Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Let's solve the given problem step-by-step:
### (a) Finding the Unit Tangent Vector [tex]\( T(t) \)[/tex] and Unit Normal Vector [tex]\( N(t) \)[/tex]
1. Compute the derivative [tex]\( \mathbf{r}'(t) \)[/tex]:
[tex]\[ \mathbf{r}(t) = \left\langle 4t^2, \sin(t) - t \cos(t), \cos(t) + t \sin(t) \right\rangle \][/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ \mathbf{r}'(t) = \left\langle 8t, \frac{d}{dt} (\sin(t) - t \cos(t)), \frac{d}{dt} (\cos(t) + t \sin(t)) \right\rangle \][/tex]
Calculate the derivatives:
[tex]\[ \frac{d}{dt}(\sin(t) - t \cos(t)) = \cos(t) - [\cos(t) - t (-\sin(t))] = -t \sin(t) \][/tex]
[tex]\[ \frac{d}{dt}(\cos(t) + t \sin(t)) = -\sin(t) + [\sin(t) + t \cos(t)] = t \cos(t) \][/tex]
Thus:
[tex]\[ \mathbf{r}'(t) = \left\langle 8t, -t \sin(t), t \cos(t) \right\rangle \][/tex]
2. Compute the magnitude of [tex]\( \mathbf{r}'(t) \)[/tex]:
[tex]\[ |\mathbf{r}'(t)| = \sqrt{(8t)^2 + (-t \sin(t))^2 + (t \cos(t))^2} = \sqrt{64t^2 + t^2 \sin^2(t) + t^2 \cos^2(t)} \][/tex]
Simplify using [tex]\( \sin^2(t) + \cos^2(t) = 1 \)[/tex]:
[tex]\[ |\mathbf{r}'(t)| = \sqrt{64t^2 + t^2} = \sqrt{65t^2} = \sqrt{65}t \][/tex]
3. Find the unit tangent vector [tex]\( T(t) \)[/tex]:
[tex]\[ T(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \frac{\left\langle 8t, -t \sin(t), t \cos(t) \right\rangle}{\sqrt{65}t} = \left\langle \frac{8t}{\sqrt{65}t}, \frac{-t \sin(t)}{\sqrt{65}t}, \frac{t \cos(t)}{\sqrt{65}t} \right\rangle \][/tex]
Simplify the components:
[tex]\[ T(t) = \left\langle \frac{8}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \right\rangle \][/tex]
4. Compute the derivative [tex]\( T'(t) \)[/tex]:
[tex]\( T(t) = \left\langle \frac{8}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \right\rangle \)[/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ T'(t) = \left\langle 0, \frac{-\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle \][/tex]
5. Compute the magnitude of [tex]\( T'(t) \)[/tex]:
[tex]\[ |T'(t)| = \sqrt{\left(0\right)^2 + \left(\frac{-\cos(t)}{\sqrt{65}}\right)^2 + \left(\frac{-\sin(t)}{\sqrt{65}}\right)^2} = \sqrt{\frac{\cos^2(t)}{65} + \frac{\sin^2(t)}{65}} \][/tex]
Simplify using [tex]\( \sin^2(t) + \cos^2(t) = 1 \)[/tex]:
[tex]\[ |T'(t)| = \sqrt{\frac{1}{65}} = \frac{1}{\sqrt{65}} \][/tex]
6. Find the unit normal vector [tex]\( N(t) \)[/tex]:
[tex]\[ N(t) = \frac{T'(t)}{|T'(t)|} = \frac{\left\langle 0, \frac{-\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle}{\frac{1}{\sqrt{65}}} = \left\langle 0, -\cos(t), -\sin(t) \right\rangle \][/tex]
### (b) Finding the Curvature [tex]\( \kappa(t) \)[/tex]
Use the formula [tex]\( \kappa(t) = \frac{|T'(t)|}{|\mathbf{r}'(t)|} \)[/tex]:
[tex]\[ \kappa(t) = \frac{\frac{1}{\sqrt{65}}}{\sqrt{65}t} = \frac{1}{\sqrt{65} \cdot \sqrt{65}t} = \frac{1}{65t} \][/tex]
### Final Answers
[tex]\[ \begin{array}{l} T (t)=\left\langle \frac{8}{\sqrt{65}}, \frac{-\sin (t)}{\sqrt{65}}, \frac{\cos (t)}{\sqrt{65}} \right\rangle \\ N (t)=\left\langle 0, -\cos (t), -\sin (t) \right\rangle \end{array} \][/tex]
