`a(t) = 10sin(t) + 3cos(t), s(0) = 0, s(2pi) = 12` A particle is moving with the given data. Find the position of the particle.

Textbook Question

Chapter 4, 4.9 - Problem 63 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

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`a(t)=10sin(t)+3cos(t)`

`a(t)=v'(t)`

`v(t)=inta(t)dt`

`v(t)=int(10sin(t)+3cos(t))dt`

`v(t)=-10cos(t)+3sin(t)+c_1`

`v(t)=s'(t)`

`s(t)=intv(t)dt`

`s(t)=int(-10cos(t)+3sin(t)+c_1)dt`

`s(t)=-10sin(t)+3(-cos(t))+c_1t+c_2`

`s(t)=-10sin(t)-3cos(t)+c_1t+c_2`

Let's find constants c_1 and c_2 , given s(0)=0 ans s(2pi)=12

`s(0)=0=-10sin(0)-3cos(0)+c_1(0)+c_2`

`0=-3+c_2`

`c_2=3`

`s(2pi)=12=-10sin(2pi)-3cos(2pi)+c_1(2pi)+3`

`12=-10(0)-3(1)+2pic_1+3`

`12=2pic_1`

`c_1=6/pi` 

`:.`  position of the particle is given by `s(t)=-10sin(t)-3cos(t)+(6t)/pi+3`

scisser's profile pic

scisser | (Level 3) Honors

Posted on

`v(t) = int(10sin(t) + 3cos(t)) dt `
`= -10cos(t) + 3sin(t) + C `

`s(t) = int(-10cos(t) + 3sin(t) + C)dt `
`= -10sin(t) - 3cos(t) + Ct + K `

Now use the known values

`s(0) = 0 `
`s(0) = -10sin(0) - 3cos(0) + C(0) + K `
`0 = -10(0) - 3(1) + K `
`0 = - 3 + K `
`K = 3 `

`s(t) = -10sin(t) - 3cos(t) + Ct + 3 `

`s(2pi) = 12 `
`s(2pi) = -10sin(2pi) - 3cos(2pi) + C(2pi) + 3 `
`12 = -10(0) - 3(1) + 2Cpi + 3 `
`12 = 2Cpi `
`6 = Cpi`
`C = 6/pi`

`s(t) = -10sin(t) - 3cos(t) + (6t)/pi + 3 `

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