Answer :
Let's start by calculating the unit tangent vector [tex]\( T(t) \)[/tex] and the unit normal vector [tex]\( N(t) \)[/tex] for the given vector function [tex]\( r(t) = \langle t, t^2, 4 \rangle \)[/tex].
### Part (a): Finding the Unit Tangent Vector [tex]\( T(t) \)[/tex]
First, we need to find the first derivative of [tex]\( r(t) \)[/tex], which we'll denote as [tex]\( r'(t) \)[/tex]:
[tex]\[ r'(t) = \frac{d}{dt} \langle t, t^2, 4 \rangle = \langle 1, 2t, 0 \rangle \][/tex]
Next, we compute the magnitude of [tex]\( r'(t) \)[/tex]:
[tex]\[ |r'(t)| = \sqrt{1^2 + (2t)^2 + 0^2} = \sqrt{1 + 4t^2} \][/tex]
The unit tangent vector [tex]\( T(t) \)[/tex] is then given by normalizing [tex]\( r'(t) \)[/tex]:
[tex]\[ T(t) = \frac{r'(t)}{|r'(t)|} = \frac{\langle 1, 2t, 0 \rangle}{\sqrt{1 + 4t^2}} = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
So, we have:
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
### Part (b): Finding the Unit Normal Vector [tex]\( N(t) \)[/tex]
Now, we need to find the second derivative of [tex]\( r(t) \)[/tex], which we'll denote as [tex]\( r''(t) \)[/tex]:
[tex]\[ r''(t) = \frac{d}{dt} r'(t) = \frac{d}{dt} \langle 1, 2t, 0 \rangle = \langle 0, 2, 0 \rangle \][/tex]
To find the unit normal vector [tex]\( N(t) \)[/tex], we need to project [tex]\( r''(t) \)[/tex] onto the unit tangent vector [tex]\( T(t) \)[/tex] and then subtract this projection from [tex]\( r''(t) \)[/tex] to get the normal component.
First, we compute the dot product [tex]\( T(t) \cdot r''(t) \)[/tex]:
[tex]\[ T(t) \cdot r''(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \cdot \langle 0, 2, 0 \rangle = \frac{2t}{\sqrt{1 + 4t^2}} \times 2 = \frac{4t}{\sqrt{1 + 4t^2}} \][/tex]
Now, we calculate the projection of [tex]\( r''(t) \)[/tex] onto [tex]\( T(t) \)[/tex]:
[tex]\[ \text{Projection} = \left( \frac{4t}{\sqrt{1 + 4t^2}} \right) T(t) = \left( \frac{4t}{\sqrt{1 + 4t^2}} \right) \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
[tex]\[ = \left\langle \frac{4t}{(1 + 4t^2)}, \frac{8t^2}{(1 + 4t^2)}, 0 \right\rangle \][/tex]
The normal component of [tex]\( r''(t) \)[/tex] is then:
[tex]\[ \text{Normal Component} = r''(t) - \text{Projection} = \langle 0, 2, 0 \rangle - \left\langle \frac{4t}{(1 + 4t^2)}, \frac{8t^2}{(1 + 4t^2)}, 0 \right\rangle = \left\langle -\frac{4t}{1 + 4t^2}, 2 - \frac{8t^2}{1 + 4t^2}, 0 \right\rangle \][/tex]
Simplify the components:
[tex]\[ \text{Normal Component} = \left\langle -\frac{4t}{1 + 4t^2}, \frac{2(1 + 4t^2) - 8t^2}{1 + 4t^2}, 0 \right\rangle = \left\langle -\frac{4t}{1 + 4t^2}, \frac{2 + 8t^2 - 8t^2}{1 + 4t^2}, 0 \right\rangle = \left\langle -\frac{4t}{1 + 4t^2}, \frac{2}{1 + 4t^2}, 0 \right\rangle \][/tex]
Next, we compute the magnitude of the normal component:
[tex]\[ |\text{Normal Component}| = \sqrt{ \left( -\frac{4t}{1 + 4t^2} \right)^2 + \left( \frac{2}{1 + 4t^2} \right)^2 + 0^2 } \][/tex]
[tex]\[ = \sqrt{ \frac{16t^2}{(1 + 4t^2)^2} + \frac{4}{(1 + 4t^2)^2} } = \sqrt{ \frac{16t^2 + 4}{(1 + 4t^2)^2} } = \sqrt{ \frac{4(4t^2 + 1)}{(1 + 4t^2)^2} } = \frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2} \][/tex]
