Question

Use theorem 10 to find the curvature f

Short answer

\(\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{4}{{25}}\)

Step-by-step solution

Step 1: Applied the theorem 10

By using the theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}}\\r(t) = 3ti + 4\sin tj + 4\cos tk\\r'(t) = 4j\cos tj - 4k\sin (tk) + 3i\\r''(t) = - 4{j^2}\sin tj - 4{k^2}\cos (tk)\end{aligned}\)

Step 2: Solution of the curvature

Let’s applied the values in theorem 10

\(\begin{aligned}{l}\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{{\left| {\left( {4j\cos tj - 4k\sin (tk) + 3i} \right)( - 4{j^2}\sin tj - 4{k^2}\cos (tk))} \right|}}{{{{\left| {\left( {4j\cos tj - 4k\sin (tk) + 3i} \right)} \right|}^3}}}\\ = \frac{4}{{25}}\end{aligned}\)

Hence, \(\frac{{\left| {r'(t) \times r''(t)} \right|}}{{{{\left| {r'(t)} \right|}^3}}} = \frac{4}{{25}}\)