18. Which is the polar form of the parametric equations x = 3t and y = t ^ 2 ?

Answer 1

To find the polar form of the parametric equations, we can use the following conversions:

$r=\sqrt{x^2+y^2}$ and $\tan \theta =\frac{y}{x}$

Substituting the given parametric equations:

$r=\sqrt{(3t)^2+(t^2)^2} = \sqrt{9t^2 + t^4} = t\sqrt{9 + t^2}$

$\tan \theta =\frac{t^2}{3t} = \frac{t}{3}$

Therefore, the polar form of the parametric equations is:

$$ \boxed{r = t\sqrt{9 + t^2}, \quad \theta = \arctan \frac{t}{3}} $$