Recent content by Donaldos

You need to find the two other rotation matrices using the exact same method. The standard matrix for T is found by simply multiplying the three matrices together.

You can use the following identity: \sin a \sin b=\frac{\cos(a-b)-\cos(a+b)} 2

u=\sqrt{1+e^{2x}} {\rm d}u=\frac{e^{2x}}{\sqrt{1+e^{2x}}}=\frac{u^2-1}{u} {\rm{d}x} and s={\int\limits_{\sqrt{2}}^{\sqrt{1+e^2^}}} {\frac{u^2}{u^2-1^}} \quad {\rm d}u ?

Try u=x-y.

Let \lambda be the linear charge distribution in the wire. Use Gauss's Law to express the electric field in terms of r and \lambda. Then use this: to determine the expression of \lambda as a function of V.

Show that A*B=B*A. Find I \in S such that A*I=I*A=A , \quad\forall A \in P. Find B \in S such that A*B=I .

Your last determinant should be \left|\begin{array}{cc} -18 & 1\\ -50 &9\end{array}\right| not \left|\begin{array}{cc} -18 & 0\\ -50 &9\end{array}\right|

The third line of your second determinant is incorrect.

In other words, we have \ker(A)\subset \ker(A^2). Yes. You could also write : \operatorname{im}(A^2)\subset \operatorname{im}(A) Now, how can you apply this to \ker(A^3), \ker(A^4)... and \operatorname{im}(A^3), \operatorname{im}(A^4)...?

a. If x \in \ker(A), what can you say about A^2 x? What does that tell you about \ker(A^2)? b. For any x, A^2x is in \operatorname{im}(A^2). If you rewrite A^2x as A(Ax)=Ay, what can you say about \operatorname{im}(A)?

Where did that expression come from? Find an expression for the radial acceleration in terms of the orbital speed and R. Then use Newton's second law.

\int \frac{{\rm d}u}{\sqrt{1-u^2}}=\arcsin u but here we have \int \frac{{\rm d}x}{\sqrt{1-\left(\frac x 3 \right)^2}} u= \frac x 3 \Rightarrow {\rm d}x= ?

1. Decompose \frac{1}{x\left(x^4+8\right)} into partial fractions 2. u=\sqrt{x} + partial fraction decomposition

Exactly.

All you need is the following property: \int\limits_{-\infty}^{+\infty} f(x)\delta\left(x-a\right)=f(a)