Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 31

$$\,\,\,\,{\left. {\frac{{dw}}{{ds}}} \right|_{s = 1/4}} = - \pi $$

$$\eqalign{ & w = {r^2} - r\tan \theta ;\,\,\,\,r = \sqrt s ,\,\,\,\,\,\theta = \pi s \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial w}}{{\partial r}}{\text{ and }}\frac{{\partial w}}{{\partial \theta }} \cr & \,\,\,\,\frac{{\partial w}}{{\partial r}} = \frac{\partial }{{\partial r}}\left( {{r^2} - r\tan \theta } \right) \cr & \,\,\,\,{\text{treat }}\theta {\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial w}}{{\partial r}} = 2r - \tan \theta \cr & \cr & \,\,\,\,\frac{{\partial w}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}\left( {{r^2} - r\tan \theta } \right) \cr & \,\,\,\,{\text{treat }}r{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial w}}{{\partial \theta }} = - r{\sec ^2}\theta \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{dr}}{{ds}}{\text{ and }}\frac{{d\theta }}{{ds}} \cr & \,\,\,\,\frac{{dr}}{{ds}} = \frac{d}{{ds}}\left[ {\sqrt s } \right] = \frac{1}{{2\sqrt s }} \cr & \,\,\,\,\frac{{d\theta }}{{ds}} = \frac{d}{{ds}}\left[ {\pi s} \right] = \pi \cr & \cr & {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr & \,\,\,\,\frac{{dw}}{{ds}} = \frac{{\partial w}}{{\partial r}}\,\frac{{dr}}{{ds}} + \frac{{\partial w}}{{\partial \theta }}\frac{{d\theta }}{{ds}} \cr & \,\,\,\,\,{\text{Substitute the derivatives}} \cr & \,\,\,\,\frac{{dw}}{{ds}} = \left( {2r - \tan \theta } \right)\,\left( {\frac{1}{{2\sqrt s }}} \right) + \left( { - r{{\sec }^2}\theta } \right)\left( \pi \right) \cr & \,\,\,\,{\text{Where }}\,r = \sqrt s ,\,\,\,\,\,\theta = \pi s \cr & \,\,\,\,\frac{{dw}}{{ds}} = \left( {2\sqrt s - \tan \pi s} \right)\,\left( {\frac{1}{{2\sqrt s }}} \right) + \left( { - \sqrt s {{\sec }^2}\pi s} \right)\left( \pi \right) \cr & \cr & {\text{Calculate }}{\left. {\frac{{dw}}{{ds}}} \right|_{s = 1/4}} \cr & \,\,\,\,{\left. {\frac{{dw}}{{ds}}} \right|_{s = 1/4}} = \left( {2\sqrt {1/4} - \tan \left( {\pi /4} \right)} \right)\,\left( {\frac{1}{{2\sqrt {1/4} }}} \right) + \left( { - \sqrt {1/4} {{\sec }^2}\left( {\pi /4} \right)} \right)\left( \pi \right) \cr & \,\,\,\,\,{\text{simplifying}} \cr & \,\,\,\,{\left. {\frac{{dw}}{{ds}}} \right|_{s = 1/4}} = \left( {1 - 1} \right)\,\left( {\frac{1}{1}} \right) + \left( { - \left( {1/2} \right)\left( 2 \right)} \right)\left( \pi \right) \cr & \,\,\,\,{\left. {\frac{{dw}}{{ds}}} \right|_{s = 1/4}} = - \pi \cr} $$