Sonia watches a painting which has its bottom edge 2 meters (m) above eye level and its top edge is…
Mathematics20256 marksCase based
Sonia watches a painting which has its bottom edge 2 meters (m) above eye level and its top edge is 3 m above eye level as shown in the diagram.
Based on the above information answer the questions that follow.
(a)[1.2]
Given $\alpha$ and $\theta$ as shown in the diagram, find $\tan\alpha$ and $\tan(\alpha+\theta)$.
(b)[1.2]
Find $\theta$ in terms of $x$ only.
(c)[1.2]
Find $\frac{d\theta}{dx}$.
(d)[1.2]
Find $x$ so that $\frac{d\theta}{dx} = 0$.
(e)[1.2]
Use 1st derivative test, find the distance Sonia should stand from the wall to maximize her viewing angle of the painting.
Final answer: $\tan^{-1}\frac{3}{x} - \tan^{-1}\frac{2}{x}$
Answer (c)
Official answer key
$\frac{d\theta}{dx} = -\frac{x^2 - 6}{(x^2+9)(x^2+4)} = -\frac{(x - \sqrt{6})(x + \sqrt{6})}{(x^2+9)(x^2+4)}$
For slightly $< \sqrt{6}$, $\frac{d\theta}{dx} > 0$; for slightly $> \sqrt{6}$, $\frac{d\theta}{dx} < 0$, i.e. $\frac{d\theta}{dx}$ is changing its sign from +ve to -ve as $x$ increases through $\sqrt{6}$. Hence, at $x = \sqrt{6}$, $\theta$ has local maximum.
Final answer: $-\frac{x^2 - 6}{(x^2+9)(x^2+4)}$
Answer (d)
Official answer key
$x = \pm\sqrt{6}$
Final answer: $x = \pm\sqrt{6}$
Answer (e)
Official answer key
Sonia should stand $\sqrt{6}$ m from the wall in order to maximize her viewing angle of the painting.
