Answer
(a) Using the Law of Cosines, we get
$||AB|| = f\left( {x,y} \right) = \sqrt {{x^2} + {y^2} - 2xy\cos \theta } $
(b) The rate at which the distance between Matthew and Jessica is changing is $3.59$ m/s.
Work Step by Step
(a) From Figure 8, we see that $APB$ is a triangle where $||AB|| = f\left( {x,y} \right)$. Using the Law of Cosines (Theorem 1 of Section 1.4), we get
$||AB|{|^2} = {x^2} + {y^2} - 2xy\cos \theta $
Hence, $||AB|| = f\left( {x,y} \right) = \sqrt {{x^2} + {y^2} - 2xy\cos \theta } $
(b) From part (a) we obtain $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2} - 2xy\cos \theta } $. The partial derivatives are
$\frac{{\partial f}}{{\partial x}} = \frac{{x - y\cos \theta }}{{\sqrt {{x^2} + {y^2} - 2xy\cos \theta } }}$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = \frac{{y - x\cos \theta }}{{\sqrt {{x^2} + {y^2} - 2xy\cos \theta } }}$
Using the Chain Rule, the rate at which the distance between Matthew and Jessica is changing:
$\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}}$
$\frac{{\partial f}}{{\partial t}} = \frac{{x - y\cos \theta }}{{\sqrt {{x^2} + {y^2} - 2xy\cos \theta } }}\frac{{dx}}{{dt}} + \frac{{y - x\cos \theta }}{{\sqrt {{x^2} + {y^2} - 2xy\cos \theta } }}\frac{{dy}}{{dt}}$
In this case we have $\frac{{dx}}{{dt}} = {v_a}$ and $\frac{{dy}}{{dt}} = {v_b}$
At the moment when $x=30$, $y=20$, ${v_a} = 4$ m/s, ${v_b} = 3$ m/s and $\theta = \pi /3$, we get
$\frac{{\partial f}}{{\partial t}}{|_{\left( {x,y} \right) = \left( {30,20} \right)}} = \frac{{30 - 20\cos \pi /3}}{{\sqrt {{{30}^2} + {{20}^2} - 2\cdot30\cdot20\cos \pi /3} }}\cdot4 + \frac{{20 - 30\cos \pi /3}}{{\sqrt {{{30}^2} + {{20}^2} - 2\cdot30\cdot20\cos \pi /3} }}\cdot3$
$\frac{{\partial f}}{{\partial t}}{|_{\left( {x,y} \right) = \left( {30,20} \right)}} = \frac{{19}}{{2\sqrt 7 }} \simeq 3.59$
So, the rate at which the distance between Matthew and Jessica is changing is $3.59$ m/s.
