Answer to Question #227719 in Calculus for Mis

Question

The position of a particle is given by r(t)= t²i + e^2tj + In(t+1)k. Find the minimum velocity of the particle

Answer 1

$$r(t)=t^2i + e^{2t}j + \ln(t+1)k$$

$$v(t)=r'(t)=2ti +2 e^{2t}j + \dfrac{1}{1+t} k$$

Let $f(t)=(|v(t)|)^2=(2t)^2+(2e^{2t})^2+(\dfrac{1}{1+t})^2$

$$f'(t)=8t+16e^{4t}-\dfrac{2}{(1+t)^3}\geq0+16-2>0, t\geq0$$

Therefore the function $f(t)$ increases for $t>0.$

Since $t\geq0$ (time), then the minimum value of $f$ for $t\geq 0$ is

$$f(0)=(2(0))^2+(2e^{2(0)})^2+(\dfrac{1}{1+0})^2=5$$

The minimum velocity of the particle is $\sqrt{5}.$