Question

Let r(t) = − e^(6t) j + e^(–6t) k . Show that r(t) is parallel to r′′(t) , for any time t.

Answer 1

r(t)=-e^(6t)j+e^(-6t)k=A
r'(t)=-6e^(6t)j-6e^(-6t)k
r''(t)=-36e^(6t)j+36e^(-6t)k=B

A*B=36e^(36t^2)+36e^(36t^2)=72e^(36t^2)
||A||=sqrt(e^36t^2)+e^(36t^2))=sqrt(2e^(36t^2))
||B||=sqrt(36^2e^(36t^2)+36^2e^(36t^2))
||B||=sqrt(2592e^(36t^2))
||A||||B||=72e^(36t^2)
LET W=36t^2
A*B/||A||||B||=cos(x)
72e^w/72e^w=1
1=cos(x)
cos^-1(1)=x
x=2pi or 0 which concludes these two are parallel

Answer 2

Thanks!

Answer 3

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