Find-the-curvature-vector-and-its-magnitude-at-any-point-r-of-the-curve-r-acos-asin-a-Show-the-locus-of-the-feet-of-the-from-the-origin-to-the-tangent-is-a-curve-that-complete

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

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

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

$$\frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{d}\theta }=-\mathrm{asin}\theta \hat{i}+\mathrm{acos}\theta \hat{\mathrm{j}}+\mathrm{a}\hat{\mathrm{k}}$$$$\frac{\mathrm{ds}}{\mathrm{d}\theta }=\mid \frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{d}\theta }\mid =\sqrt{2}\mathrm{a}$$$$\overrightarrow{\tau }=\frac{\frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{d}\theta }}{\frac{\mathrm{ds}}{\mathrm{d}\theta }}=\frac{1}{\sqrt{2}}(-\sin \theta \hat{\mathrm{i}}+\cos \theta \hat{\mathrm{j}}+\hat{\mathrm{k}})$$$$\frac{\mathrm{d}\overrightarrow{\tau }}{\mathrm{d}\theta }=\frac{1}{\sqrt{2}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}})$$$$\mathrm{curvature}\mathrm{vector},\overrightarrow{\mathrm{K}}=\frac{\mathrm{d}\overrightarrow{\tau }}{\mathrm{ds}}$$$$\overrightarrow{\mathrm{K}}=\frac{\mathrm{d}\overrightarrow{\tau }/\mathrm{d}\theta }{\mathrm{ds}/\mathrm{d}\theta }=\frac{1}{\mathrm{a}\sqrt{2}}\times \frac{1}{\sqrt{2}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}})$$$$=-\frac{\cos \theta \hat{\mathrm{i}}}{2\mathrm{a}}-\frac{\sin \theta \hat{\mathrm{j}}}{2\mathrm{a}}$$$$\mathrm{magnitude}\mid \mathrm{K}\mid =\frac{1}{2\mathrm{a}}.$$

Commented by bemath last updated on 13/Jun/20

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

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