Recent content by chickendude

A parody of Crank that Soulja boy... about math =). There was very little choreographing done, so bear with it. I hope you like it. The lyrics are in the "about this video".

The range equation is R = \frac{v^2}{g}\sin{2\theta} rearranging \sin{2\theta} = \frac{Rg}{v^2} There are two solutions for theta \theta = \frac{1}{2} \sin^{-1}\frac{Rg}{v^2} or \theta = \frac{1}{2}[180 - \sin^{-1}\frac{Rg}{v^2}] Plugging in all the numbers, you get \theta = 4.39^\circ or...

ah, yes, I just got it through that method.Yeah, I confused angular impulse with work. When we were doing linear momentum and impulse, I would make that mistake as well (integrate with respect to x rather than t), but now I see the problem. Thanks a lot!

Ah yes, that makes complete sense Now I understand why the energy conservation method didn't work through the inelastic collision. The second method uses torque and conservation of angular momentum. It should work in an inelastic collision yet it doesn't :(

Homework Statement Find the maximum theta given that h (initial height of block) m (mass of block) M (mass of rod) L (length of rod) Homework Equations The Attempt at a Solution I did it three different ways and none of them worked. What is wrong with these ideas...

I'm not going to do out all the algebra, but basically here's what you get if you do it out f'(x) = 2\cos^2 x g'(x) = e^{\sin x}[2\cos^2 x + \cos x(x + \sin x \cos x)] Dividing them, one of the cos will cancel, leaving \lim_{x\to\infty} \frac{2\cos x}{e^{\sin x}[2\cos x + x + \sin x...

Using the shell method, volume is V = 2\pi\int_a^b x * f(x) dx plugging in y=f(x) and the bounds determined by the x and y intercepts = 2\pi k \int_0^2 x * (x-2)^2 dx = 2\pi k \int_0^2 x^3 - 4x^2 + 4x dx = 2\pi k \left[ \frac{x^4}{4} - \frac{4x^3}{3} + 2x^2 \right]_0^2 = 2\pi k \left[ 4...

Alternatively, you can use a couple of cases Once you prove it for nonnegative integers, you can do this For negative integers: let n be a positive integer x^{-n} = \frac{1}{x^n} taking the derivative of x^-n is the same as taking the derivative on the right via the quotient rule...

I'll solve in terms of variables, then you can see how it works It's all based around the formula d = v_0t + \frac{1}{2}at^2 In the vertical direction h_f - h_i = (v_0\sin\theta) t + \frac{1}{2}gt^2 Where h_f is the final height (the height of the hoop) h_i is the initial height (8 feet) v_0...

Derivatives are only defined for points in the domain of a function if c is not in the domain of f(x), then f'(c) by definition does not exist

It is pretty obvious that the function is smooth whenever x \neq 0 At x=0, we need to prove that f(0) = \lim_{x\to 0}f(x) by the definition of smoothness f(0)=0 by the def of f(x) \lim_{x\to 0}e^{-x^{-2}} =\lim_{x\to 0}(1/e)^{\frac{1}{x^2}} =\lim_{x\to \infty}(1/e)^{x^2} since 1/e...

For example, when you are finding the area under a curve, each slice is basically a vertical line whose width is approaching zero. Each slice has height f(x) and width dx. The dx here represents that tiny "ghost of a departed quantity", meaning the width that has shrunk toward zero. The...

How about this approach: (sorry for the double post) Draw segments from P to each vertex of the triangle. You have split the triangle into three new triangles. The area of each triangle is given by \frac{1}{2}sh_i where s is the side length of the equilateral triangle and h_i is the height of...

They are not perpendicular to each other. They are perpendicular to each side of the triangle.

The way I like to think of an integral is as a sum of slices. For example \int_0^5 x^2 dx You are finding the area under the curve x^2 from 0 to 5. This is done by finding the sum of a bunch of infinitesimal vertical slices. Think of dx as the infinitesimal width of each slice. x^2...