Recent content by PiRho31416

Simon, Thanks so much! :) Adrian

I just want to check to see if I did the problem correctly. A person pushes a 2.0 kg box across a flat, horizontal surface with a force of 5.0 N for 4.0 m. Determine the acceleration of a box if the friction coefficient is 0.2. Determine the acceleration of the box if the friction coefficient...

Oh crap. The negative sign! Sorry it was late last night and somehow I thought he was saying \mathcal{L}\{y(t)\}\neq Y Sorry about that. He even says it's +4y not -4y. Geez :-(

Pretty sure it is.

Thanks! Duh!

[Solved] Laplace Transform Homework Statement \frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0 Homework Equations Laplace transform is defined as: \mathcal{L}\{f(t)\} = \int_{-\infty}^{\infty}f(t)e^{st}dt The Attempt at a Solution...

Thanks for correcting me. It was really late last night :-)

Yup. No problem. Feel free to ask. I'm struggling with math myself :-)

As you mentioned f''(x)=-6x-6=-6(x-1). So there is a root at x=1. Let's look at when f''(0) and when f''(2). f''(0)=-6(0)-6=-6 which means concave up. f''(2)=-6(2)-6=-12-6=-6 which means concave down.

Notice my f(x) is different from your problem. I'm making up a whole new problem. I was hoping you take take my example and figure out your problem.

Let's take the last part of your equation for an example. f(x)=-3x^{2}-4x-2 f'(x)=-6x-4 The root of the first derivative is when x=-2/3. So let's plug in -1 and 0 since it's to the left and right of the critical point. f'(-1)=-6(-1)-4=2 So from -\infty to -1 it is increasing. Now let's plug in...

Sorry latex fail :-P It's fixed now. If you need more examples, feel free to ask.

It is concave down when the second derivative is negative. It is concave up when the second derivative is positive. Let's say f(x)=x^{3}. f'(x)=3x^{2} f''(x)=6x First derivative tells you whether the function is increasing or decreasing. Second derivative tells you the concavity. If the...

I'm using lyx and I'm having difficulty trying to find the code for this vertical line. Any suggestions? I use the | but it's extremely small.

Got it! Dang those minus signs :-P Thanks!