/sci/ - Science & Math

>>11544729
>>11544741
Instead of being a total piece of shit, you should at least put on a classic movie, or start a good book. You may go for a walk.
>>11544443
>what does the laplace transform actually do?
It maps a function of time to a function of [math] s=\sigma+j\omega [/math]; AKA time domain to frequency domain.
>but what, geometrically/intuitively, does that help accomplish?
Say [math] y [/math] is a real function of time. Its graph is a depiction of how magnitude changes with time. [math] \mathcal{L}\{y\} [/math] is a complex function, however, with both a magnitude and phase. Its graph shows how the magnitude and "direction" of a signal changes with frequency. If you are trying to figure out how the voltage over a passive component changes with the frequency of an input voltage somewhere else in a circuit, a log plot of the ratio of the Laplace transforms of output to input gives you a very quick and handy way to visualize the relationship. Much, much easier than working with ODEs in the time domain.
>What is it about multiplying a function with exp(-st) and integrating
The only functions that can be transformed are of exponential order. This means that for [math] y(t) [/math] there exists some large [math] T [/math] and constants [math] k, c [/math] such that [math] |y(t)|<M\exp ct [/math]. It can be shown that polynomials, rational functions, trig functions, exponential functions themselves, etc. are all of exponential order [math] -s [/math].

>>11544729
>>11544741
Instead of being a total piece of shit, you should at least put on a classic movie or start a good novel

>>11496474
v cannot be c though. I don't see any division by zero either.
>>11496476
The basic form of newton's second law is for a point mass only. Mass is assumed to be constant so [math] \dot{p}=m\dot{v} [/math]. Expressions like [math] F=\dot{m}v [/math] follow from applying the conservation of mass and Reynolds transport theorem to a control volume, and not just from Newton's law alone.
>>11497856
Use a table or some other reference (it's probably in the appendix of your textbook. If not, usually just assume 999 kg/m^3) to look up the density of water [math] \rho [/math] at that temperature. You are given the volumetric flow rate [math] Q=...\text{m}^3\text{/s}. [/math] Mass flow rate is [math]
\dot{m}=\rho Q [/math] and weight flow rate is [math] \dot{w}=\gamma Q=\rho g Q [/math] where g is gravitational acceleration.
>>11497863
The fossil record.

>>11414105
The acceleration of a particle moving in a circular path (like the bob of a pendulum) is given by [math] \mathbf{a}=-R\omega^2\mathbf{\hat{r}}+R\dot{\omega}\mathbf{\hat{\phi}} [/math] in polar coordinates. But at the pendulum's apex, [math] \dot{\phi}=\omega=0 [/math] so [eqn] \mathbf{F}_\text{net}=m\mathbf{a}=mR\dot{\omega}\mathbf{\hat{\phi}} [/eqn] which is a vector that points in a direction perpendicular to the cable that supports the bob (the [math]\mathbf{\hat{\phi}} [/math] direction).

hope u weren't the one embarrassing urself

>>11414318
ChemE if you don't want to trap yourself. Petroleum industry still needs lots of chemical engineers.

>>11414349
[math] \text{d}u=1\ \text{d}u=(u-u+2-1)\ text{d}u=(u+2)-(u+1)\ \text{d}u [/math]

I have an AA in math from community college. Does that mean anything?

>>11116773
let the engines go out at t=0. From basic kinematics, you know the velocity vi at t=0 and you also obviously know the height at t=0. The acceleration is 9.81 m/s/s downward. Now, you can apply some other kinematic equation to get the height where vf=0.