Question

Find the curvature ofat the point (1,0,0)

Short answer

\({\left| {\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}} \right|_{(1,0,0)}} = \frac{{\sqrt {30} }}{{18}}\)

Step-by-step solution

Step 1: Applied the theorem 10

By using the theorem 10

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

Step 2: Solution of the curvature

Let’s applied the values in theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{{\left| {\frac{1}{{{t^2}}}\left( {2 + \ln t} \right)i + 2\ln tj4/tk} \right|}}{{{{\left( {4{t^2} + \frac{1}{{{t^2}}} + {{(\ln t + 1)}^2}} \right)}^{\frac{3}{2}}}}}\\\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{{\left| {2i + 0j - 4k} \right|}}{{{{\left| 6 \right|}^{\frac{3}{2}}}}}\\ = \frac{{\sqrt {30} }}{{18}}\end{aligned}\)

Hence \({\left| {\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}} \right|_{(1,0,0)}} = \frac{{\sqrt {30} }}{{18}}\)