Answer :
Let's break down the detailed solution step-by-step.
### Part (a): Finding the unit tangent vector [tex]\( T(t) \)[/tex] and unit normal vector [tex]\( N(t) \)[/tex]
#### 1. Find [tex]\( r'(t) \)[/tex]
Given the vector function:
[tex]\[ r(t) = \langle t, t^2, 4 \rangle \][/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ r'(t) = \langle 1, 2t, 0 \rangle \][/tex]
#### 2. Find the magnitude of [tex]\( r'(t) \)[/tex]
[tex]\[ |r'(t)| = \sqrt{1^2 + (2t)^2 + 0^2} = \sqrt{1 + 4t^2} \][/tex]
#### 3. Find the unit tangent vector [tex]\( T(t) \)[/tex]
[tex]\[ T(t) = \frac{r'(t)}{|r'(t)|} = \frac{\langle 1, 2t, 0 \rangle}{\sqrt{1 + 4t^2}} \][/tex]
So,
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
#### 4. Differentiate [tex]\( T(t) \)[/tex] to find [tex]\( T'(t) \)[/tex]
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ \left( \frac{1}{\sqrt{1 + 4t^2}} \right)' = -\frac{4t}{(1 + 4t^2)^{3/2}} \][/tex]
[tex]\[ \left( \frac{2t}{\sqrt{1 + 4t^2}} \right)' = \frac{2\sqrt{1 + 4t^2} - 2t \cdot \left( \frac{4t}{\sqrt{1 + 4t^2}} \right)}{(1 + 4t^2)^{3/2}} = \frac{2(1 + 4t^2) - 8t^2}{(1 + 4t^2)^{3/2}} = \frac{2 - 4t^2}{(1 + 4t^2)^{3/2}} \][/tex]
So,
[tex]\[ T'(t) = \left\langle -\frac{4t}{(1 + 4t^2)^{3/2}}, \frac{2 - 4t^2}{(1 + 4t^2)^{3/2}}, 0 \right\rangle \][/tex]
#### 5. Find the magnitude of [tex]\( T'(t) \)[/tex]
[tex]\[ |T'(t)| = \sqrt{\left( -\frac{4t}{(1 + 4t^2)^{3/2}} \right)^2 + \left( \frac{2 - 4t^2}{(1 + 4t^2)^{3/2}} \right)^2} \][/tex]
[tex]\[ |T'(t)| = \sqrt{\frac{16t^2}{(1 + 4t^2)^3} + \frac{(2 - 4t^2)^2}{(1 + 4t^2)^3}} \][/tex]
[tex]\[ |T'(t)| = \sqrt{\frac{16t^2 + (2 - 4t^2)^2}{(1 + 4t^2)^3}} \][/tex]
Simplify the numerator:
[tex]\[ (2 - 4t^2)^2 = 4 - 16t^2 + 16t^4 \][/tex]
[tex]\[ 16t^2 + 4 - 16t^2 + 16t^4 = 4 + 16t^4 \][/tex]
Therefore,
[tex]\[ |T'(t)| = \sqrt{\frac{4 + 16t^4}{(1 + 4t^2)^3}} \][/tex]
[tex]\[ |T'(t)| = \frac{\sqrt{4 + 16t^4}}{(1 + 4t^2)^{3/2}} \][/tex]
#### 6. Find the unit normal vector [tex]\( N(t) \)[/tex]
[tex]\[ N(t) = \frac{T'(t)}{|T'(t)|} \][/tex]
Using the computed values:
[tex]\[ N(t) = \frac{1}{\sqrt{4 + 16t^4}} \left\langle -4t, 2 - 4t^2, 0 \right\rangle \][/tex]
Or simplifying:
[tex]\[ N(t) = \left\langle \frac{-4t}{2 \sqrt{4t^2 + 1}}, \frac{2 - 4t^2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
So the unit tangent and normal vectors are:
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
[tex]\[ N(t) = \left\langle \frac{-4t}{2 \sqrt{4t^2 + 1}}, \frac{2 - 4t^2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
### Part (b): Finding the curvature [tex]\( \kappa(t) \)[/tex]
Using the formula for curvature:
[tex]\[ \kappa(t) = \frac{|T'(t)|}{|r'(t)|} \][/tex]
We already have:
[tex]\[ |T'(t)| = \sqrt{16t^2/(4t^2 + 1)^3 + (-8t^2/(4t^2 + 1)^{3/2} + 2/\sqrt(4t^2 + 1))^2} \][/tex]
[tex]\[ |r'(t)| = \sqrt{1 + 4t^2} \][/tex]
So,
[tex]\[ \kappa(t) = \frac{\sqrt{16t^2/(4t^2 + 1)^3 + (-8t^2/(4t^2 + 1)^{3/2} + 2/\sqrt(4t^2 + 1))^2}}{\sqrt{1 + 4t^2}} \][/tex]
Hence,
[tex]\[ \kappa(t) = \frac{\sqrt{16t^2/(4t^2 + 1)^3 + (-8t^2/(4t^2 + 1)^{3/2} + 2/\sqrt(4t^2 + 1))^2}}{\sqrt{4t^2 + 1}} \][/tex]
