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...
Homework Statement
This is not a book problem but rather just a question about how the book got here.
\frac{d^2u}{dx^2}+u=e^{x} \quad \mbox{with} \quad u(0)=0 \quad \mbox{and} \quad u(\pi)=0
\Rightarrow \frac{d^2\phi}{dx^2}=-\lambda \phi \quad \mbox{with} \quad \phi(0)=0 \quad \mbox{and} \quad...
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 :-(
[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...
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.
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...
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...