Maximum/Minimum Value Problem Involving Trigonometry (1 Viewer)

Sunyata · Nov 3, 2009

Hi

I was wondering if someone could please demonstrate to me how to do this question. I've spent quite some time on it without any luck.

Thank you.

Thecorey0 · Nov 3, 2009

I remember doing this one a while back, took me a bit to realise what I had to do. Could you do part i? For part ii you have to differentiate t(bus) -t(girl) and maximise it.

Thecorey0 · Nov 3, 2009

There is probably a mistake somewhere in here. All times are in minutes.

$I\ assume\ you\ could\ derive\ these.\\ T_{2}=\frac{25sec\theta}{24} \ \ T_{1}= \frac{40 +5tan\theta }{18}\\ Let\ T = T_{1} -T_{2}\\ \frac{\mathrm{dT} }{\mathrm{d} \theta }=\frac{5sec^2\theta }{6} -\frac{25sec\theta tan\theta }{24}\\ Let \frac{\mathrm{dT} }{\mathrm{d} \theta }=0\\ \therefore 0 =\frac{5sec^2\theta }{18} -\frac{25sec\theta tan\theta }{24}\\ =\frac{5sec\theta }{6}(\frac{sec\theta }{3} -\frac{5tan\theta }{4})\\ \therefore sec\theta =0 \ , which\ has \ no \ solutions.\\ Or \frac{sec\theta }{3} = \frac{5tan\theta }{4}\\ \therefore sin\theta = \frac{4}{15}\\ \theta =0.267 radians.\\ Then \ test \ this \ point \ to \ prove \ it \ is \ a \ maximum.\\ Sub \ this \ value \ into \ T \ to \ find \ out \ she \ can \ leave \ 1 \ minute \ and \ 13 \ seconds \ after \ 8:00.$

Sunyata · Nov 3, 2009

Could you please tell me what the expressions are for part (i) and I'll c if i can work it out?

Thecorey0 · Nov 3, 2009

T(1) and T(2) in the above post are the two expressions. T(1) is the time taken for the bus to go from D to P. T(2) is the time taken for Claire to run from F to P. BTW, this is a 2-unit question as you most probably know.

Maximum/Minimum Value Problem Involving Trigonometry (1 Viewer)

Sunyata

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