I'm going over the photoelectric effect and have run into a conceptual problem, and was hoping for some help in resolving it. In particular, I am looking at the frequencies below the threshold frequency of whatever metal is being examined.
So, because of the lower frequency, there will be no...
Wow, interesting. So \dot{z}=iωz → \dot{z}-iωz=0 because iz=ix-y. Thank you! Sorry, one more question. Now that I have that, when solving it I get z=z_{0}e^{ωt}, but the answer has an additional phase angle \phi in it, so I was just wondering if there was a quick explanation for that.
It's...
Yes. I then get \dot{z}=-ωy+iωx=ω(-y+ix). So I can sort of see the relation here now, except my variables seem to be the wrong way, and I have no idea where the \dot{z}-iωz=o comes from still, because i would be gone if the variables were the proper way and I subbed in z for the x-iy.
Hi. I'm not really sure what to do. As far as I know, taking \dot{z}=\dot{x}+i\dot{y}, and \dot{y}=ωx and \dot{x}=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.
Homework Statement
I'm given two equations for coordinates of a certain particle in the xy plane, \dot{x}+ωy=0
and \dot{y}-ωx=0.
Then using the complex variable z=x+iy, find the differential for z, and solve it. Hence give x and y as functions of time.
Homework Equations
The...
Homework Statement
I'm trying to define a vector space over Q. Does this make any sense?
Homework Equations
The properties of a vector space
The Attempt at a Solution
Let V=Q^2 over Q. It seems to me that everything would be defined and I shouldn't be able to do anything to a...
Homework Statement
Let D be the triangular domain given by 0\leq y \leq3, (y/3)-1 \leq 1-(y/3). Then
\int\int (e-x^{5}e^(sqrt(1+y^2))
Homework Equations
The Attempt at a Solution
There is a quick way to solve it by breaking apart the double integral and then, apparently the x^5...
Thank you both very much! Those two posts cleared up my problems with it. I was thinking of it in an accumulating way :uhh: (I have no idea why, series and sequence mix up I guess :blushing:), so now it makes sense.
:cool: Thanks again!
Sorry, I wasn't clear enough. I'm just not seeing how it shows that. I've memorized it and am not really worried about getting it wrong, but it's really not helpful if I don't understand that it keeps under the limit. I guess my induction skills are lacking.
Ok, so if ak < 3, then ak+1 =...
Homework Statement
Let a_n be defined recursively by
a_{1}=1, a_{n+1}=sqrt(6+a_{n}) (n=1,2,3,...).
Show that lim n->infinity a_{n} exists and find its value
The Attempt at a Solution
Observe that a_{2}=\sqrt{6+1}=\sqrt{7} > a_{1}. If a_{k+1} > a_{k}, then a_{k+2} = \sqrt{6+a_{k+1}} >...
Perhaps I'm going about this completely wrong then. How would I start out solving y''+25y=cot(5x).
(I should be more specfic, when I say solve, I mean to find A solution, so y_h and y_p. I'm just not sure about how to find the y_p=u1(y_1)+u2(y_2))
Homework Statement
y''+25y=cot(5x)
Find one possible solution
The Attempt at a Solution
I don't have any background in linear algebra, so I can't use cramers rule as a heads up, so I have to solve the system of equations (no linear algebra for this course is needed).
Ok, so I take...
Homework Statement
y''-49y=7cos(7x)+7+e^(7x)
The Attempt at a Solution
I have no idea how to solve this Differential equation. I could solve one that has y''-49y=one term, but I'm stumped with more than one.
First, I get the homogeneous equation, y''-49y=0 and fine y_c, then use the...
Homework Statement
\frac{dy}{dx} = \frac{3xy}{3x^2+7y^2}, y(1)=1
Express it in the form F(x,y)=0
The Attempt at a Solution
I'm not sure where I'm going wrong. I let v=y/x,
v+x\frac{dv}{dx} = \frac{3x^2v}{3x^2+7x^2v^2}=\frac{x^2(3v)}{x^2(3+7v^2)}= \frac{3v}{3+7v^2}
\Rightarrow...
Homework Statement
Determine if the following is improper and convergent, improper and divergent, or proper
\int \frac{dx}{\sqrt[3]{x^2 - 7}}
from 8 to infinity
The Attempt at a Solution
Since I don't know how to...
Homework Statement
I'm confused about integrating something like \int\frac{1}{u}
Sometimes the answer seems to be ln|u| and sometimes just ln(u), and I wasn't sure why it is different from problem to problem. (after subbing u back in; I'm using the u-sub method)
It looks like the answer should...
Ok, so
\frac{ds}{dh} = \frac{10}{4-h} + \frac{10h}{(4-h)^2}, when the height, h, is 3, is equal to 40. So, then \frac{dh}{dt} * 40, and solving the dh/dt for 1m/hr, makes the shadow have an increase of 40 m/hr at h=3. Hopefully I'm starting to get it.
Sorry, I'm not quite sure what I'm doing. I got the 3 for the height from the question, asking the shadows rate of increase at 3. It's a one variable calculus course, so I was avoiding t to keep it down to one variable, I'm not sure at all how to do differentiation with two variables. I could...
Interesting. So,
\frac{ds}{dh} = \frac{10}{4-h} + \frac{10h}{(4-h)^2}
where h is 3, the shadows rate is increasing at 40.... metres every 3 hours? That has to be wrong, since it doesn't really make any sense to me. I'm unsure of what the units would be.
And, when it approaches 4, the...
I think I understand what you mean (for a single variable course). And if I have 3 variables, that is one too many. So, since h is variable, then 1 must be variable t, and I can state it as 4-h. So, \frac{4-h}{10} = \frac{h}{s}, so s=\frac{10h}{4-h}, and s'(h)=\frac{10}{4-h} +...
Ok, so taking that into account, I have \frac{1}{10} = \frac{h}{s}, which then can be solved to become s=10h. s'=10, so the rate at which the length s is increasing is 10 metres per hour. I'm still not sure where the stalk growth of 1 metre per hour comes in at then.
Thanks, I'm studying for a final, and on the finals from prior years there seems to be questions like this one asked.
So, after drawing the other line, I can see two triangles in the diagram. I know the details of the one triangle, so if I remember correctly, I can find the details of the other...
Homework Statement
A magic stalk starts growing at a spot 10 metres from a 4 metre lamppost. The stalk grows 1 metre per hour. Find the rate at which the length 's' of the stalk's shadow is increasing at the instant when the height h of the stalk is 3 metres.
The Attempt at a Solution...
Interesting. Let me see if I understand.
x(y)-2y=1-2x <----Am I to factor(not sure if that's the proper word) the left side to get that final answer then, similar to what you showed?
And then I get y(x-2) after the factor
Which then is y(x-2)=1-2x
And then I can get my final answer of...