Answer
$\sqrt{5}e$
Work Step by Step
Recall: The speed of the particle is formulated by $v(t)=\sqrt{(dx/dt)^2+(dy/dt)^2}$.
Given: $x=e^t$ and $y=te^t$
Find the parameter $t$ corresponding to the point $(e,e)$:
$(x,y)=(e,e)$
$(e^t,te^t)=(e,e)$
$e^t=e$ and $te^t=e$
$e^t=e^1$ and $te=e$
$t=1$ and $t=1$
So, $t=1$
Find $dx/dt$ and $dy/dt$:
$dx/dt=e^t$
$dy/dt=dt/dt\cdot e^t+t\cdot de^t/dt=1\cdot e^t+t\cdot e^t=e^t+te^t$
Evaluate the speed of the particle at $t=1$:
$v(t)=\sqrt{(e^t)^2+(e^t+te^t)^2}$
$v(1)=\sqrt{((e^1)^2+(e^1+1\cdot e^1)^2}$
$v(1)=\sqrt{e^2+(2e)^2}$
$v(1)=\sqrt{5e^2}$
$v(1)=\sqrt{5}e$
Thus, the speed of the particle is $\sqrt{5}e$.
