## University Calculus: Early Transcendentals (3rd Edition)

$r=\dfrac{1}{t}+2t-2$
We need to find the anti-derivative for $\dfrac{d^2r}{dt^2}=\dfrac{2}{t^3}$ We have: $\dfrac{dr}{dt}=\dfrac{-1}{t^2}+c$; $r=\dfrac{1}{t}+ct+c'$ Apply the initial conditions $r'(1)=1 ; r(1)=1$ Thus, $-1+c=1 \implies c=2$ and $1+c+c'=1 \implies c'=-2$ Hence, $r=\dfrac{1}{t}+2t-2$