5c-13.1-2.math

13.1 19

Find a vector equation and parametric equations for the line segment that joins P to Q.
P(0, −1, 4),    Q(1/2,1/3,1/4)

r(t) = (1-t)r0+tr1

r(t) = (1-t)<0,-1,4>+t<1/2,1/3,1/4>
r(t) = <0,-1,4> -t<0,-1,4>+ t<1/2,1/3,1/4>
r(t) = <0 -0t+1/2t, -1 +t+1/3t, 4-4t+1/4t>
r(t) = <1/2t, -1 +4/3t,4-15/4t>

r(t) = (1/2t, -1 +4/3t, 4 - 15/4t)
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13.2 22

r  <4e^(3t),2e^(-3t),4te^(3t)>

r' <4e^(3t)*3,2e^(-3t)*-3,4*e^(3t)+e^(3t)*3*4t>
r' <12e^(3t),-6e^(-3t),4e^(3t)+12te^(3t)>
r' <12e^(3t),-6e^(-3t),e^(3t)(4+12t)>

r" <12e^(3t)*3,-6e^(-3t)*-3,e^(3t)*3(4+12t)+12e^(3t)>
r" <36e^(3t),18e^(-3t),e^(3t)(24+36t)>
r" <36e^(3t),18e^(-3t),12e^(3t)(2+3t)>
r"(0) <36,18,24>

r'.r"  <12e^(3t)+-6e^(-3t),e^(3t)(4+12t)>.<36e^(3t),18e^(-3t),12e^(3t)(2+3t)>
r'.r"  12e^(3t)*36e^(3t)+-6e^(-3t)*18e^(-3t)+e^(3t)(4+12t)*12e^(3t)(2+3t)>
r'.r"  432e^(6t)+-108e^(-6t)+(4e^(3t)+12te^(3t))*(2*12e^(3t)+3t*12e^(3t))>
r'.r"  432e^(6t)+-108e^(-6t)+(4e^(3t)+12te^(3t))*(2*12e^(3t)+3t*12e^(3t))>
r'.r"  4e^(3t)*2*12e^(3t)+12te^(3t)*2*12e^(3t)+4e^(3t)*3t*12e^(3t)+12te^(3t)*3t*12e^(3t)>
r'.r"  432e^(6t)+-108e^(-6t)+96e^(6t)+288te^(6t)+144te^(6t)+432t^2e^(6t)
r'.r"  432e^(6t)+-108e^(-6t)+96e^(6t)+432te^(6t)+432t^2e^(6t)
r'.r"  432e^(6t)-108e^(-6t)+48e^(6t)(3t+1)(3t+2)
r'.r"  528e^(6t)-108e^(-6t)+288te^(6t)+144te^(6t)+432t^2e^(6t)
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13.2 25

x = e^(−6t)cos(3t), y = e^(−6t)sin(3t), z = e^(−6t); (1, 0, 1)
let x=1,y=0,z=1

=> 1 = e^(-6t)cos(3t) 0 = e^(−6t)sin(3t) 1 =  e^(−6t)
=> 1 = e^(-6t) or 1 = cos(3t) and 0 = e^(-6t) or 0 = sin(3t) and 1 = e^(-6t)
=> ln(1) = -6t or t = 0 and false or t = 0 and ln(1) = -6t
=> t = 0

x'= -6e^(-6t)cos(3t)-3sin(3t)e^(-6t)
y'= -6e^(-6t)sin(3t)+3cos(3t)e^(-6t)
z'= -6e^(-6t)
r'(0) = <-6,3,-6>

s(t) = (1-0t)<1,0,1>+t<-6,3,-6>
s(t) = <1-6t,3t,1-6t>
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13.2 39
integral t^3lnt dt 
u=t^3 dv=ln(t)
du=3t^2 v= ( u= ln(t) dv = dt
           ( du = dt/t v = t
           ( tlnt -integral tdt/t)
           (tlnt-t)= t(lnt-1)
integral t^3lnt dt  = t^4(lnt -1) -3 integral t^2t(lnt-1) dt
integral t^3lnt dt = t^4(lnt-1) -3 integral t^3lnt dt + 3 integral t^3
4integral t^3lnt dt = t^4(lnt-1) + 3 integral t^3
4integral t^3lnt dt = t^4(lnt-1) + 3t^4/4
integral t^3lnt dt = t^4(lnt-1)/4 + 3t^4/16
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13.2 41

r(t)= t^6 i +t^7 j+2/3 t^(3/2)}k+C
  -k = +C
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13.2 33
r1(t) = (5t, t^2, t^4)
and 
r2(t) = (sin(t), sin(2t), 4t)

r1'=(5,2t,4t^3)
r1'(0)=<5,0,0>

r2'=(cos(t),2cos(2t),4)
r2'(0)=<1,2,4>

1/5*sqrt(21) <5,0,0>.<1,2,4>
5/5sqrt(21)