Chapter 15 - Differentiation in Several Variables - 15.6 The Chain Rule - Exercises - Page 809: 20

(a) $\frac{{\partial \theta }}{{\partial a}} = - \frac{{a - b\cos \theta }}{{ab\sin \theta }}$ $\frac{{\partial \theta }}{{\partial b}} = - \frac{{b - a\cos \theta }}{{ab\sin \theta }}$ $\frac{{\partial \theta }}{{\partial c}} = \frac{c}{{ab\sin \theta }}$ (b) the angle $\theta $ will increase by approximately $0.052$ radians.

(a) Recall the Law of Cosines: ${c^2} = {a^2} + {b^2} - 2ab\cos \theta $. Write $F\left( {a,b,c,\theta } \right) \equiv {a^2} + {b^2} - {c^2} - 2ab\cos \theta $. So, the Law of Cosines becomes $F\left( {a,b,c,\theta } \right) = 0$ Suppose $\theta $ is defined implicitly such that $\theta = \theta \left( {a,b,c} \right)$, a function of $a$, $b$, and $c$. Using the Chain Rule we compute the following: 1. differentiate $F$ with respect to $a$: $\frac{{\partial F}}{{\partial a}}\frac{{\partial a}}{{\partial a}} + \frac{{\partial F}}{{\partial b}}\frac{{\partial b}}{{\partial a}} + \frac{{\partial F}}{{\partial c}}\frac{{\partial c}}{{\partial a}} + \frac{{\partial F}}{{\partial \theta }}\frac{{\partial \theta }}{{\partial a}} = 0$ We have $\frac{{\partial a}}{{\partial a}} = 1$, ${\ }$ $\frac{{\partial b}}{{\partial a}} = 0$ and ${\ }$ $\frac{{\partial c}}{{\partial a}} = 0$. So, $\frac{{\partial F}}{{\partial a}} + \frac{{\partial F}}{{\partial \theta }}\frac{{\partial \theta }}{{\partial a}} = 0$ $2a - 2b\cos \theta + \left( {2ab\sin \theta } \right)\frac{{\partial \theta }}{{\partial a}} = 0$ $\frac{{\partial \theta }}{{\partial a}} = - \frac{{a - b\cos \theta }}{{ab\sin \theta }}$ 2. differentiate $F$ with respect to $b$: $\frac{{\partial F}}{{\partial a}}\frac{{\partial a}}{{\partial b}} + \frac{{\partial F}}{{\partial b}}\frac{{\partial b}}{{\partial b}} + \frac{{\partial F}}{{\partial c}}\frac{{\partial c}}{{\partial b}} + \frac{{\partial F}}{{\partial \theta }}\frac{{\partial \theta }}{{\partial b}} = 0$ We have $\frac{{\partial b}}{{\partial b}} = 1$, ${\ }$ $\frac{{\partial a}}{{\partial b}} = 0$ and ${\ }$ $\frac{{\partial c}}{{\partial b}} = 0$. So, $\frac{{\partial F}}{{\partial b}} + \frac{{\partial F}}{{\partial \theta }}\frac{{\partial \theta }}{{\partial b}} = 0$ $2b - 2a\cos \theta + \left( {2ab\sin \theta } \right)\frac{{\partial \theta }}{{\partial b}} = 0$ $\frac{{\partial \theta }}{{\partial b}} = - \frac{{b - a\cos \theta }}{{ab\sin \theta }}$ 3. differentiate $F$ with respect to $c$: $\frac{{\partial F}}{{\partial a}}\frac{{\partial a}}{{\partial c}} + \frac{{\partial F}}{{\partial b}}\frac{{\partial b}}{{\partial c}} + \frac{{\partial F}}{{\partial c}}\frac{{\partial c}}{{\partial c}} + \frac{{\partial F}}{{\partial \theta }}\frac{{\partial \theta }}{{\partial c}} = 0$ We have $\frac{{\partial c}}{{\partial c}} = 1$, ${\ }$ $\frac{{\partial a}}{{\partial c}} = 0$ and ${\ }$ $\frac{{\partial b}}{{\partial c}} = 0$. So, $\frac{{\partial F}}{{\partial c}} + \frac{{\partial F}}{{\partial \theta }}\frac{{\partial \theta }}{{\partial c}} = 0$ $ - 2c + \left( {2ab\sin \theta } \right)\frac{{\partial \theta }}{{\partial a}} = 0$ $\frac{{\partial \theta }}{{\partial c}} = \frac{c}{{ab\sin \theta }}$ (b) Let $a=10$, $b=16$, $c=22$. Using the Law of Cosines we compute ${22^2} = {10^2} + {16^2} - 320\cos \theta $ $\cos \theta = - \frac{2}{5}$, ${\ \ }$ $\theta \simeq 1.98$ rad If $a$ and $b$ are increased by 1 and $c$ is increased by 2, we have $\Delta a = 1$, ${\ \ }$ $\Delta b = 1$, ${\ \ }$ $\Delta c = 2$, Using the linear approximation, Eq. (5) of Section 15.4, the change in $\theta $ is estimated to be $\Delta \theta = \frac{{\partial \theta }}{{\partial a}}\Delta a + \frac{{\partial \theta }}{{\partial b}}\Delta b + \frac{{\partial \theta }}{{\partial c}}\Delta c$ $ = \left( { - \frac{{a - b\cos \theta }}{{ab\sin \theta }}} \right)\Delta a + \left( { - \frac{{b - a\cos \theta }}{{ab\sin \theta }}} \right)\Delta b + \left( {\frac{c}{{ab\sin \theta }}} \right)\Delta c$ Substituting $a=10$, $b=16$, $c=22$, $\Delta a = 1$, $\Delta b = 1$, $\Delta c = 2$ and $\theta \simeq 1.98$ in $\Delta \theta $ gives $\Delta \theta = \left( { - \frac{{10 - 16\cos 1.98}}{{160\sin 1.98}}} \right) + \left( { - \frac{{16 - 10\cos 1.98}}{{160\sin 1.98}}} \right) + 2\left( {\frac{{22}}{{160\sin 1.98}}} \right)$ $\Delta \theta \simeq 0.052$ So, the angle $\theta $ will increase by approximately $0.052$ radians.