Recent content by Deadleg

You should get that, -\sin\left(\frac{\pi/\sqrt{2}}{\sqrt{2}}\right) = -\sin(\pi/2) = -1. t = pi/2 comes from r(t) = (0, 1, pi/2) = (cos t, sin t, t) and looking at the last entry.

Ah, so for 1. h(x)=[f(x)]^{g(x)}=e^{g(x)\ln[f(x)]} h'(x)=e^{g(x)\ln[f(x)]}.(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)}) h'(x)=[f(x)]^{g(x)}.(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)}) And for 2. A(P-x)+Bx=1 AP+(B-A)x=1 AP=1,\ A=\frac{1}{P} B-A=0,\ B=A=\frac{1}{P}

Homework Statement 1. If f(x), g(x) and h(x) are real functions of x, show that when h(x)=[f(x)]^{g(x)} then h'(x)=[f(x)]^{g(x)}(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)}) 2. \frac{A}{x}+\frac{B}{P-x}=\frac{1}{x(P-x)} where x is a variable, and P is a constant. Find A and B in terms of...

Oh that makes sense. Thanks for explaining :)

Homework Statement y=\frac{x^2}{1+x^2} where -1\leq x \leq 1 The gradient at the point x=1 is \frac{1}{2} Hence show that there is a point with \frac{1}{4} < x < \frac{1}{2} where the gradient is also \frac{1}{2} The Attempt at a Solution I differentiated, set the derivative to...

And for an electron, q is negative, so in F=Eq, F will be negative ie against the electric field :) I don't think proportionalities work like that... I haven't done a lot of work with them, but by looking at the relationship, if r decreases, F will increase yes? So if r decreases by 80%, r^2...

Q is not force, it is charge. They are simply constants in the equation in this context. Try using F\propto\frac{1}{r^2} Shouldn't the second question be downwards, because if the field lines goes upwards, that means a negative charge will be at the top, therefore repelling the electron downwards?

bleh edziura would be right.

V_f=V_i+at Acceleration\ due\ to\ gravity=-10 ms^{-2} Look at the time when the rocket stops propelling itself, and the time its velocity is zero, then it should click :)

v=f\lambda Using that relationship you can change the frequency in wavelength (speed\ of\ sound=330 ms^{-1} if not specified.) So from there it's should be pretty straight forward to solve for L

Oh yeah whoops :S But span=2a so a=59, so I believe the equation is \frac{x^2}{59^2}+ \frac{y^2}{b^2}= 1.

I think the statement "The height of the arch 25 feet from the center is to be 8 feet" means that at x=\pm{25}, y=8 giving the coordinates (\pm{25},8). So try putting those coodinates into the equation {\frac{x^2}{25^2}}+{\frac{y^2}{b^2}}=1 and solve for b^2.

Ah I get it now, thanks!

No, we don't learn that this year. This is last year high school stuff, and I was sick when the class was taught it so I'm trying to get through it myself. The notes up to this exercise in the book simply goes over what a limit is, evaluating by algebraic manipulation or solve for values of x...

Homework Statement 1. Find the limit of \lim_{x\rightarrow 0} \frac{1}{xe^{\frac{1}{x}}} 2. " " " " \lim_{x\rightarrow\infty} \frac {x}{\log_e x} Homework Equations \lim_{x\rightarrow\infty} \frac{N}{x} = 0 \lim_{x\rightarrow n} x+a = \lim_{x\rightarrow n} x + \lim_{x\rightarrow n}...