Recent content by jrcdude

D'oh I think the form I was looking for was: x^2+2x(y+z)+(y-z)^2 which is clearly greater than zero. Thanks for the insight.

Homework Statement Show that the descriminant of the characteristic polynomial of K is greater than 0. K=\begin{pmatrix}-k_{01}-k_{21} & k_{12}\\ k_{21} & -k_{12} \end{pmatrix} And k_i > 0 Homework Equations b^2-4ac>0 The Attempt at a Solution I have tried the following...

Yeah, we too let Maple do the grunt work. Funny thing is Maple was developed at my university so they push us to use it.

The next part of the question was actually to graph the forcing function and solution on the same set of coordinates. So you are interpreting it correctly.

Yes, that subscript notation is definitely horrible as it caused me a lot of grief haha. And yes, by "it works" I meant I solved the DE resulting in the solution of, y(t)=h(t)+2\sum_{k=1}^{n}(-1)^{k}h(t-k\pi) where...

Courtesy of Brannan, The only other possibility is that c=k\pi. Edit: Yes I think that's actually the problem faced here.. I believe it works if I let u_{k\pi}(t)=u(t-k\pi)

I'm sorry, I thought it was common notation that u_{k\pi}(t) is the indicator function denoting the unit step function: u_{k\pi}(t)=u_{k}(t)-u_{\pi}(t)=\begin{cases} 0, & t<k\quad\text{or}\quad t\ge\pi\\ 1, & k\le t<\pi \end{cases} As for what I got for the Laplace transform of the right...

Homework Statement y''+0.1y'+y=1+2\sum_{k=1}^{n}(-1)^{k}u_{k\pi}(t) and quiescent initial conditions. Homework Equations None. The Attempt at a Solution (s^{2}+0.1s+1)Y(s)=\mathcal{L}\{1\}+2\sum_{k=1}^{n}(-1)^{k}\mathcal{L}\big\{ u_{k\pi}(t)\big\} I'm not sure if this step was...

My main problem with understanding this, is shouldn't the line that I added to your picture: be perpendicular, resulting in 25N of force on each side? I do not understand why you cannot do this.

Looking at it like this: Now I'm wondering if it's possible to have a 90 degree angle for the vertical, while still connect to make the parallelogram. Wouldn't the diagonal line represent the 50 N force, and the 2 lines going to the horizontal add up to be 50, but are different values?

What is stopping this from happening then? (Where the 2 points do not connect)

So is it fair to say you cannot use the parallelogram method, because it would have a different vertical tension on each side of the parallelogram?

Do you think you could please draw a proper diagram for me? I am not understanding what you are saying.

Based on the diagram provided above, wouldn't they have to? Looking at the top triangle: cos65 = adj/T1 and looking at the bottom triangle: cos65 = adj/T1 Therefore they should be same, as based on the fact that parallelograms' diagonals bisect.

I actually mean that the 50N downward force was divided into 2, so there would be a 25N force on each side for the vertical. This is based off of parallelogram properties.