Finally, we normalize the normal component to get the unit normal vector [tex]\( N(t) \)[/tex]:
[tex]\[ N(t) = \frac{\left\langle -\frac{4t}{1 + 4t^2}, \frac{2}{1 + 4t^2}, 0 \right\rangle}{\frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2}} = \left\langle -\frac{4t}{(1 + 4t^2) \cdot \frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2}}, \frac{2}{(1 + 4t^2) \cdot \frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2}}, 0 \right\rangle \][/tex]
[tex]\[ = \left\langle -\frac{4t}{2 \sqrt{4t^2 + 1}}, \frac{2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle = \left\langle -\frac{4t}{2 \sqrt{4t^2 + 1}}, \frac{1}{\sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
[tex]\[ = \left\langle -\frac{4t}{2 \sqrt{4t^2 + 1}}, \frac{2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
Thus, we have:
[tex]\[ N(t) = \left\langle -\frac{4t}{(4t^2 + 1)\sqrt{\left(\frac{16t^2}{(4t^2 +1)^2}\right)+ \left(\frac{2 - 8t^2}{4t^2+1}\right)}}, \frac{2 - 8t^2}{(4t^2 + 1)\sqrt{\left(\frac{16t^2}{(4t^2 +1)^2}\right)+ \left(\frac{2 - 8t^2}{4t^2 +1}\right)}}, 0 \right\rangle]\][/tex]
So, the unit normal vector N(t)=
[tex]\[ [ -4t/((4t^{2}+1)\sqrt{16t^{2}/(4t^{2}+1)^{2} + ( -8t^{2}/(4t^{2} +1)+2)^{2}}), \ ( -8t^{2}/(4t^{2}+1)+2)/\sqrt{16t^{2} /(4t^{2} +1)^{2}+(-8t^{2}/4t^{2} +1)^{2}}, \ 0 \][/tex]
Now, we have the unit tangent vector [tex]\(T(t)\)[/tex] and the unit normal vector[tex]\(N(t)\)[/tex]:
[tex]\[ \begin {array}{l} T(t)=\left \langle \frac{1} {\sqrt {4t^{2}+1}}, \ \frac {2t}{\sqrt {4t^{2}+1}}, \ 0 \right \rangle \\ \\ N(t )=[-4t/((4t^{2}+1)\sqrt{16t^{2} /(4t^{2} +1)^{2} + (-8t^{2}/4t^{2}+1) + 2)^{2 }), \ ( -8t^{2}/(4t^{2}+1)+2) /\sqrt{(4t^{2}+2)+ \left ((-8t^{2})^2 \right) }/(4t^{2}+1)), 0] \\ \][/tex]
### Part (a): Finding the Unit Tangent Vector [tex]\( T(t) \)[/tex]
First, we need to find the first derivative of [tex]\( r(t) \)[/tex], which we'll denote as [tex]\( r'(t) \)[/tex]:
[tex]\[ r'(t) = \frac{d}{dt} \langle t, t^2, 4 \rangle = \langle 1, 2t, 0 \rangle \][/tex]
Next, we compute the magnitude of [tex]\( r'(t) \)[/tex]:
[tex]\[ |r'(t)| = \sqrt{1^2 + (2t)^2 + 0^2} = \sqrt{1 + 4t^2} \][/tex]
The unit tangent vector [tex]\( T(t) \)[/tex] is then given by normalizing [tex]\( r'(t) \)[/tex]:
[tex]\[ T(t) = \frac{r'(t)}{|r'(t)|} = \frac{\langle 1, 2t, 0 \rangle}{\sqrt{1 + 4t^2}} = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
So, we have:
[tex]\[ T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
### Part (b): Finding the Unit Normal Vector [tex]\( N(t) \)[/tex]
Now, we need to find the second derivative of [tex]\( r(t) \)[/tex], which we'll denote as [tex]\( r''(t) \)[/tex]:
[tex]\[ r''(t) = \frac{d}{dt} r'(t) = \frac{d}{dt} \langle 1, 2t, 0 \rangle = \langle 0, 2, 0 \rangle \][/tex]
To find the unit normal vector [tex]\( N(t) \)[/tex], we need to project [tex]\( r''(t) \)[/tex] onto the unit tangent vector [tex]\( T(t) \)[/tex] and then subtract this projection from [tex]\( r''(t) \)[/tex] to get the normal component.
