose of B followed by taking the complex conjugate of each entry. For an n × n matrix A with complex entries, prove that if u*Au = 0 for all n × 1 column...
[PLAIN]http://img815.imageshack.us/img815/9894/fourier5313.png [Broken]
there was a minus sign. oops.
In any case, relevant equations:
F_{c}(f(x,t)) = \sqrt{\frac{2}{\pi}} \int^{\infty}_{0} cos (\lambda x) f(x,t) dx
F_{c}(f''(x,t)) = -\sqrt{\frac{2}{\pi}} f'(0,t) - \lambda^{2} F_{c}(f(x,t))...
Homework Statement
Find the solution u, via the Fourier sine/cosine transform, given:
u_{tt}-c^{2}u_{xx}=0
IC: u(x,0) = u_{t}(x,0)=0
BC: u(x,t) bounded as x\rightarrow \infty , u_{x}(0,t) = g(t)
2. The attempt at a solution
Taking the Fourier transform of the PDE, IC and BC...
Well, then I would have
m_{2}\ddot{x_{2}}-m_{1}\ddot{x_{1}} = -kq
...which, from what I see, doesn't do a whole lot because I still can't factor the left side to do anything.
Homework Statement
Two masses, m1 and m2, are connected to each other by a spring with a spring constant k. The system moves freely on a horizontal frictionless plane. Find the natural frequency of oscillation.
Homework Equations
F = -kx
F = ma
The Attempt at a Solution
Let m1 be the mass...
Homework Statement
Given u_tt = F(x,t,u,u_x, u_xx), give the finite difference approximation of the pde (ie using u_x = (u(x + dx; t) - u(x - dx; t))/(2dx) etc.)
Homework Equations
Well, clearly, u_x = (u(x + dx; t) - u(x - dx; t))/(2dx)
The Attempt at a Solution
I really have no idea how...
Homework Statement
u_t = -{{u_{x}}_{x}}
u(x,0) = e^{-x^2}
Homework Equations
The Attempt at a Solution
The initial state is a bell curve centred at x=0. The second partial derivative of u at t=0 is {4x^2}{e^{-x^2}}, which is a Gaussian function, which means nothing to me other than its...
tiny-tim, thanks for those. desperately needed.
hallsofivy, I did draw the diagram, and found that it was rather inconveniently symmetrical, which was why I got stumped. going by your suggestion, from θ = 0 to θ = π/2, I would be integrating along the right side of the curve, where the...
Homework Statement
Find
\int{\int_{D}x dA}
where D is the region in Q1 between the circles x2+y2=4 and x2+y2=2x using only polar coordinates.
The Attempt at a Solution
Well, the two circles give me r=2 and r=2 cos \theta, and the integrand is going to be r2cos \theta, but I have no...
*backtracks*
ooooooooooookay. so.
when x=0, f_x = -2, but f_x = 0, so x=0 is not a solution.
which brings me back to the points (-1,0) and (1,2). looking at my calculations for (-1,0), turns out that it's horrendously wrong.
At (-1,0), f_xx = -2(6-2) = -8, D = (-8)(-2) - (0-3+2)^2 = 16 - 1...
well, from f_y, I can deduce that x=0 or yx^2 - x - 1 = 0.
If x=0, from f_x, y= -2.
If yx^2 - x - 1 = 0,
y = (x+1)/(x^2)
Subbing into f_x,
-2(x^2 - 1)(2x) - 2((x+1)-x-1)(2x(x+1)/(x^2) - 1) = 0
-2(x^2 - 1)(2x) = 0
x(x^2 - 1) = 0
x = 0 (as above), x = 1 or x = -1
If x = 1, y = 2
If x = -1...
Homework Statement
Show that f(x,y) = -(x^2 - 1)^2 - (yx^2-x-1)^2 has only two critical points, and both are maxima.
The Attempt at a Solution
Set partial derivatives (wrt x and y) to zero to find critical pts.
f_x = -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1) = 0
f_y = -2(yx^2 - x -...
ahh. of course.
r^2 = x^2 + y^2, and sin theta = y/r, where r is the square root of r^2. I indeed was making it too complicated for myself. thanks everyone.
I want to convert del g into cartesians so I can find a directional derivative, considering the direction vector is in cartesians. I know how to calculate the directional derivative; I just don't know how to convert del g into cartesians from polars.
