Question

Find the curvature ofat the point (1,1,1)

Short answer

\({\left| {\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}} \right|_{(1,1,1)}} = \frac{1}{7}\sqrt {\frac{{19}}{{14}}} \)

Step-by-step solution

Step 1: Applied the theorem 10

By using the theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}\\r(t) = (t,{t^2},{t^3})\\r'(t) = (1,2t,3{t^2})\\r''(t) = (0,2,6t)\end{aligned}\)

Step 2: Solution of the curvature

Let’s applied the values in theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{{\left| {\sqrt {36{t^4} + 36{t^2} + 4} } \right|}}{{{{\left( {1 + 4{t^2} + 9{t^4}} \right)}^{\frac{3}{2}}}}}\\ = \frac{{\left| {\sqrt {36 + 36 + 4} } \right|}}{{{{\left( {1 + 4 + 9} \right)}^{\frac{3}{2}}}}}\\ = \frac{{\sqrt {76} }}{{{{\left( {14} \right)}^{\frac{3}{2}}}}}\\ = \frac{1}{7}\sqrt {\frac{{19}}{{14}}} \end{aligned}\)

Hence \({\left| {\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}} \right|_{(1,1,1)}} = \frac{1}{7}\sqrt {\frac{{19}}{{14}}} \)