IBDP Maths SL 3.3 Applications of trigonometry AA HL Paper 1- Exam Style Questions- New Syllabus

(a)

Time taken = Distance ÷ Speed.

Walking across the beach:

\( \frac{RX}{0.8}=\frac{400\sec\theta}{0.8}=500\sec\theta. \)

Jogging along the promenade:

\( \frac{XQ}{1.2} =\frac{1000-400\tan\theta}{1.2} =\frac{2500-1000\tan\theta}{3}. \)

Hence,

\( T =500\sec\theta +\frac{2500-1000\tan\theta}{3}. \)

✅ Answer:

\( \boxed{T=500\sec\theta+\frac{2500-1000\tan\theta}{3}} \)

(b)

Differentiate each term with respect to \(\theta\).

\( \frac{d}{d\theta}(500\sec\theta) =500\sec\theta\tan\theta, \)

\( \frac{d}{d\theta}\left(\frac{2500-1000\tan\theta}{3}\right) =-\frac{1000}{3}\sec^2\theta. \)

Therefore,

\( \boxed{\frac{dT}{d\theta} = 500\sec\theta\tan\theta -\frac{1000}{3}\sec^2\theta.} \)

This derivative measures how the total travel time changes as the walking angle changes.

(c)

At the minimum travel time,

\( \frac{dT}{d\theta}=0. \)

Hence,

\( 500\sec\theta\tan\theta = \frac{1000}{3}\sec^2\theta. \)

Divide both sides by \(\sec\theta\):

\( 500\tan\theta = \frac{1000}{3}\sec\theta. \)

Using \( \tan\theta=\frac{\sin\theta}{\cos\theta}, \qquad \sec\theta=\frac1{\cos\theta}, \)

gives \( 500\sin\theta = \frac{1000}{3}, \)

so \( \sin\theta=\frac23. \)

Then \( \cos\theta=\frac{\sqrt5}{3}, \qquad \tan\theta=\frac2{\sqrt5}. \)

Since

\( PX=400\tan\theta, \)

we obtain

\( PX = 400\left(\frac2{\sqrt5}\right) = \frac{800}{\sqrt5} = 160\sqrt5\text{ m}. \)

Thus,

\( \boxed{PX=160\sqrt5\text{ m}.} \)

The optimization step uses the first derivative to determine the angle that minimizes the total journey time.