Question:If $\vec{r}=3t\hat{i} + (4t-5t^2)\hat{j}$, find the angle between $\vec{r}$ and the x-axis?
My attempt: Let the angle between $\vec{r}$ and the x-axis be $\theta$
$\vec{r}.\hat{i}=r\cdot i\cdot \cos(\theta)$
$\vec{r}.\hat{i}=r_x \cdot i_x+r_y \cdot i_y$
Therefore, $r\cdot i \cdot \cos(\theta)=r_x\cdot i_x+r_y \cdot i_y$
$\cos(\theta)=\frac{r_x \cdot i_x+r_y \cdot i_y}{r \cdot i}$
$\cos(\theta)=\frac{r_x \cdot i_x}{r*i}$
$\cos(\theta)=\frac{r_x}{r}$
$\cos(\theta)=\frac{3t}{\sqrt{(3t)^2+(4t-5t^2)^2}}$
$\cos(\theta)=\frac{3t}{\sqrt{9t^2+16t^2+25t^4-40t^3}}$
$\cos(\theta)=\frac{3t}{\sqrt{25t^2+25t^4-40t^3}}$
$\cos(\theta)=\frac{3t}{\sqrt{5t^2(5+5t^2-8t)}}$
$\cos(\theta)=\frac{3t}{t\sqrt{5(5+5t^2-8t)}}$
$\cos(\theta)=\frac{3}{\sqrt{5(5+5t^2-8t)}}$
My problem: I am unable to simplify it further. Please help.
Best Answer
As noted already, the angle $\theta$ satisfies $\displaystyle\tan \theta = \frac{4t-5t^2}{3t}=\frac{4-5t}{3}\,$, and therefore $\theta$ depends on $t\,$. Consider for example:
$\displaystyle \;t=\frac{4}{5}\;$: $\displaystyle\;\;\;\;\vec{r}=\frac{12}{5}\hat{i} \;\;\implies\;\; \theta = 0$
$\displaystyle \;t=\frac{1}{5}\;$: $\displaystyle\;\;\;\;\vec{r}=\frac{3}{5}\hat{i} + \frac{3}{5}\hat{j} \;\;\implies\;\; \theta = \frac{\pi}{4}$