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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]
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