[tex]\[ \kappa(t)= \frac{1}{65t} \][/tex]
### (a) Finding the Unit Tangent Vector [tex]\( T(t) \)[/tex] and Unit Normal Vector [tex]\( N(t) \)[/tex]
1. Compute the derivative [tex]\( \mathbf{r}'(t) \)[/tex]:
[tex]\[ \mathbf{r}(t) = \left\langle 4t^2, \sin(t) - t \cos(t), \cos(t) + t \sin(t) \right\rangle \][/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ \mathbf{r}'(t) = \left\langle 8t, \frac{d}{dt} (\sin(t) - t \cos(t)), \frac{d}{dt} (\cos(t) + t \sin(t)) \right\rangle \][/tex]
Calculate the derivatives:
[tex]\[ \frac{d}{dt}(\sin(t) - t \cos(t)) = \cos(t) - [\cos(t) - t (-\sin(t))] = -t \sin(t) \][/tex]
[tex]\[ \frac{d}{dt}(\cos(t) + t \sin(t)) = -\sin(t) + [\sin(t) + t \cos(t)] = t \cos(t) \][/tex]
Thus:
[tex]\[ \mathbf{r}'(t) = \left\langle 8t, -t \sin(t), t \cos(t) \right\rangle \][/tex]
2. Compute the magnitude of [tex]\( \mathbf{r}'(t) \)[/tex]:
[tex]\[ |\mathbf{r}'(t)| = \sqrt{(8t)^2 + (-t \sin(t))^2 + (t \cos(t))^2} = \sqrt{64t^2 + t^2 \sin^2(t) + t^2 \cos^2(t)} \][/tex]
Simplify using [tex]\( \sin^2(t) + \cos^2(t) = 1 \)[/tex]:
[tex]\[ |\mathbf{r}'(t)| = \sqrt{64t^2 + t^2} = \sqrt{65t^2} = \sqrt{65}t \][/tex]
3. Find the unit tangent vector [tex]\( T(t) \)[/tex]:
[tex]\[ T(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \frac{\left\langle 8t, -t \sin(t), t \cos(t) \right\rangle}{\sqrt{65}t} = \left\langle \frac{8t}{\sqrt{65}t}, \frac{-t \sin(t)}{\sqrt{65}t}, \frac{t \cos(t)}{\sqrt{65}t} \right\rangle \][/tex]
Simplify the components:
[tex]\[ T(t) = \left\langle \frac{8}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \right\rangle \][/tex]
4. Compute the derivative [tex]\( T'(t) \)[/tex]:
[tex]\( T(t) = \left\langle \frac{8}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \right\rangle \)[/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ T'(t) = \left\langle 0, \frac{-\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle \][/tex]
5. Compute the magnitude of [tex]\( T'(t) \)[/tex]:
[tex]\[ |T'(t)| = \sqrt{\left(0\right)^2 + \left(\frac{-\cos(t)}{\sqrt{65}}\right)^2 + \left(\frac{-\sin(t)}{\sqrt{65}}\right)^2} = \sqrt{\frac{\cos^2(t)}{65} + \frac{\sin^2(t)}{65}} \][/tex]
Simplify using [tex]\( \sin^2(t) + \cos^2(t) = 1 \)[/tex]:
[tex]\[ |T'(t)| = \sqrt{\frac{1}{65}} = \frac{1}{\sqrt{65}} \][/tex]
6. Find the unit normal vector [tex]\( N(t) \)[/tex]:
[tex]\[ N(t) = \frac{T'(t)}{|T'(t)|} = \frac{\left\langle 0, \frac{-\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle}{\frac{1}{\sqrt{65}}} = \left\langle 0, -\cos(t), -\sin(t) \right\rangle \][/tex]
### (b) Finding the Curvature [tex]\( \kappa(t) \)[/tex]
Use the formula [tex]\( \kappa(t) = \frac{|T'(t)|}{|\mathbf{r}'(t)|} \)[/tex]:
[tex]\[ \kappa(t) = \frac{\frac{1}{\sqrt{65}}}{\sqrt{65}t} = \frac{1}{\sqrt{65} \cdot \sqrt{65}t} = \frac{1}{65t} \][/tex]
### Final Answers
[tex]\[ \begin{array}{l} T (t)=\left\langle \frac{8}{\sqrt{65}}, \frac{-\sin (t)}{\sqrt{65}}, \frac{\cos (t)}{\sqrt{65}} \right\rangle \\ N (t)=\left\langle 0, -\cos (t), -\sin (t) \right\rangle \end{array} \][/tex]
[tex]\[ \kappa(t)= \frac{1}{65t} \][/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.