Science:Math Exam Resources/Courses/MATH104/December 2012/Question 05 (b)/Solution 1

The diagram from part a has been recopied for ease of reading

Proceeding via the hints, we are interested in solving for ${\frac {dy}{dt}}$. We first note that triangles $T_{2}$ and triangle BPL are similar (they share an angle and both contain a 90 degree angle). Thus, we have that

${\frac {y}{9}}={\frac {D}{15}}$.

Multiplying by 15, reducing and differentiating with respect to time gives

${\frac {5}{3}}{\frac {dy}{dt}}={\frac {dD}{dt}}$.

To solve for ${\frac {dy}{dt}}$ we will need ${\frac {dD}{dt}}$, the speed at which Bob's shadow tip is changing. To do this we consider that the distance between the two tips is decreasing by 11ft/s and so

${\frac {dC}{dt}}+{\frac {dD}{dt}}=-11$.

To determine ${\frac {dC}{dt}}$ we turn to the similar triangle problem for Alice. Triangles $T_{1}$ and triangle APL are similar (they share an angle and both contain a 90 degree angle). Thus, we have that

${\frac {x}{10}}={\frac {C}{15}}$

Multiplying by 15, reducing and differentiating with respect to time gives

${\frac {3}{2}}{\frac {dx}{dt}}={\frac {dC}{dt}}$

We are given that ${\frac {dx}{dt}}=-4$ and so we have that

${\frac {dC}{dt}}=-6$.

Therefore we have that

${\frac {dD}{dt}}=-11-{\frac {dC}{dt}}=-5$

and finally then that

${\frac {dy}{dt}}={\frac {3}{5}}{\frac {dD}{dt}}=-{\frac {3}{5}}(5)=-3$

and so Bob is walking towards the lamppost at 3ft/s. Notice the sign we get makes sense because we are told that each person is walking towards the lamp in the question!