Cc Coordinate Algebra: Fill & Download for Free

Here is an example from applied differential equations in physics: Suppose an object of mass [math]m_1[/math] is suspended vertically from a spring with force constant [math]k_1[/math], and from that object a second spring is hanging with force constant [math]k_2[/math], at the end of which is attached a second object of mass [math]m_2[/math]. Given the initial positions and (vertical) speeds of the two objects, predict the positions of both at any given time [math]t[/math].The equations of motion for the system are[math]\frac{d^2 x_1}{dt^2} = m_1 g - k_1x_1 [/math][math][/math][math]+ k_2(x_2-x_1),[/math][math]\frac{d^2 x_2}{dt^2} = m_2 g - k_2(x_2-x_1).[/math]We can get rid of the gravitational terms by changing the coordinates to [math]x’_1 = x_1 - (m_1+m_2)g/k_1,[/math] and [math]x’_2 = x_2 - m_1g/k_1 - m_2g(k_1+k_2)/k_1k_2[/math]. The equations now become[math]\frac{d^2 x_1}{dt^2} = -(k_1+k_2)x’_1 [/math][math][/math][math]+ k_2x’_2,[/math][math]\frac{d^2 x_2}{dt^2} = k_2x’_1 - k_2x’_2.[/math]We may write this as[math]\frac{d^2 \vec x}{dt^2} = \left(\begin{array}{cc} -k_1-k_2 & k_2 [/math][math][/math][math]\\ k_2 & -k_2 \end{array}\right) \vec x.[/math]Linear algebra helps us find a basis of eigenvectors [math](u_1,u_2)[/math] so that these equations may be rewritten as[math]\frac{d^2 \vec u}{dt^2} = \left(\begin{array}{cc} -\omega_1^2 & 0 [/math][math][/math][math]\\ 0 & -\omega_2^2 \end{array}\right) \vec u;[/math]since the two coordinates [math]u_1, u_2[/math] are no longer coupled, it is easy to find the solutions[math]u_i = A_i \sin \omega_i t [/math][math][/math][math]+ B_i \cos \omega_i t,[/math]where the amplitudes [math]A_i,B_i[/math] depend on the given initial conditions.

When you work out dimensions, all you are doing is playing around with algebraic blobs to which we attach numbers to.So the fastes way to pressure is the barometer of metric water, as1 Pa = 0.1 mm * 1 g/cc * 1 dm/ds². That is, it’s a tenth of a millimetre of metric water under the gravity of 10 m/s².Don’t try to make too much out of the dimensions themselves. They are coordinates that things of pressure and energy density point to. The actual choice of base units and the number of dimensional bases is purely arbitary. I use density-velocity-time as base units.

There have been pretty good answers. Here's how one might do it mathematically (and very quickly).You're given 4 points, [math]p_1, p_2, p_3, p_4[/math]. Each pair of points (assuming 2D) is going to give us an equation of a line:[math]y-y_a = m_a (x - x_a)[/math][math]y-y_b = m_b (x - x_b)[/math]where [math]a,b[/math] denote a pair of the following possible combinations:[math]{\{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}}[/math]It's clear that of these, you only want subsets that use all 4 points once, so you want two lines such as[math]a = (p_1, p_2) \qquad b = (p_3, p_4)[/math]The slope is easily calculated by[math]m_a = \frac{p_{a,1,y} - p_{a,2,y}}{p_{a,1,x} - p_{a,2,x}}[/math]where [math]p_{a,1,y}[/math] denotes the [math]y[/math]-component of the first point of the pair of points [math]a[/math], etcetera. Going back to the above equations, there exists a solution if we can find a solution to[math]A \vec{x} = \vec{b}[/math]which is just a linear algebra problem. Here,[math]A = \left[\begin{array}{cc}-m_a & 1\\-m_b & 1\end{array}\right][/math] [math]\vec{x} = \left[\begin{array}{c}x [/math][math][/math][math]\\ y\end{array}\right][/math] [math]\vec{b} = \left[\begin{array}{c}p_{a,i,y} [/math][math][/math][math]+ m_a p_{a,i,x} [/math][math][/math][math]\\ p_{b,i,y} [/math][math][/math][math]+ m_b p_{b,i,x}\end{array}\right][/math]which has a solution assuming that [math]A[/math] is invertible which means the determinant of [math]A[/math] is non-zero. Therefore, two lines intersect if [math]\det(A) \neq 0[/math].*There are special cases to consider, such as when [math]m_a[/math] or [math]m_b[/math] is undefined (this occurs when two of the points have the same [math]x[/math]-coordinate; a vertical line). Unless both [math]m_a[/math] and [math]m_b[/math] are undefined -- then there will always be an intersection (why?).