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2 years ago

Please help due right now

Mathematics

1 answer:

nlexa [21]2 years ago

6 0

Answer:

It is a function

Step-by-step explanation:

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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

I NEED HELP WITH THIS ASAP

grin007 [14]

hopefully someone helps u with this tho sorry

3 0

2 years ago

Which number is a perfect cube?<br> O9<br> O 18<br> 0 27<br> O 36

Arturiano [62]

Answer:

Step-by-step explanation:

9 is not a perfect cube.

18 is not a perfect cube.

27 is a perfect cube.

∛27 = 3

36 is not a perfect cube.

8 0

3 years ago

Read 2 more answers

Which expression is equivalent to the given expression?

djverab [1.8K]

Answer:

a. 5q+5q+5

Step-by-step explanation:

5q+5q=10q

10q+5 is equivalent to 10q+5

8 0

3 years ago

Help Find the lateral area of this pyramid whose base is an equilateral triangle. Its slant height is 14 ft and the length of ea

kenny6666 [7]

Answer:

168 ft²

Step-by-step explanation:

The equation for the lateral area of a square pyramid is LA = (Pl/2)

Plug in the values LA = (24(14)) / 2

LA = 336/2

LA = 168 ft²

8 0

3 years ago