# Search results

1. ### Calculating Extrema on Surface of Sphere

.. 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)...
2. ### Calculating Extrema on Surface of Sphere

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...
3. ### Calculating Extrema on Surface of Sphere

.. did you post a reply? I got a notification email but it's not showing up here :confused:
4. ### Initial Number Density of Ink Particles

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...
5. ### Calculating Extrema on Surface of Sphere

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...
6. ### Calculating Extrema on Surface of Sphere

.. 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...
7. ### Calculating Extrema on Surface of Sphere

so: 3x + 2\lambda = 0 \implies x = \frac{-2\lambda}{3} ?
8. ### Calculating Extrema on Surface of Sphere

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?
9. ### Calculating Extrema on Surface of Sphere

\frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0 \frac{\partial F}{\partial y} = 3y^{2} + 2y \lambda = 0 \frac{\partial F}{\partial z} = 3z^{2} + 2z \lambda = 0 \frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0
10. ### Calculating Extrema on Surface of Sphere

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:
11. ### Calculating Extrema on Surface of Sphere

Ok, good. Yes, I just missed that, so I should have: \frac{\partial F}{\partial x} = 3x^{2} + y^{3} + z^{3} + \lambda \left(2x + y^{2} + z^{2} - 1\right) = 0 \frac{\partial F}{\partial y} = 3y^{2} + x^{3} + z^{3} + \lambda \left(2y + x^{2} + z^{2} - 1\right) = 0 \frac{\partial...
12. ### Calculating Extrema on Surface of Sphere

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...
13. ### Rectangle Inside An Ellipse

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 ...
14. ### Calculating Stationary Points of a Function

.. 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...
15. ### Calculating Extrema on Surface of Sphere

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...
16. ### Calculating Stationary Points of a Function

OH.. I just forgot them! :redface: ok, so: f(x,y) = xy - \left(\frac{y^{3}}{3}\right) \frac{\partial f(x,y)}{\partial xy} = 1 - y^{2} \frac{\partial f(x,y)}{\partial xy} = 2y .. correct? (I'm not too good on partial derivations :frown:) So then: 2y = 2\left(-1 -...
17. ### Rectangle Inside An Ellipse

? :confused: ?
18. ### Calculating Stationary Points of a Function

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...
19. ### Rectangle Inside An Ellipse

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...
20. ### Calculating Stationary Points of a Function

.. calculate second derivatives? negative = max point, positive = min point?
21. ### Calculating Stationary Points of a Function

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...
22. ### Calculating Stationary Points of a 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...
23. ### Calculating Stationary Points of a Function

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...
24. ### Yukawa Potential

.. 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...

..huh?
26. ### Yukawa Potential

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 =...
27. ### Yukawa Potential

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} =...
28. ### Yukawa Potential

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?
29. ### Yukawa Potential

.. 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?
30. ### Yukawa Potential

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...