### Part (a): Finding the unit tangent vector [tex]\( T(t) \)[/tex] and unit normal vector [tex]\( N(t) \)[/tex]
#### 1. Find [tex]\( r'(t) \)[/tex]
Given the vector function:
[tex]\[ r(t) = \langle t, t^2, 4 \rangle \][/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ r'(t) = \langle 1, 2t, 0 \rangle \][/tex]
#### 2. Find the magnitude of [tex]\( r'(t) \)[/tex]
[tex]\[ |r'(t)| = \sqrt{1^2 + (2t)^2 + 0^2} = \sqrt{1 + 4t^2} \][/tex]
#### 3. Find the unit tangent vector [tex]\( T(t) \)[/tex]
[tex]\[ T(t) = \frac{r'(t)}{|r'(t)|} = \frac{\langle 1, 2t, 0 \rangle}{\sqrt{1 + 4t^2}} \][/tex]
So,
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
#### 4. Differentiate [tex]\( T(t) \)[/tex] to find [tex]\( T'(t) \)[/tex]
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
Differentiate each component with respect to [tex]\( t \)[/tex]:
[tex]\[ \left( \frac{1}{\sqrt{1 + 4t^2}} \right)' = -\frac{4t}{(1 + 4t^2)^{3/2}} \][/tex]
[tex]\[ \left( \frac{2t}{\sqrt{1 + 4t^2}} \right)' = \frac{2\sqrt{1 + 4t^2} - 2t \cdot \left( \frac{4t}{\sqrt{1 + 4t^2}} \right)}{(1 + 4t^2)^{3/2}} = \frac{2(1 + 4t^2) - 8t^2}{(1 + 4t^2)^{3/2}} = \frac{2 - 4t^2}{(1 + 4t^2)^{3/2}} \][/tex]
So,
[tex]\[ T'(t) = \left\langle -\frac{4t}{(1 + 4t^2)^{3/2}}, \frac{2 - 4t^2}{(1 + 4t^2)^{3/2}}, 0 \right\rangle \][/tex]
#### 5. Find the magnitude of [tex]\( T'(t) \)[/tex]
[tex]\[ |T'(t)| = \sqrt{\left( -\frac{4t}{(1 + 4t^2)^{3/2}} \right)^2 + \left( \frac{2 - 4t^2}{(1 + 4t^2)^{3/2}} \right)^2} \][/tex]
[tex]\[ |T'(t)| = \sqrt{\frac{16t^2}{(1 + 4t^2)^3} + \frac{(2 - 4t^2)^2}{(1 + 4t^2)^3}} \][/tex]
[tex]\[ |T'(t)| = \sqrt{\frac{16t^2 + (2 - 4t^2)^2}{(1 + 4t^2)^3}} \][/tex]
Simplify the numerator:
[tex]\[ (2 - 4t^2)^2 = 4 - 16t^2 + 16t^4 \][/tex]
[tex]\[ 16t^2 + 4 - 16t^2 + 16t^4 = 4 + 16t^4 \][/tex]
Therefore,
[tex]\[ |T'(t)| = \sqrt{\frac{4 + 16t^4}{(1 + 4t^2)^3}} \][/tex]
[tex]\[ |T'(t)| = \frac{\sqrt{4 + 16t^4}}{(1 + 4t^2)^{3/2}} \][/tex]
#### 6. Find the unit normal vector [tex]\( N(t) \)[/tex]
[tex]\[ N(t) = \frac{T'(t)}{|T'(t)|} \][/tex]
Using the computed values:
[tex]\[ N(t) = \frac{1}{\sqrt{4 + 16t^4}} \left\langle -4t, 2 - 4t^2, 0 \right\rangle \][/tex]
Or simplifying:
[tex]\[ N(t) = \left\langle \frac{-4t}{2 \sqrt{4t^2 + 1}}, \frac{2 - 4t^2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
So the unit tangent and normal vectors are:
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
[tex]\[ N(t) = \left\langle \frac{-4t}{2 \sqrt{4t^2 + 1}}, \frac{2 - 4t^2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
### Part (b): Finding the curvature [tex]\( \kappa(t) \)[/tex]
Using the formula for curvature:
[tex]\[ \kappa(t) = \frac{|T'(t)|}{|r'(t)|} \][/tex]
We already have:
[tex]\[ |T'(t)| = \sqrt{16t^2/(4t^2 + 1)^3 + (-8t^2/(4t^2 + 1)^{3/2} + 2/\sqrt(4t^2 + 1))^2} \][/tex]
[tex]\[ |r'(t)| = \sqrt{1 + 4t^2} \][/tex]
So,
[tex]\[ \kappa(t) = \frac{\sqrt{16t^2/(4t^2 + 1)^3 + (-8t^2/(4t^2 + 1)^{3/2} + 2/\sqrt(4t^2 + 1))^2}}{\sqrt{1 + 4t^2}} \][/tex]
Hence,
[tex]\[ \kappa(t) = \frac{\sqrt{16t^2/(4t^2 + 1)^3 + (-8t^2/(4t^2 + 1)^{3/2} + 2/\sqrt(4t^2 + 1))^2}}{\sqrt{4t^2 + 1}} \][/tex]
