Chapter 10, Problem 44RE

To determine

To find: The position vector.

Answer to Problem 44RE

The position vector is $〈{\tan }^{-1}t+1,\sqrt{t^2+1}〉$ .

Explanation of Solution

Given information:

The velocity is and .

Formula used:

If the velocity of particle is then the particle’s position is .

Calculation:

Consider the velocity .

Substitute $\frac{1}{t^2+1}$ for $v_1\left(t\right)$ and $\frac{t}{\sqrt{t^2+1}}$ for $v_2\left(t\right)$ in the formula .

   …… (1)

It is given that . Substitute $0$ for $t$ in the equation (1).

  $\begin{array}{c}〈{\tan }^{-1}0+a,\sqrt{0+1}+b〉=〈1,1〉\\ 〈a,1+b〉=〈1,1〉\end{array}$

Simplifying the above equation.

  $a=1,b=0$

Substituting $1$ for $a$ and $0$ for $b$ in equation (1).

  

Thus, the position vector is .