2015 TWO(d)

Excellence
Question
A street light is 5 m above the ground, which is flat.

A boy, who is 1.5 m tall, is walking away from the point directly below the streetlight at
2 metres per second.

At what rate is the length of his shadow changing when the boy is 8 m away from the point
directly under the light?

*You must use calculus and show any derivatives that you need to find when solving this
problem.*
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Official Answer
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x+L5=L1.5\dfrac{x+L}{5}=\dfrac{L}{1.5}

1.5x+1.5L=5L1.5x+1.5L=5L

1.5x=3.5L1.5x=3.5L

x=7L3x=\dfrac{7L}{3}

dxdL=73\dfrac{dx}{dL}=\dfrac{7}{3}

dxdt=2\dfrac{dx}{dt}=2

dLdt=dLdx×dxdt\dfrac{dL}{dt}=\dfrac{dL}{dx}\times\dfrac{dx}{dt}

=37×2=\dfrac{3}{7}\times2

=67=0.857 m s1=\dfrac{6}{7}=0.857\ \mathrm{m\ s^{-1}}
Grading Criteria

Achievement (u)

  • dxdL\dfrac{dx}{dL} correct.

Merit (r)

-

Excellence T1

  • Correct solution with correct derivatives. (Units not required.)

Excellence T2

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