2015 THREE(e)

Excellence
Question
A corridor is 2 m wide.

At the end it turns 90° into another corridor.
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What is the minimum width, ww, of the second corridor if a ladder of length 5 m can be carried
horizontally around the corner?

*You must use calculus and show any derivatives that you need to find when solving this
problem.*
Official Answer
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cos⁡A=2f\cos A=\dfrac{2}{f}

f=2cos⁡Af=\dfrac{2}{\cos A}

sin⁡A=w5−f\sin A=\dfrac{w}{5-f}

w=(5−f)sin⁡Aw=(5-f)\sin A

=(5−2cos⁡A)sin⁡A=\left(5-\dfrac{2}{\cos A}\right)\sin A

=5sin⁡A−2tan⁡A=5\sin A-2\tan A

dwdA=5cos⁡A−2sec⁡2A\dfrac{dw}{dA}=5\cos A-2\sec^2 A

dwdA=0⇒5cos⁡A−2sec⁡2A=0\dfrac{dw}{dA}=0\Rightarrow 5\cos A-2\sec^2 A=0

5cos⁡3A−2=05\cos^3 A-2=0

cos⁡3A=25\cos^3 A=\dfrac{2}{5}

A=42.5∘A=42.5^\circ

w=1.55 mw=1.55\ \text{m}
Grading Criteria

Achievement (u)

  • Differentiate correctly relatedly but incorrect expression for ww.

Merit (r)

  • Correct dwdA\dfrac{dw}{dA}

Excellence T1

  • Correct solution.

Excellence T2

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Exam Paper (2015)Assessment Schedule (2015)
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