.. oh wait, for say the first one do I get:
f_{1}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}
f_{2}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}
f_{3}(x,y,z) = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}
f_{4}(x,y,z)...
So as an answer to the first part of the question, I get that the extrema are:
f(x,y,z) = \left(\frac{-2\lambda}{3} , \frac{-2\lambda}{3} , \frac{-2\lambda}{3}\right)
since \lambda = \frac{\sqrt{3}}{2}
.. is this sufficient to state just this? I thought I'd get a list of points maybe...
Homework Statement
Suppose an initial number density of ink particles (i.e. number per unit length) is
given by:
f(x) = 2Nx ; for 0 < x < 1
f(x) = 0 ; otherwise.
Suppose also add a point source containing N molecules at the point x = − 1.
(a) Showing that the initial total...
OK, so I now know the values of x , y , z and of \lambda which should be:
\lambda = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}
SO that answers part i of the question (where x, y, and z are all non-zero), yep?
Now for part ii (where x = 0), I just do the same but with x = 0? I.e...
.. so from the first three equations I get:
x = \frac{-2\lambda}{3}
y = \frac{-2\lambda}{3}
z = \frac{-2\lambda}{3}
.. then square these values and input into the fourth equation, which gives:
\frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} - 1 = 0...
I can only see how to solve those first three equations in terms of a quadratic? I.e. that:
\frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0 \implies x = \frac{-2\lambda \pm \sqrt{(2\lambda)^{2}}}{6} \implies x = -3\lambda \pm 2\lambda
.. is this the right way to go?
Ah ok, so it should be:
\frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0
yes?
.. and then how to I go about getting values of x, y, and \lambda? Not sure how to equate those 4 simulataneous equations :confused:
Ermm.. so need to do this:
f(x,y) = x^{3} + y^{3} + z^{3}
g(x,y) = x^{2} + y^{2} + z^{2} = 1
Hence need to then extremise:
F(x,y,\lambda) = f + \lambda y = x^{3} + y^{3} + z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 1\right)
So need to then calculate partial derivatives of this...
Thought I may have made a mistake with calculations of the partial derivatives :frown:
So, should be then:
\frac{\partial F}{\partial x} = 2 + 2x\lambda = 0
\frac{\partial F}{\partial x} = 2 + 2y\lambda = 0
\frac{\partial F}{\partial x} = \left(x^{2} + 4y^{2} - 1\right) = 0
...
.. I was doing in the \lambda way (um.. lagrange multiplier?!) as this is how I was taught a similar problem and hence should really use this way.
Looking back through my notes on this, once the values of x and y have been found (as they have been) then just put these into the original...
Homework Statement
Considering the surface of a sphere of radius 1 with its centre at coordinates
(0,0,0).
For the function:
f(x,y,z) = x^{3} + y^{3} + z^{3}
Need to find the following:
(i) All the extrema on the surface which have x, y and z all non-zero simultaneously...
erm.. so for the first set of values I have:
f(x,y) = xy - \left(\frac{y^{3}}{3}\right) = \left(\left(2+\sqrt{2}\right)\left(-1-\sqrt{2}\right)\right) - \left(\frac{\left(-1 - \sqrt{2}\right)}{3}\right) = \left(-4 - 3\sqrt{2}\right) - \left(-7 + 5\sqrt{2}\right) = 3 + 2\sqrt{2}
which is...
Homework Statement
A rectangle is placed symmetrically inside an ellipse (i.e. with all four corners
touching the ellipse) which is defined by:
x^{2} + 4y^{2} = 1
Find the length of the longest perimeter possible for such a rectangle.
Homework Equations
Within the problem...
um.. nothing! I don't know why I thought there was an issue there earlier! :redface: .. maybe just going a bit mad! :wink:
Right, so since now have:
y = -1 - \sqrt{2}, x = 2 + \sqrt{2}
y = -1 + \sqrt{2}, x = 2 - \sqrt{2}
.. then there are just 2 stationary points of the function...
OK.. So from the first partial derivative equation I get that:
y = -\lambda
.. then put this into the second partial derivative equation, which gives:
x - \lambda^{2} + \lambda = 0
.. hence:
x = \lambda^{2} - \lambda
.. then put this into the third partial derivative equation...
Homework Statement
Finding the stationary point(s) of the function:
f(x,y) = xy - \frac{y^{3}}{3}
.. on the line defined by x+y = -1.
For each point, state whether it is a minimum or maximum.
Homework Equations
.. within the problem statement and solutions.
The Attempt at a Solution...
.. so you mean for the second part it will be this:
\int_{0}^{\infty} r e^{-\left(\frac{2}{a}\right)r}dr = \left(\frac{r}{\left(\frac{2}{a}\right)}-\frac{1}{\left(\frac{2}{a})^{2}}\right)\right|^{\infty}_{0} = \frac{4\pi}{\left(\frac{2}{a}\right)^{2}} = \pi a^{2}
??
Going back to the...
um, don't get how I can rearrange that how you said :confused:
.. also, so to get the final answer I need also caclulate this:
\int_{0}^{\infty} r e^{-\left(\frac{2}{a_{0}}\right)r}dr =...
erm.. yep. so should be:
\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(\frac{r}{\left(\frac{2}{a}+\gamma\right)}-\frac{1}{\left(\frac{2}{a}+\gamma\right)^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0} =...
so it should be:
\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(-\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)
better?
.. that's what I've used, obviously with the value of \alpha substituted.
.. OH wait, it's \alpha^{2} not \alpha, so then the result should be:
=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)^{2}}
correct now?
These are my exact full calculations:
V_{\gamma}(r) = -\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}
\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> = 4\pi...