Recent content by adriaat

I am working on a problem of electrostatics, and I am having troubles in trying to figure out one part of it. 1. Homework Statement It consists of an inner solid cylinder of radius ##a## with a voltage ##V_A##, and an outer coaxial cylindrical shell of inner radius ##b## and outer radius ##c##...

I know that the differentiation of a power series is \sum n a_{n} x^{n-1}, so if by differentiating f(x) = \frac{1}{\sqrt{4+x^{2}}} I got rid of the square root, I would know how to find a power series: integrating the result found by the power series of the differentiation. But I think it...

Homework Statement Find a power series for f(x) = \frac{1}{\sqrt{4+x^{2}}}, at x=0. 2. The attempt at a solution I have looked up the Taylor series of \frac{1}{\sqrt{4+x^{2}}}, but I don't find any similarity with a power serie like \sum_{n\geq 0} a_{n} x^{n} I don't know how to start...

I've been trying to figure out what a negative area means, but I can't. Homework Statement Calculate the area between f(x) = 3^{x} \, , \, g(x)=2x+1 The attempt to a solution The intersections are located in x=0 and x=1. So I do the integral from 0 to 1. \int_{0}^{1} (g(x)-f(x))dx =...

I found a solution which agrees with my text's. Here is what I did: 1st equation: \frac{dy_2}{dt}=x_2·h·\frac{1}{x_1}=x_2·h·\frac{1}{v·t} I derivated the above equation, which is the velocity of the shadow's variation, and found time for x_1 2nd equation: t=\frac{x_1}{v} I...

Thank you, I didn't write that [...]calculate the velocity of the shadow variation when the person is at 9.14 m from the spotlight. Yes, it is a calculus based class, but I just started calculus classes, so I'm not familiar with differentiation. --------------------- To combine the...

Helo, I've been working out this exercise, but my solution and the text's aren't the same. Homework Statement We have a spotlight on the floor located at a distance of 30.5 m from a wall of a building. There is a person 1.83 m tall between the spotlight and the wall moving away from the...

Anyway, the hypotenuse of a right triangle can't be never shorter than any cathetus, so you must have wrong lengths.