Mathematics - Application of Derivative Question with Solution | TestHub

(a) $P(1,1)$: point of intersection. Slopes of tangents are $\frac{1}{2},2$.

$$\tan \theta =\frac{2-\frac{1}{2}}{1+2\cdot \frac{1}{2}}⟹\theta ={\tan }^{-1}\frac{3}{4}$$

(b) Point of intersection: $P(4,4)$. Slopes of tangents: $\frac{1}{2},-1$.

$$\tan \theta =\frac{\frac{1}{2}+1}{1+\frac{1}{2}(-1)}⟹\theta ={\tan }^{-1}3$$

(c) Sufficient to consider $x\in [0,2\pi )$. Points of intersection: $P_1\left(\frac{\pi }{4},\frac{1}{\sqrt{2}}\right)$; $P_2\left(\frac{5\pi }{4},\frac{-1}{\sqrt{2}}\right)$. Slopes of tangents at $P_1:\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$. Slopes of tangents at $P_2:-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$.

$\tan \theta =2\sqrt{2}$ in each case $⟹\theta ={\tan }^{-1}2\sqrt{2}$