Velocity Vector for Parametric Motion

Question

The position of a particle is given by the parametric equations $$x(t) = \arctan(t)$$ and $$y(t) = \displaystyle\frac{t}{\sqrt{t^2 + 4}}$$. What is the velocity vector of the particle at time $$t = 0$$?

Accepted Answer

$$\langle 1, \displaystyle\frac{1}{2} \rangle$$

Answer

$$\langle 1, 0 \rangle$$

Answer

$$\langle \displaystyle\frac{1}{2}, 1 \rangle$$

Answer

$$\langle 0, \displaystyle\frac{1}{2} \rangle$$

Velocity Vector for Parametric Motion

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