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Question-164639




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

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