Find r(d if r\"(r).(e.\"(1-2,-(1-2A) and r(0)-(-1,2,-3) 2) 3) Find the definite

Question

Find r(d if r"(r).(e."(1-2,-(1-2A) and r(0)-(-1,2,-3) 2) 3) Find the definite integral Use the projection of vectors to find the distance from the point (1, 5) the line The acceleration of a particle is a)-(1.2.2). If the particle starts at the point (0, 1,-3) with an initial velocity of .(0)-/-/-2k 4) a/ Find the position r(t) of the particle at t = 1. b/ Find the unit tangent T and the principal normal vector N to the curve at c/ Find the components of the acceleration on Tand N 1 d/ what is the curvature of the trajectory of the particle at t = 1 5) Find the curvature k of the curve given by y-U2x at x=4

Explanation / Answer

posted multiple questions. please post each question seperately

1)

given r'(t)=(e-2t,(1-2e-t),-(1-2et)) ,r(0)=(-1,2,-3)

r(t)=r'(t) dt

r(t)=(e-2t,(1-2e-t),-(1-2et)) dt

r(t)=(e-2t,1-2e-t,-1+2et) dt

r(t)=(e-2tdt,(1-2e-t)dt,(-1+2et) dt)

r(t)=((-1/2)e-2t,t+2e-t,-t+2et) +C

r(0)=(-1,2,-3)

=>((-1/2)e-2*0,0+2e-0,-0+2e0) +C=(-1,2,-3)

=>((-1/2),0+2,-0+2) +C=(-1,2,-3)

=>((-1/2),2,2) +C=(-1,2,-3)

=>C=(-1,2,-3)-((-1/2),2,2)

=>C=(-1-(-1/2),2-2,-3-2)

=>C=(-(1/2),0,-5)

r(t)=((-1/2)e-2t,t+2e-t,-t+2et) +C

=>r(t)=((-1/2)e-2t,t+2e-t,-t+2et) +(-(1/2),0,-5)

=>r(t)=((-1/2)e-2t-(1/2),t+2e-t+0,-t+2et-5)

=>r(t)=((-1/2)(e-2t+1),t+2e-t,-t+2et-5)

========================================================

2)

[0 to 2] te2t(3i +4j -k) dt

=(3([0 to 2] te2tdt)i +4([0 to 2] te2tdt)j -([0 to 2] te2tdt)k)

integration by parts: u =t , dv =e2tdt , du =dt , v=(1/2)e2t,u dv =uv -v du

=(3([0 to 2] te2tdt)i +4([0 to 2] te2tdt)j -([0 to 2] te2tdt)k)

=(3(|[0 to 2]t(1/2)e2t -[0 to 2] (1/2)e2t dt)i +4(|[0 to 2]t(1/2)e2t -[0 to 2] (1/2)e2t dt)j -(|[0 to 2]t(1/2)e2t -[0 to 2] (1/2)e2t dt)k)

=(3(|[0 to 2]t(1/2)e2t -(1/4)e2t )i +4(|[0 to 2]t(1/2)e2t -(1/4)e2t )j -(|[0 to 2]t(1/2)e2t -(1/4)e2t )k)

=(3((2(1/2)e2*2 -(1/4)e2*2)-(0(1/2)e2*0 -(1/4)e2*0))i +4((2(1/2)e2*2 -(1/4)e2*2)-(0(1/2)e2*0 -(1/4)e2*0))j -((2(1/2)e2*2 -(1/4)e2*2)-(0(1/2)e2*0 -(1/4)e2*0))k)

=(3((e4 -(1/4)e4)-(0-(1/4)))i +4((e4 -(1/4)e4)-(0-(1/4)))j -((e4 -(1/4)e4)-(0-(1/4)))k)

=(3((3/4)e4-0+(1/4))i +4((3/4)e4-0+(1/4))j -((3/4)e4-0+(1/4))k)

=(3(1/4)(3e4+1)i +4(1/4)(3e4+1)j -(1/4)(3e4+1)k)

=(3/4)(3e4+1)i +(3e4+1)j -(1/4)(3e4+1)k

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