Find the curvature of r(t)=&lt;7t,t2,t3&gt; at the point (7,1,1)

Given,r(t)=&lt;7t,t2,t3&gt;at the point (7,1,1)γ(t)=&lt;7t, t2, t3&gt;γ′(t)=&lt;7, 2t, 3t2&gt;γ″(t)=&lt;0, 2, 6t&gt;γ′(t)×γ″(t)=[i^j^k^t2t3t2026t]=(12t2−6t2)i^−(42t−0)j^+(14−0)k^=(6t2)i^−42tj^+14k^|r′(t)×γ″(t)|=(6t2)2+(−42t)2+(14)2at (7,1,1)⇒t=1|γ′(1)×γ″(1)|=(6)2+(42)2+(14)2=36+1764+196=1996|γ′(1)|=(7)2+(2)2+(3)2=45+4+9=62k=4992(31)3/2≈0.091515

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pro4ph5e4q2

Answered question

2021-11-12

Find the curvature of

r (t) = < 7 t, t^{2}, t^{3} >

at the point

(7, 1, 1)

Answer & Explanation

Befory

Beginner2021-11-13Added 19 answers

Given,
$r (t) = < 7 t, t^{2}, t^{3} >$
at the point $(7, 1, 1)$
$γ (t) = < 7 t, t^{2}, t^{3} >$
$γ^{'} (t) = < 7, 2 t, 3 t^{2} >$
$γ^{″} (t) = < 0, 2, 6 t >$
$γ^{'} (t) \times γ^{″} (t) = [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ t & 2 t & 3 t^{2} \\ 0 & 2 & 6 t \end{matrix}]$
$= (12 t^{2} - 6 t^{2}) \hat{i} - (42 t - 0) \hat{j} + (14 - 0) \hat{k}$
$= (6 t^{2}) \hat{i} - 42 t \hat{j} + 14 \hat{k}$
$| r^{'} (t) \times γ^{″} (t) | = \sqrt{{(6 t^{2})}^{2} + {(- 42 t)}^{2} + {(14)}^{2}}$
at $(7, 1, 1) \Rightarrow t = 1$
$| γ^{'} (1) \times γ^{″} (1) | = \sqrt{{(6)}^{2} + {(4^{2})}^{2} + {(14)}^{2}}$
$= \sqrt{36 + 1764 + 196}$
$= \sqrt{1996}$
$| γ^{'} (1) | = \sqrt{{(7)}^{2} + {(2)}^{2} + {(3)}^{2}} = \sqrt{45 + 4 + 9} = \sqrt{62}$
$k = \frac{\sqrt{499}}{\sqrt{2} (31)^{3 / 2}} \approx 0.091515$

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