Answer
$2\sqrt{17}$
Work Step by Step
Recall: The speed of the particle is formulated by $v(t)=\sqrt{(dx/dt)^2+(dy/dt)^2}$.
Given: $x=t^2+1$ and $y=t^4+2t^2+1$
Find $t$ corresponding the point $(2,4)$:
$(x,y)=(2,4)$
$(t^2+1,t^4+2t^2+1)=(2,4)$
$t^2+1=2$ and $t^4+2t^2+1=4$
$t^2-1=0$ and $t^4+2t^2-3=0$
$t^2-1=0$ and $(t^2+3)(t^2-1)=0$
$t^2-1=0$
$t^2=1$
$t=\pm 1$
Find $dx/dt$ and $dy/dt$:
$dx/dt=2t$
$dy/dt=4t^3+4t$
Evaluate the speed of the particle at $t=\pm 1$:
$v(t)=\sqrt{(2t)^2+(4t^3+4t)^2}$
$v(t)=\sqrt{4t^2+16t^6+32t^4+16t^2}$
$v(t)=\sqrt{16t^6+32t^4+20t^2}$
$v(\pm 1)=\sqrt{16(\pm1)^6+32(\pm 1)^4+20(\pm 1)^2}$
$v(\pm 1)=\sqrt{16+32+20}$
$v(\pm 1)=\sqrt{68}$
$v(\pm 1)=2\sqrt{17}$
Thus, the speed of the particle is $2\sqrt{17}$.