Homework Statement
Find the gradient vector of:
g(r, \theta) = e^{-r} sin \theta
Homework Equations
The Attempt at a Solution
I know how to get gradients for Cartesian - partially derive the equation of the surface wrt each variable. But I have no idea how to do it for...
let's start with the second inequality. xy goes from 1 to 3. this forms one dimension of a rectangle - along the xy-axis. but instead of using an xy-axis, you could use a u-axis or v-axis, if you let u or v equal to xy. Hint hint.
now for the first inequality, if y goes from x^2 and 2x^2, is...
Ok, so now I have
3t^2 \frac {\partial s}{\partial y} = -6s \frac{\partial t}{\partial y}
\frac {\partial s}{\partial y} = \frac{-2s}{t^2} \frac{\partial t}{\partial y}
I sub this back into the first equations to get \frac{\partial t}{\partial y}. However, I get a different value for...
But how would I get the partial derivative of t wrt y from there? I mean, aren't we assuming t is a function of x and y (and with s as well?) ?
So let me try this out:
8y*s^2*t + 4y^2(2 \partial s / \partial y)t + 4y^2*s^2*( \partial t / \partial y) = -7( \partial s / \partial y)*t^2 - 7s(2...
Homework Statement
If the equations
x^2 - 2(y^2)(s^2)t - 2st^2 = 1
x^2 + 2(y^2)(s^2)t + 5st^2 = 1
define s and t as functions of x and y, find \partial^2 t / \partial y^2
The Attempt at a Solution
Equating the two, we get 4y^2*s^2*t = -7s*t^2. My main problem is, as simple as this...
How would I adjust for the "cube of the distance" though ? Looking at the numerator of that unit vector, there isn't much, I think, that one can do with (x\hat{i} + y\hat{j})^3
Homework Statement
A particle is attracted towards the origin by a force proportional to the cube of its distance from the origin. How much work is done in moving the particle from the origin to the point (2,4) along the path y = x^2 assuming a coefficient of friction \mu between the particle...
The factor is \sqrt{1+(dy/dx)^2}, which I'm sure I accounted for above.
dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}
It might not be "dr", but I'm 99% sure that's the calculation of the factor.
Homework Statement
Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.
The Attempt at a Solution
x-axis means y = 0
When y = 0, x = 0, -1 or 1.
Since this curve is "the infinity symbol", the curve has symmetry at x = 0.
Isolating y,
8y^2 =...
Yes, but unforunately, my Cartesian answer not the same.
I see now why though:
c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} \int_{-\sqrt{2c^2 - x^2}}^{\sqrt{2c^2 - x^2}} ({{\sqrt{3c^2 - x^2 - y^2}})^{-{\frac{1}{2}}}}dy dx}
= -c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} ({\sqrt{3c^2 - x^2 -...
But now I have a new problem with the surface area done in Cartesians. When z=c,
2c^2 = x^2 + y^2
where
-\sqrt{2c^2 - x^2} \leq y \leq \sqrt{2c^2 - x^2}
and
-\sqrt{2c^2} \leq x \leq \sqrt{2c^2}
dS for the sphere is {\sqrt{\frac{3c^2}{3c^2 - x^2 - y^2}}}dy dx
Surface area =...
Well, the other part of the question (which I had done already) was doing it in Cartesians, so I know the two surfaces intersect at z = c, as you stated.
So,
(c \sqrt{3}) cos(\varphi) = c
cos (\varphi) = 1/{\sqrt{3}}
\varphi = a really ugly number.
Should I just keep this as cos^{-1}...
Homework Statement
Find the surface area of the portion of the sphere x^2 + y^2 + z^2 = 3c^2 within the paraboloid 2cz = x^2 + y^2 using spherical coordinates. (c is a constant)
Homework Equations
The Attempt at a Solution
I converted all the x's to \rho sin\phi cos\theta, y's to \rho...
That was what I was looking for, yes. >_> Because I wasn't sure if trying to get away with three separate answers would be a good idea. (One for each region.)
Yes, but then it leads to my next question: are forces on the hemisphere equal to forces on the surface? (If so, I can probably just equate the surface integrals to each other. But then what happens to the hidden disk?)
Homework Statement
A uniform surface charge lies in the region z = 0 for x^2 + y^2 > a^2, and z = \sqrt{a^2-x^2-y^2} for x^2 + y^2 \leq a^2. Find the force on a unit charge placed at the point (0,0,b)
Homework Equations
dF = G (\delta)(vector of point to surface) / (magnitude of surface)^3...
Homework Statement
Given a curve C that starts from the origin, goes to (1,0) then goes to (0,1), then back to the origin, find the centroid of the enclosed area D.
Homework Equations
\bar{x} = {1/(2A)}*\int_C {x^2 dy}
\bar{y} = -{1/(2A)}*\int_C {y^2 dx}
The Attempt at a Solution
Well...