This is a calculus 3 problem. Please explain each step clearly, no cursive writing. c) Find the maximum curvature of this curve.\[ \kappa_{\text{max}} = \]The response given is: 0.17. A red cross indicates the response is incorrect. The function $ r(t) = \langle x(t), y(t), z(t) \rangle $ is defined where\[x(t) = -2t^2 + 3t + 3\]\[y(t) = 2t - 2\]\[z(t) = 2t^2 + 5t\]Since all components are of no higher degree than quadratic, $ r'''(t) = \langle 0, 0, 0 \rangle $ and so the torsion of this curve is zero. This means the binormal vector $ B = B(t) $ is constant, and that the curve lies in a single plane.

Name: This is a calculus 3 problem. Please explain each step clearly, no cur
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Description: This is a calculus 3 problem. Please explain each step clearly, no cursive writing. c) Find the maximum curvature of this curve.\[ \kappa_{\text{max}} = \]The response given is: 0.17. A red cross indicates the response is incorrect. The function \( r

Step:- 1 Curvature κ=T'(t)r'(t); T is tangent vector, k=r'(t)×r''(t)r'(t)3 Given that, r(t)=x(t),…

Answered: r(t) = (x(t), y(t), z(t)) where x(t) =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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This is a calculus 3 problem. Please explain each step clearly, no cursive writing.

c) Find the maximum curvature of this curve.

\[ \kappa_{\text{max}} = \]

The response given is: 0.17. A red cross indicates the response is incorrect.

The function $ r(t) = \langle x(t), y(t), z(t) \rangle $ is defined where

\[
x(t) = -2t^2 + 3t + 3
\]
\[
y(t) = 2t - 2
\]
\[
z(t) = 2t^2 + 5t
\]

Since all components are of no higher degree than quadratic, $ r'''(t) = \langle 0, 0, 0 \rangle $ and so the torsion of this curve is zero. This means the binormal vector $ B = B(t) $ is constant, and that the curve lies in a single plane.

Expert Solution

Step 1

Step:- 1

Curvature

$κ = \frac{||T' (t)||}{||r' (t)||}; T i s \tan g e n t v e c t o r, k = \frac{||r' (t) \times r'' (t)||}{{||r' (t)||}^{3}}$

Given that, $r (t) = <x (t), y (t), z (t)> H e r e, x (t) = - 2 t^{2} + 3 t + 3 y (t) = 2 t - 2 z (t) = 2 t^{2} + 5 t, t h e n r' (t) = <x' (t), y' (t), z' (t)> H e r e, x' (t) = - 4 t + 3 y' (t) = 2 z' (t) = 4 t + 5, A n d r'' (t) = <x'' (t), y'' (t), z'' (t)> H e r e, x'' (t) = - 4 y'' (t) = 0 z'' (t) = 4, A n d, ||r' (t)|| = \sqrt{{(x')}^{2} + {(y')}^{2} + {(z')}^{2}} ||r' (t)|| = \sqrt{{(- 4 t + 3)}^{2} + 2^{2} + {(4 t + 5)}^{2}} ||r' (t)|| = \sqrt{32 t^{2} + 38 + 16 t} {||r' (t)||}^{3} = {(32 t^{2} + 38 + 16 t)}^{\frac{3}{2}} - - - (1) (r' (t) \times r'' (t)) = |\begin{matrix} i & j & k \\ - 4 t + 3 & 2 & 4 t + 5 \\ - 4 & 0 & 4 \end{matrix}| (r' (t) \times r'' (t)) = i (8 - 0) - j (- 16 t + 12 + 16 t + 20) + k (0 + 8) (r' (t) \times r'' (t)) = 8 i - 32 j + 8 k = <8, - 32, 8> ||(r' (t) \times r'' (t))|| = \sqrt{8^{2} + {(- 32)}^{2} + 8^{2}} = \sqrt{8^{2} + {(- 4 \times 8)}^{2} + 8^{2}} = \sqrt{8^{2} (1 + {(- 4)}^{2} + 1)} = \sqrt{8^{2} (1 + 16 + 1)} = \sqrt{8^{2} \times 18} ||(r' (t) \times r'' (t))|| = \sqrt{8^{2} \times 9 \times 2} \sqrt{8^{2} \times 3^{2} \times 2} = 8 \times 3 \sqrt{2} = 24 \sqrt{2} ||(r' (t) \times r'' (t))|| = 24 \sqrt{2} - - - (2) c u r v a t u r e k = \frac{||r' (t) \times r'' (t)||}{{||r' (t)||}^{3}} = \frac{24 \sqrt{2}}{{(32 t^{2} + 38 + 16 t)}^{\frac{3}{2}}} [b y (1) a n d (2)] c u r v a t u r e k = \frac{24 \sqrt{2}}{{(32 t^{2} + 38 + 16 t)}^{\frac{3}{2}}} - - - (3)$

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