Recent content by sqrt(-1)

Homework Statement I have a two part question, the first part involves solving Laplace's equation u_{xx} + u_{yy} = 0 for the boundary conditions u_x(0,y) = u_x(2,y) = 0 u(x,0) = 0 u(x,1) = \sin(\pi x) for 0 < x < 2, 0 < y < 1. The second part now states a new boundary problem...

Thanks for the reply - now that I've looked at it with the information you provided I realize I made a howler of a mistake in the first part of my working.

I'm slightly confused with the following function so I was wondering if anybody could give me some hints as to the next step. A function f is defined as f:\mathbb{C} \longrightarrow \mathbb{C} \\ ~~z \longmapsto |z| where \mathbb{C} = (\mathbb{C},+) assuming the function is...

I thought Roger Penrose did most of his math for him.

Remember that the piano is kept from accelerating not from moving - the man is making sure the piano moves with a constant velocity.

He's already moving at 4m/s so his KE won't be zero at the top. Next you should try and work out the magnitude of the normal reaction of the slope so that you can work out how much energy is lost to friction.

Use the relation \ddot{x} = v \frac{dv}{dx} on the left hand side of your equation and you should find things work out :)

First find the total energy for the system in terms of m. At the top of the slope we have E_{tot} = \frac{1}{2}mv^2 + mgh = \frac{1}{2}m(4)^2 + m \times 9.81 \times 195 \sin(12) now think about what's happened to the system in the course of the sled moving to the bottom of the slope.

First ask yourself how much further the light has to travel when the Earth is furthest away from Jupiter compared to when it was closest to Jupiter. (Note that your value for the average radius of the Earth's orbit around the sun is given in kilometres)

Try using the quadratic formula and thinking about what comes underneath the square root in relation to Im(a) = 2

Using the given parameterisation we have a curve with position vector \mathbf{\rm{r}}(t) = (2 - t^3)\mathbf{\rm{i}} + (2t - 1)\mathbf{\rm{j}} + \ln(t)\mathbf{\rm{k}} Clue : If y = 0 then what value of t should you use to find the point of intersection?

Remember the integral on the RHS is asking for the derivative of f, so we have \int t\cos(\pi t) dt = \frac{t}{\pi}\sin(\pi t) - \frac{1}{\pi} \int \sin( \pi t ) dt

Because the book is asking for the maximum height reached by the projectile over its entire flight and not the height the projectile reached after being in the air for 2 seconds.

Your integral for dg has found a factor of t for some reason, your integral should be: \int \cos (\pi t ) dt = \frac{1}{\pi} \sin(\pi t)

Sorry should have said, u is the initial speed of the particle and t is the time in seconds. So we have: y(2) = 7 + 43*2*sin(45) - 0.5*32*2^2 = 3.8 ft to 2 s.f.