Please help with both questions

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

Michael thought he could read 23 pages of his book in class. He was only able to read 20. what is his percent error

Cerrena [4.2K]

Percent error is calculated using

((Actual - predicted) / actual) * 100

In this case, the actual value of the number of pages he read is 20. The expected value was 23. So simply plug in:

((20 - 23) / 20) * 100 = -15%

Please don’t forget to like and mark brainliest

8 0

2 years ago

What is the inverse function of d(x) = 2x - 4?

zysi [14]

Answer:

$d^{-1}$ (x) = $\frac{x+4}{2}$

Step-by-step explanation:

let d(x) = y and rearrange making x the subject, that is

2x - 4 = y ( add 4 to both sides )

2x = y + 4 ( divide both sides by 2 )

x = $\frac{y+4}{2}$

Change y back into terms of x, so

$d^{-1}$ (x ) = $\frac{x+4}{2}$

7 0

3 years ago

If g(x) = -2x - 2, find g(-2x).

USPshnik [31]

Answer:

2

Step-by-step explanation:

You would replace the x with -2x so you would end up with -2×-2 and that is 4 then minus 2 is 2.

7 0

3 years ago

Graph + > -1.

andrew-mc [135]

You have the right idea but the endpoint is at the wrong location. Instead, the green dot should go at -1. The shading is to the right. We use a closed filled in circle here (instead of an open hole) to tell the reader "include the endpoint". So -1 is part of the solution set.

In short, the graph consists of a closed filled in circle at -1 with shading to the right. This visually describes all values that are larger than -1, or equal to -1.

6 0

3 years ago

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