Recent content by LostInSpace

Hi! I have a bezier curve defined by: \vec{b}(t) = (x(t), y(t)) where \begin{array}{lcl} x(t) &=& a_xt^3 + b_xt^2 + c_xt + x_0 \\ y(t) &=& a_yt^3 + b_yt^2 + c_yt + y_0 \end{array} for t \in \lbrack 0, 1 \rbrack. All constants are computed from vertices on the curve and control...

Hi I have been assigned a problem that I can't solve. I have a rotating axis to which a thread is connected in one end. The thread is perpendicular to the rotating axis. The thread has a charge density \lambda and a length L. First of all, I need a mean value of the charge density of...

Thanks for responding! Your interpretation appears to be correct. I agree with your solution, but I have now been instructed to use a Carnot-process to solve it, which really doesn't help me. Any ideas? Oh, and just a question regarding your formula: What does \Delta Q and T_c represent...

Hi! I have a problem that I simply can't solve. I was hoping you could push me in the right direction. English isn't my native language, so please excuse errors relating to that. A pump fetches water from the bottom of a sea where the water has a temperature of 4C (277K). It transfers this...

Thanks again for your help! Lets say I split the pipe into section of 5 metres. In a section I would get something like T_i = f(t) * T_{i-1} where f(t) is the decrease in temperatur as a function of time. Given the number of sections, the time can easily be found. But how to I found the...

Hi! Thanks for responding! And I'm sorry about the typo... it should be 60mm. What about the time factor? If you check at any given point in the pipe, the temperatur is always the same, right? I would like to know the temperatur after 1km, and I know the speed of the water so I know the time...

Hi! I have a problem that I need to solve. I was really hoping someone could help me get started. It should be simple... I have a pipe made out of copper with an inner diameter of 50mm and a thickness of 5mm (i.e. outer diameter is 55mm). Around this pipe is an isolating layer with a thermal...

Hi! I'm trying to solve an equation system \vec{\pi}\mathbb{P} = \vec{\pi} where \vec{\pi} = (\pi_1, \pi_2, \pi_3, \pi_4, \pi_5) and \mathbb{P} is a 5x5 matrix (constants). The problem is that the equation system is a bit to large to handle, at least for me. I remember that linear equation...

How come a signal modulated using Amplitud Shift Keying and/or Phase Shift Keying (such as QAM) require a bandwidth? I mean, since the frequence is constant; only the amplitude and phase is changed. Thanks in advance!

Are you sure about this? The height of the cylinder \Delta z \rightarrow 0, so the curve shouldn't be any problem. Or should it? Anyway, I assume you are right, but I just wanted to test this. But how do you integrate \int_a^b \sqrt{2t^2+1}\mathrm{d}t I tried to set t = \sinh s...

Ok... I was thinking about something like this as well: The hyperboloid can be defined as \lbrace (x,y,z) \mid x^2 + y^2 - z^2 = 1 \rbrace This set can be approximated by cylinders as \lim_{n\rightarrow\infty}\bigcup_{i=0}^n \lbrace (x,y) \mid x^2 + y^2 = 1 + z_i^2, z_i \le z \le...

Hi! I'm supposed to find the area of a hyperboloid within a sphere with the radius 2. The hyperboloid and the sphere intersect at \sqrt{\frac{3}{2}} and the intersecting curve is a circle with the radius \sqrt{\frac{5}{2}}. The hyperboloid is defined as \left\lbrace\begin{array}{lcl} x &=&...

I am a bit confused about taylor approximation. Taylor around x_0 yields f(x) = f(x_0) + f'(x_0)(x-x_0) + O(x^2) which is the tangent of f in x_0, where f'(x) = f'(x_0) + f''(x_0)(x-x_0) + O(x^2) which adds up to f(x) &=& f(x_0) + (f'(x_0) + f''(x_0)(x-x_0) +...

How about this: \lim_{x\rightarrow 0}\frac{\sin x}{x} = \lim_{x\rightarrow 0}\frac{x + O(x^2)}{x} = 1 + \lim_{x\rightarrow 0} O(x) = 1

Hi all! I'm having some problems with parametrization. I read somewhere that you should locate circles, ellipses, hyperboloides, paraboloides etc and use these elements to express a parametric function. But someone must have figured out how to do it! The way I see it, there's nothing logical...