A particle moves in the -plane according to the parametric equations and .An expression for the length of the path of the particle from to is

Question

​A particle moves in the  -plane according to the parametric equations  and  .An expression for the length of the path of the particle from  to  is ​

Quizplus · Accepted Answer

Using the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$d=\sqrt{((-t)^2+(t^2+1)^2}-(1^2+0^2)}$

$d=\sqrt{t^4+2t^2+1+t^4} - 1$

$d=\sqrt{2t^4+2t^2+1} - 1$

To minimize $d$, we need to minimize the expression under the radical sign. This is a calculus problem, and we take the derivative:

$\frac{d}{dt}\sqrt{2t^4+2t^2+1} = \frac{4t^3+2t}{2\sqrt{2t^4+2t^2+1}}$

Setting the derivative equal to zero, we have:

$4t^3+2t=0$

$2t(2t^2+1)=0$

Since $2t^2+1$ is always positive, the only solution is $t=0$.

Thus, the minimum distance is:

$d=\sqrt{2(0)^4+2(0)^2+1} - 1 = 0$

This means that the particle's path goes directly from $(1,0)$ to $(0,1)$, which is the line $y=1-x$.

Now we just need to find the distance between $(1,0)$ and $(0,1)$:

$d=\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}-1$

Therefore, the answer is B.

​A Particle Moves in the -Plane According to the Parametric