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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 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

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{curvature}\:\mathrm{vector}\:\mathrm{and} \\ $$$$\mathrm{its}\:\mathrm{magnitude}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point}\: \\ $$$$\overset{\rightarrow} {\mathrm{r}}\:=\:\left(\theta\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\overset{\rightarrow} {\mathrm{r}}=\:\left(\mathrm{acos}\:\theta,\mathrm{asin}\:\theta,\mathrm{a}\theta\right) \\ $$$$.\mathrm{Show}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{feet}\:\mathrm{of}\:\mathrm{the} \\ $$$$\bot\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{to}\:\mathrm{the}\:\mathrm{tangent}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{curve}\:\mathrm{that}\:\mathrm{completely}\:\mathrm{lies} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{hyperbolic}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} =\:\mathrm{a}^{\mathrm{2}} \\ $$

Answered by john santu last updated on 13/Jun/20

(dr^→ /dθ) = −asin θ i^� +acos θ j^� +ak^� (ds/dθ) = ∣(dr^→ /dθ)∣ = (√2) a τ^→ = ((dr^→ /dθ)/(ds/dθ)) = (1/((√2) )) (−sin θ i^� +cos θ j^� + k^� ) (dτ^→ /dθ) = (1/(√2)) (−cos θ i^� −sin θ j^� ) curvature vector , K^→ = (dτ^→ /ds) K^→ = ((dτ^→ /dθ)/(ds/dθ)) = (1/(a(√2))) ×(1/(√2)) (−cos θ i^� −sin θ j^� ) = −((cos θ i^� )/(2a)) −((sin θ j^� )/(2a)) magnitude ∣K∣ = (1/(2a)) .

$$\frac{\mathrm{d}\overset{\rightarrow} {\mathrm{r}}}{\mathrm{d}\theta}\:=\:−\mathrm{asin}\:\theta\:\hat {{i}}\:+\mathrm{acos}\:\theta\:\hat {\mathrm{j}}+\mathrm{a}\hat {\mathrm{k}} \\ $$$$\frac{\mathrm{ds}}{\mathrm{d}\theta}\:=\:\mid\frac{\mathrm{d}\overset{\rightarrow} {\mathrm{r}}}{\mathrm{d}\theta}\mid\:=\:\sqrt{\mathrm{2}}\:\mathrm{a} \\ $$$$\overset{\rightarrow} {\tau}\:=\:\frac{\frac{\mathrm{d}\overset{\rightarrow} {\mathrm{r}}}{\mathrm{d}\theta}}{\frac{\mathrm{ds}}{\mathrm{d}\theta}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\:\left(−\mathrm{sin}\:\theta\:\hat {\mathrm{i}}+\mathrm{cos}\:\theta\:\hat {\mathrm{j}}+\:\hat {\mathrm{k}}\right) \\ $$$$\frac{\mathrm{d}\overset{\rightarrow} {\tau}}{\mathrm{d}\theta}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left(−\mathrm{cos}\:\theta\:\hat {\mathrm{i}}\:−\mathrm{sin}\:\theta\:\hat {\mathrm{j}}\:\right) \\ $$$$\mathrm{curvature}\:\mathrm{vector}\:,\:\overset{\rightarrow} {\mathrm{K}}\:=\:\frac{\mathrm{d}\overset{\rightarrow} {\tau}}{\mathrm{ds}}\: \\ $$$$\overset{\rightarrow} {\mathrm{K}}\:=\:\frac{\mathrm{d}\overset{\rightarrow} {\tau}/\mathrm{d}\theta}{\mathrm{ds}/\mathrm{d}\theta}\:=\:\frac{\mathrm{1}}{\mathrm{a}\sqrt{\mathrm{2}}}\:×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left(−\mathrm{cos}\:\theta\:\hat {\mathrm{i}}−\mathrm{sin}\:\theta\:\hat {\mathrm{j}}\right) \\ $$$$=\:−\frac{\mathrm{cos}\:\theta\:\hat {\mathrm{i}}}{\mathrm{2a}}\:−\frac{\mathrm{sin}\:\theta\:\hat {\mathrm{j}}}{\mathrm{2a}} \\ $$$$\mathrm{magnitude}\:\mid\mathrm{K}\mid\:=\:\frac{\mathrm{1}}{\mathrm{2a}}\:. \\ $$

Commented by bemath last updated on 13/Jun/20

thank you

$$\mathrm{thank}\:\mathrm{you}\: \\ $$

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