Science:Math Exam Resources/Courses/MATH104/December 2015/Question 11/Solution 1

Let $x$ be the distance of the woman from the spotlight and let $\ell$ be the length of her shadow on the wall (both in metres). Note that the distance from the spotlight to the wall (15 m) and the height of the woman (2 m) remain constant throughout the problem.

The quantity ${\tfrac {dx}{dt}}$ (where $t$ represents time) therefore represents the rate of change in the distance of the woman from the spotlight, i.e., her speed. As the woman walks towards the wall, her distance from the spotlight increases; hence ${\tfrac {dx}{dt}}=+{\color {OliveGreen}0.7}$ (metres per second). The quantity that we seek is ${\tfrac {d\ell }{dt}},$ the rate of change in the length of her shadow.

Next, we observe that there are two similar right-angled triangles in the problem (outlined below):

so the ratios of their leg lengths are equal, i.e., ${\frac {\ell }{15}}={\frac {2}{x}}.$ Rearranging this, we obtain $\ell ={\frac {30}{x}}.$

(Intuitively, this makes sense because her shadow should get smaller as she moves further away from the spotlight. A question to ponder: according to our equation, how tall will her shadow be when she is right in front of the wall? Does this agree with your intuition?)

We now have an equation relating $\ell$ to $x,$ which we can differentiate with respect to time to find an equation relating ${\tfrac {d\ell }{dt}}$ (which we seek) to ${\tfrac {dx}{dt}}$ (which is known).

${\begin{aligned}{\frac {d\ell }{dt}}&={\frac {d}{dt}}\left({\frac {30}{x}}\right)\\&=-{\frac {30}{x^{2}}}\cdot {\frac {dx}{dt}}\end{aligned}}$

At the instant in the question, the woman is 8 m from the wall, so she is $x=15-8={\color {OrangeRed}7}$ metres from the spotlight. We now simply substitute in all the information we know:

${\frac {d\ell }{dt}}=-{\frac {30}{{\color {OrangeRed}7}^{2}}}\cdot {\color {OliveGreen}0.7}=-{\frac {21}{49}}=-{\frac {3}{7}}\approx -0.43$

Hence the length of her shadow on the wall is decreasing at a rate of ${\color {blue}0.43\,\mathrm {m/s} }.$