Recent content by ehild

Well, whar do the letters m, g, h mean?

You have to find the orthogonal projection of u onto the subspace of R4 spanned by the vectors v1, v2, and v3. x1, x2, x3 you got are the componets of this projection vector P with respect to the basis v1, v2, v3, that is P= x1 v1 +x2 v2+x3 v3 , a linear combination of three four-dimensional...

You have p2xt1/p1=v1xt2/v2 . Divide both sides of the equation by (t1t2) and muliply by (p1v2).

A picture may help.(two points connected with a viire are the same point in the circuit. ) Which resistors are connected to point A ? And to point B?

Two particles conneted by a spring corresponds to a single oscillator, a single vibrational mode, but it has two degrees of vibrational freedom,. A molecule of N atoms can perform 3N independent motions. From these , 3 are translations of the whole molecule: and 3 are rotations of the whole...

Good! A two-atomic molecle has a single vibrational mode. When performig that motion, the molecule has both kinetic and potential energy. If you want to specify thevibrational state of the molecule, you have to give both the velocity and the phase of the vibrationThese are two degrees of freedom...

What kind of energies belong to a single vibration?

Subtracting the two equations, you got y2- 3y-(x2 -3x )=0 . That does not mean y2- 3y = x2 -3x = 0. Show what you did. That method should work.

What exactly is the vector-vector function you have to integrate over the surface of the sphere?

Factor out y from the first equation . As it is zero, teither y must be zero, or the other factor. Work with both cases separateli, and substitute what you got into the secomd equation. Solve.

The way to solve a two-variable systen of eequations is to eliminate one variable. You eliminated xy but neither x nor y. You got a single two-variable equation with infinite many solutions.

Can you subtract fractions? Write the second term as a fraction with denominato m+M. Show your working.

Some currents can be zero in a circuit...

Very intuitive! So you did not differentiated ar all, and you guessed that the function has minimum where |x-a| does, that is at x=a for some a values. and found that intervall of a where that happens.

You gave the minimum value of the function correctly, but you can not get x=a for the position of minimum by differentiazing the function : it is not differentiable at x=a. "The absolute value function is continuous, but fails to be differentiable at x = 0 since the tangent slopes do not...