Answered: "Find the curvature of the curve r(t)=ti+t2j+(t22)k"

IMLOG10ct

Answered question

2021-11-16

Find the curvature of the curve

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

James Kilian

Beginner2021-11-17Added 20 answers

Concept:
The amount of deviation from the straight line is called curve of curvature.
Given:

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

Formula used:

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

The curvature of the curve,
On differentiating,

r^{'} (t) = (1, 2 t, t)

On differentiating again,

r^{″} (t) = (1, 2 t, t)

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

On simplifying,

| | r^{'} (t) | | = \sqrt{1 + 5 t^{2}}

Calculating

r^{'} (t) \times r^{″} (t) = [\begin{matrix} i & j & k \\ 1 & 2 t & t \\ 0 & 2 & 1 \end{matrix}]

On simplifying,

r^{'} (t) \times r^{″} (t) = (2 t - 2 t) i - (1 - 0) j + (2 - 0) k

= (0, - 1, 2)

| | r^{'} (t) \times r^{″} (t) | | = \sqrt{0 + {(- 1)}^{2} + 2^{2}} = \sqrt{5}

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

On solving

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

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