Mathematics, 17.10.2019 11:30 walfrerr

Use definition 7.1.1, definition 7.1.1 laplace transform let f be a function defined for t ≥ 0. then the integral ℒ{f(t)} = ∞ e−stf(t) dt 0 is said to be the laplace transform of f, provided that the integral converges. to find ℒ{f(t)}. (write your answer as a function of s.) f(t) = cos t, 0 ≤ t < π 0, t ≥ π

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The Laplace Transform of the piecewise Function is $\mathcal {L}\{f(t)\} = \frac{s\cdot (e^{-s\cdot \pi + 1})}{s^{2}+1}$ .

Given that Function is continuous and integrable. The Laplace Transform is defined by the following expression:

$\mathcal {L} \{f(t)\} = \int\limits^{+\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt$ , $t \ge 0$ (1)

We have the following piecewise Function:

$f(t) = \left \{ {{\cos t,\,0 \le t \le \pi} \atop {0,\,t\ge \pi}} \right.$ (2)

By (1) and using Integral Properties, the Laplace Transform of the piecewise Function is:

$\mathcal{L}\{f(t)\} = \int\limits^{\pi}_{0} {e^{-s\cdot t}\cdot \cos t} \, dt + \int\limits^{+\infty}_{\pi} {e^{-s\cdot t}\cdot 0} \, dt$

$\mathcal{L}\{f(t)\} = \int\limits^{\pi}_{0} {e^{-s\cdot t}\cdot \cos t} \, dt$ (3)

By Part Integration, we have the following outcome:

$u = e^{-s\cdot t}$ , $dv = \cos t\,dt$ , $du = -s\cdot e^{-s\cdot t}$ , $v = \sin t$

$\mathcal {L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |\limits ^{+\infty}_{0} + s\int\limits^\pi_0 {e^{-s\cdot t}\cdot \sin t} \, dt$ (4)

$m = e^{-s\cdot t},\, dn = \sin t\,dt,\, dm = -s\cdot e^{-s\cdot t},\,n = -\cos t$

$\mathcal{L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |_{0}^{\pi}-s\cdot e^{-s\cdot t}\cdot \cos t|_{0}^{\pi}-s^{2}\int\limits^\pi_0 {e^{-s\cdot t}\cdot \cos t} \, dt$

$\mathcal{L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |_{0}^{\pi}-s\cdot e^{-s\cdot t}\cdot \cos t|_{0}^{\pi}-s^{2}\cdot \mathcal{L}\{f(t)\}$

$(s^{2}+1)\cdot \mathcal{L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |_{0}^{\pi}-s\cdot e^{-s\cdot t}\cdot \cos t|_{0}^{\pi}$

$\mathcal {L}\{f(t)\} = \frac{s\cdot (e^{-s\cdot \pi + 1})}{s^{2}+1}$

The Laplace Transform of the piecewise Function is $\mathcal {L}\{f(t)\} = \frac{s\cdot (e^{-s\cdot \pi + 1})}{s^{2}+1}$ .

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Answer from: kkelley9223

$f(t)=\begin{cases}\cos t&\text{for }0\le t$

The Laplace transform is then

$\mathcal L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_0^\pi e^{-st}\cos t\,\mathrm dt$

Let denote the integral we want to compute. Integrating by parts, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=\cos t\,\mathrm dt\implies v=\sin t$

gives

$\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\int_0^\pi e^{-st}\sin t\,\mathrm dt$

Integrate by parts again, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=\sin t\,\mathrm dt\implies v=-\cos t$

Then

$\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\left(-e^{-st}\cos t\bigg|_{t=0}^{t=\pi}-s\int_0^\pi e^{-st}\cos t\,\mathrm dt\right)$

$I=e^{-st}(\sin t-s\cos t)\bigg|_{t=0}^{t=\pi}-s^2I$

$(s^2+1)I=s(e^{-\pi s}+1)$

$I=\dfrac s{s^2+1}(e^{-\pi s}+1)$

Answer from: Quest

0.5 meters per second

step-by-step explanation:

the average speed between 2 seconds and 8 seconds is 0.5 meters per second.

hope this ! mark brainliest!

Answer from: Quest

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab in the y coordinate

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

x coordinate

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

y coordinate

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

x coordinate

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

y coordinate

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

x coordinate

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

y coordinate

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

x coordinate

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

y coordinate

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

x coordinate

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

y coordinate

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

x coordinate

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

y coordinate

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

x coordinate

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

y coordinate

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕

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Use definition 7.1.1, definition 7.1.1 laplace transform let f be a function defined for t ≥ 0. then...