## Calculus (3rd Edition)

The minimum speed of the particle is $3.57131$.
We have $x\left( t \right) = {t^3}$, ${\ \ \ }$ $x'\left( t \right) = 3{t^2}$, $y\left( t \right) = {t^{ - 2}}$, ${\ \ \ }$ $y'\left( t \right) = - 2{t^{ - 3}}$. By Theorem 2, the speed of the particle is $s'\left( t \right) = \sqrt {{{\left( {3{t^2}} \right)}^2} + {{\left( { - 2{t^{ - 3}}} \right)}^2}} = \sqrt {9{t^4} + 4{t^{ - 6}}}$. Write $f\left( t \right) = s'\left( t \right)$. First, we find the critical point of $f\left( t \right)$ for $t \ge 0.5$ by solving $f'\left( t \right) = 0$: $f'\left( t \right) = \frac{1}{2}{\left( {9{t^4} + 4{t^{ - 6}}} \right)^{ - 1/2}}\left( {36{t^3} - 24{t^{ - 7}}} \right) = \frac{{18{t^3} - 12{t^{ - 7}}}}{{{{\left( {9{t^4} + 4{t^{ - 6}}} \right)}^{1/2}}}}$ $f'\left( t \right) = 0$ gives $18{t^3} - \frac{{12}}{{{t^7}}} = 0$, $t = {\left( {\frac{2}{3}} \right)^{1/10}} \simeq 0.96$. So, the critical point is at $t \simeq 0.96$. Next, we check if the critical point $t = {\left( {\frac{2}{3}} \right)^{1/10}} \simeq 0.96$ is a minimum by examining the second derivative of $f\left( t \right)$. $f{\rm{''}}\left( t \right) = \frac{{6\left( {27{t^{12}} + 234{t^2} + 32{t^{ - 8}}} \right)}}{{\sqrt {9{t^4} + 4{t^{ - 6}}} \left( {9{t^{10}} + 4} \right)}}$ Since $f{\rm{''}}\left( t \right) > 0$, the critical point $t = {\left( {\frac{2}{3}} \right)^{1/10}} \simeq 0.96$ is a minimum. So, the minimum speed of the particle is $s'\left( {0.96} \right) = 3.57131$.