Check the solution to "Find the curvature K of the curve. r(t)=ti+t2j+t22k" - Plainmath

Cabiolab

Answered question

2021-10-22

Find the curvature K of the curve.

r (t) = t i + t^{2} j + \frac{t^{2}}{2 k}

Answer & Explanation

Layton

Skilled2021-10-23Added 89 answers

r (t) = t i + t^{2} j + \frac{t^{2}}{2} k

At first we have

{\vec{r}}^{'} (t) - \hat{i} + 2 \hat{t} j + t \hat{k}

\vec{r^{″}} = 2 \hat{j} + \hat{k}

\vec{r^{'}} (t) \times \vec{r^{″}} (t) = - \hat{j} + 2 \hat{k}

Then is

| | r^{'} (t) | | = \sqrt{1 + 4 t^{2} + t^{2}}

= \sqrt{1 + 5 t^{2}}

The curvature is

k = \frac{| | \vec{r^{'}} (t) \times \vec{r^{″}} (t) | |}{{| | \vec{r^{'}} (t) | |}^{3}}

= \frac{\sqrt{5}}{{(1 + 5 t^{2})}^{\frac{3}{2}}}

k = \frac{\sqrt{5}}{{(1 + 5 t^{2})}^{\frac{3}{2}}}

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