Question

Graph the curve with parametric equationsand find the curvature at the point

Short answer

\({\left| {\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}} \right|_{(1,0,0)}} = \frac{1}{{26}}\)

Step-by-step solution

Step 1: Graph of the curvature

Graph of the curve

Step 2: Applied the theorem 10

By using the theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}\\r(t) = (\cos ti + \sin tj + \sin 5tk)\\r'(t) = ( - \sin ti - \sin tj + 5\cos 5tk)\\r''(t) = ( - \cos ti - \sin tj - 5\cos 5tk)\end{aligned}\)

Step 3: Solution of the curvature

Let’s applied the values in theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{{\left| {j + 5k} \right|}}{{{{\left| {j + 5k} \right|}^3}}}\\ = \frac{{\sqrt {26} }}{{{{\left( {\sqrt {26} } \right)}^3}}}\\ = \frac{1}{{26}}\end{aligned}\)

Hence \({\left| {\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}} \right|_{(1,0,0)}} = \frac{1}{{26}}\)