Question Number 87130 by jagoll last updated on 03/Apr/20
$$\mathrm{find}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{for}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$$\mathrm{r}\:=\:\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{at}\:\theta\:=\frac{\pi}{\mathrm{4}}\:? \\ $$
Commented by mr W last updated on 03/Apr/20
$${y}={r}\:\mathrm{sin}\:\theta=\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta \\ $$$${x}={r}\:\mathrm{cos}\:\theta=\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{d}\theta}×\frac{\mathrm{1}}{\frac{{dx}}{{d}\theta}}=\frac{\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta+\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta}{−\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta+\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta} \\ $$$$=\frac{\mathrm{1}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2}×\mathrm{0}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{−\mathrm{1}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2}×\mathrm{0}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=−\mathrm{1} \\ $$
Commented by mr W last updated on 03/Apr/20
Commented by john santu last updated on 03/Apr/20
$$\mathrm{sir}\:\mathrm{joggal}.\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{in}\:\mathrm{polar}\:\mathrm{coordinate}\:? \\ $$
Commented by jagoll last updated on 03/Apr/20
$$\mathrm{yes}\:\mathrm{mister}\:\mathrm{john} \\ $$
Commented by jagoll last updated on 03/Apr/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\:\mathrm{w} \\ $$