Question Number 164639 by cortano1 last updated on 20/Jan/22
Answered by mr W last updated on 20/Jan/22
$${x}={a}\:\mathrm{cos}\:\mathrm{2}\pi{t}\:\Rightarrow\frac{{dx}}{{dt}}=−\mathrm{2}\pi{a}\:\mathrm{sin}\:\mathrm{2}\pi{t} \\ $$$${y}={a}\:\mathrm{sin}\:\mathrm{2}\pi{t}\:\Rightarrow\frac{{dy}}{{dt}}=\mathrm{2}\pi{a}\:\mathrm{cos}\:\mathrm{2}\pi{t} \\ $$$${z}={bt}\:\Rightarrow\frac{{dz}}{{dt}}={b} \\ $$$${at}\:{t}=\mathrm{1}: \\ $$$${x}={a},\:{y}=\mathrm{0},\:{z}={b} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{0},\:\frac{{dy}}{{dt}}=\mathrm{2}\pi{a},\:\frac{{dz}}{{dt}}={b} \\ $$$$ \\ $$$${tangent}\:{vector}: \\ $$$$\left(\mathrm{0},\:\mathrm{2}\pi{a},\:{b}\right) \\ $$$$ \\ $$$${tangent}\:{line}: \\ $$$$\frac{{x}−{a}}{\mathrm{0}}=\frac{{y}−\mathrm{0}}{\mathrm{2}\pi{a}}=\frac{{z}−{b}}{{b}} \\ $$$$\Rightarrow\frac{{a}−{x}}{\mathrm{0}}=\frac{{y}}{\mathrm{2}\pi{a}}=\frac{{z}−{b}}{{b}} \\ $$$$ \\ $$$${speed}: \\ $$$${v}=\sqrt{\left(\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{2}\pi{a}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }=\sqrt{\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$${normal}\:{plane}: \\ $$$$\mathrm{0}\left({x}−{a}\right)+\mathrm{2}\pi{a}\left({y}−\mathrm{0}\right)+{b}\left({z}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\pi{ay}+{bz}={b}^{\mathrm{2}} \\ $$