NCEA Calculus 2014 THREE(d - Solution & Explanation

Question

A container is winched up vertically from a point P at a constant rate of

1.5\ \mathrm{m\ s^{-1}}

.
It is being observed from point Q, which is

20\ \mathrm{m}

horizontally from point P.

\theta

is the angle of elevation of the container from point Q.

Loading diagram...

At what rate is the angle of elevation increasing when the object is

20\ \mathrm{m}

above point P?

*Show any derivatives that you need to find when solving this problem.*

Official Answer

\dfrac{dh}{dt}=1.5\ \text{m s}^{-1}

\tan\theta=\dfrac{h}{20}

h=20\tan\theta

\dfrac{dh}{d\theta}=20\sec^2\theta

\dfrac{d\theta}{dt}=\dfrac{d\theta}{dh}\times\dfrac{dh}{dt}

=\dfrac{1.5}{20\sec^2\theta}

When

h=20

,

\theta=\dfrac{\pi}{4}

,

\sec^2\theta=2

\dfrac{d\theta}{dt}=\dfrac{1.5}{40}=0.0375\ \text{radians s}^{-1}

Grading Criteria

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2014 THREE(d)