Recent content by cwrn

I got a quick question regarding the potential energy of a fluid (see picture). If I want to express the potential energy of a fluid (with a given volume) above the reference line x, using nothing but the given variables/constants, would this be the correct way to do so? picture

Using the same method in 4) as in 3) gives me $$ \begin{align} f_Y(y) = \frac{1}{\sqrt{2\pi y\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} \end{align} $$

I see, I must've misinterpreted, my bad! Thanks again.

Of course. It makes a lot more sense to look at the substitution in an integration. Thanks for the help. Will try part 4) and post the result later. Edit: What do you mean by "yours will integrate to 1/2"? Do I need to change the limits according to z(y)?

I gave part 3 another try using the so called "change of variable" technique. Although, instead of finding and taking the derivative of the CDF I used the definition of normal distribution (pdf) and just substituted z. I set z=g(y)=\sqrt{y} \Rightarrow \frac{dg}{dy}=\frac{1}{2\sqrt{y}}...

Alright, according to the definition, the distribution should be $$ \begin{align} F_Z(u) = 1-(1-F_X(u))(1-F_Y(u)) = 1-P(X>u)P(Y>u) \end{align} $$ where F_Y(u)=\sum_{k:k\leq u}q^k and F_X(u)=\sum_{k:k\leq u}p^k.

Homework Statement Two independent series of experiments are performed. The probability of a positive result (independent of each other) in the respective series are given by p and q. Let X and Y be be the amount of experiments before the first negative result occur in the respective series...

Yes, I'm aware of this. Solving this with ode45 should yield the x and y vectors that I want to plot. I'm just not sure how to define the function itself since it depends on B which in turn depends on y. $$ \begin{align} B(y) = B_{0}e^{-y/y_{0}} \end{align} $$

I'm working on a little project where I want to plot the motion of a projectile with air resistance. The air resistance can be assumed to be proportional to the velocity squared. F_{f}=-Bv^{2} F_{f,x}=F_{f}\frac{v_{x}}{v}, \ \ F_{f,y}=F_{f}\frac{v_{y}}{v} where B depends on the height...

I understand your point, but the particular solution is still confusing to me. Would you mind further explaining how G(u) becomes G(v)?

Plugging in the values for the partial derivatives gives \frac{\partial}{\partial x} \left[4\frac{\partial F}{\partial v}+F(u,v) \right] = 0 = \frac{\partial}{\partial x}g(y). Does that mean the general solution is F(u,v) = g(y)-4\frac{\partial F}{\partial v} where g(y) and...

But how would I go about determining the function g(y) from this? f(x,y)=g(y)-4\frac{∂f}{∂v} The equation above doesn't seem like a sufficient answer.

That makes sense. The only thing I don't quite understand is why it equals a function depending only on y.

Homework Statement Find the general solution f = f(x,y) of class C2 to the partial differential equation \frac{\partial^2 f}{\partial x^2}+4\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial f}{\partial x}=0 by introducing the new variables u = 4x - y, v = y. Homework Equations...

I am soon to apply for university and I am not quite sure what I want to study. Theoretical physics and astrophysics/astronomy/cosmology all seem like very interesting fields of physics to me. How do I go about when I choose my career path? Which one of the fields has been more prospering than...