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3 years ago

How are periodic phenomena relatively easy to model using trig functions? What makes them difficult to model this way?

Mathematics

1 answer:

Klio2033 [76]3 years ago

6 0

Answer:

Periodic phenomena means that the phenomena has a (almost) constant time period or space period.

As you know, the trigonometric functions cos(x) and sin(x) also have a constant period of 2*pi, so these functions are really useful to model periodic phenomena.

Now, the problem may be that the trigonometric functions may be useful to describe the "periodic" part, but not to describe the actual phenomena.

An example of this can be a square alternating current.

While it has a constant period like a trigonometric function, the trigonometric functions can not really model the "square" part of this current (you know that the sinusoidal functions actually are curves and continuous)

Here comes something called the Fourier Series, that are series of the form:

F(x) = a₀ + ∑(aₙ*cos(nx) + bₙ*sin(nx))

That can be used to model almost any periodic phenomena, but the actual Fourier Series may be hard to construct.

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Solve the inequality. -54<2 1/4x

Leona [35]

Divide each side by 2 1/4 to get x by itself. The answer is -24 < x and the sign didn't change because we didn't divide by a negative.

5 0

3 years ago

Pls help with math ASAP

9966 [12]

I think 6,1 and the upper point is -6,11

7 0

3 years ago

If man's salary is 20 dollors a week and he spends 14 dollors a week how long will it take him to save 300 dollors

Contact [7]

Solutions

⇒ Given

Salary - $ 20

Spends - $ 14

The first step is to subtract 14 from 20. When we do this, we will know how much he saves.

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Calculations

20 - 14 = 6

He saves $ 6 each week.

Now we have to find how many times 6 goes into 300. To do this we have to divide.

300 ÷ 6 = 50

<span>It is going to take him 50 weeks. </span>

7 0

3 years ago

Read 2 more answers

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

3 years ago

First and second place in a contest use a ratio to share a cash prize. When first place pays $100 , second place pays $60. How m

torisob [31]

Answer:

$60

Step-by-step explanation:

the ratio is 100:60 or 5/3

if second place pays 36, to find out how much first place pays, multiply 36 by 5/3

36 * 5/3 = 60

6 0

3 years ago