Find the curvature vector and its magnitude at any point r^→ = (θ) of the curve r^→ = (acos θ,asin θ,aθ) .Show the locus of the feet of the ⊥ from the origin to the tangent is a curve that completely lies on the hyperbolic x^2 +y^2 −z^2 = a^2


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Question Number 98325 by bemath last updated on 13/Jun/20

$Find the curvature vector and$ $its magnitude at any point$ $\vec{r} = (θ) of the curve \vec{r} = (acos θ, asin θ, a θ)$ $. Show the locus of the feet of the$ $⊥ from the origin to the tangent$ $is a curve that completely lies$ $on the hyperbolic x^{2} + y^{2} - z^{2} = a^{2}$
	Answered by john santu last updated on 13/Jun/20

	$\frac{d \vec{r}}{d θ} = - asin θ \hat{i} + acos θ \hat{j} + a \hat{k}$ $\frac{ds}{d θ} = ∣ \frac{d \vec{r}}{d θ} ∣ = \sqrt{2} a$ $\vec{τ} = \frac{\frac{d \vec{r}}{d θ}}{\frac{ds}{d θ}} = \frac{1}{\sqrt{2}} (- \sin θ \hat{i} + \cos θ \hat{j} + \hat{k})$ $\frac{d \vec{τ}}{d θ} = \frac{1}{\sqrt{2}} (- \cos θ \hat{i} - \sin θ \hat{j})$ $curvature vector, \vec{K} = \frac{d \vec{τ}}{ds}$ $\vec{K} = \frac{d \vec{τ} / d θ}{ds / d θ} = \frac{1}{a \sqrt{2}} \times \frac{1}{\sqrt{2}} (- \cos θ \hat{i} - \sin θ \hat{j})$ $= - \frac{\cos θ \hat{i}}{2 a} - \frac{\sin θ \hat{j}}{2 a}$ $magnitude ∣ K ∣ = \frac{1}{2 a} .$
	Commented by bemath last updated on 13/Jun/20

	$thank you$
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