First, we compute the dot product [tex]\( T(t) \cdot r''(t) \)[/tex]:
[tex]\[ T(t) \cdot r''(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \cdot \langle 0, 2, 0 \rangle = \frac{2t}{\sqrt{1 + 4t^2}} \times 2 = \frac{4t}{\sqrt{1 + 4t^2}} \][/tex]
Now, we calculate the projection of [tex]\( r''(t) \)[/tex] onto [tex]\( T(t) \)[/tex]:
[tex]\[ \text{Projection} = \left( \frac{4t}{\sqrt{1 + 4t^2}} \right) T(t) = \left( \frac{4t}{\sqrt{1 + 4t^2}} \right) \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}}, 0 \right\rangle \][/tex]
[tex]\[ = \left\langle \frac{4t}{(1 + 4t^2)}, \frac{8t^2}{(1 + 4t^2)}, 0 \right\rangle \][/tex]
The normal component of [tex]\( r''(t) \)[/tex] is then:
[tex]\[ \text{Normal Component} = r''(t) - \text{Projection} = \langle 0, 2, 0 \rangle - \left\langle \frac{4t}{(1 + 4t^2)}, \frac{8t^2}{(1 + 4t^2)}, 0 \right\rangle = \left\langle -\frac{4t}{1 + 4t^2}, 2 - \frac{8t^2}{1 + 4t^2}, 0 \right\rangle \][/tex]
Simplify the components:
[tex]\[ \text{Normal Component} = \left\langle -\frac{4t}{1 + 4t^2}, \frac{2(1 + 4t^2) - 8t^2}{1 + 4t^2}, 0 \right\rangle = \left\langle -\frac{4t}{1 + 4t^2}, \frac{2 + 8t^2 - 8t^2}{1 + 4t^2}, 0 \right\rangle = \left\langle -\frac{4t}{1 + 4t^2}, \frac{2}{1 + 4t^2}, 0 \right\rangle \][/tex]
Next, we compute the magnitude of the normal component:
[tex]\[ |\text{Normal Component}| = \sqrt{ \left( -\frac{4t}{1 + 4t^2} \right)^2 + \left( \frac{2}{1 + 4t^2} \right)^2 + 0^2 } \][/tex]
[tex]\[ = \sqrt{ \frac{16t^2}{(1 + 4t^2)^2} + \frac{4}{(1 + 4t^2)^2} } = \sqrt{ \frac{16t^2 + 4}{(1 + 4t^2)^2} } = \sqrt{ \frac{4(4t^2 + 1)}{(1 + 4t^2)^2} } = \frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2} \][/tex]
Finally, we normalize the normal component to get the unit normal vector [tex]\( N(t) \)[/tex]:
[tex]\[ N(t) = \frac{\left\langle -\frac{4t}{1 + 4t^2}, \frac{2}{1 + 4t^2}, 0 \right\rangle}{\frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2}} = \left\langle -\frac{4t}{(1 + 4t^2) \cdot \frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2}}, \frac{2}{(1 + 4t^2) \cdot \frac{2 \sqrt{4t^2 + 1}}{1 + 4t^2}}, 0 \right\rangle \][/tex]
[tex]\[ = \left\langle -\frac{4t}{2 \sqrt{4t^2 + 1}}, \frac{2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle = \left\langle -\frac{4t}{2 \sqrt{4t^2 + 1}}, \frac{1}{\sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
[tex]\[ = \left\langle -\frac{4t}{2 \sqrt{4t^2 + 1}}, \frac{2}{2 \sqrt{4t^2 + 1}}, 0 \right\rangle \][/tex]
Thus, we have:
[tex]\[ N(t) = \left\langle -\frac{4t}{(4t^2 + 1)\sqrt{\left(\frac{16t^2}{(4t^2 +1)^2}\right)+ \left(\frac{2 - 8t^2}{4t^2+1}\right)}}, \frac{2 - 8t^2}{(4t^2 + 1)\sqrt{\left(\frac{16t^2}{(4t^2 +1)^2}\right)+ \left(\frac{2 - 8t^2}{4t^2 +1}\right)}}, 0 \right\rangle]\][/tex]
So, the unit normal vector N(t)=
[tex]\[ [ -4t/((4t^{2}+1)\sqrt{16t^{2}/(4t^{2}+1)^{2} + ( -8t^{2}/(4t^{2} +1)+2)^{2}}), \ ( -8t^{2}/(4t^{2}+1)+2)/\sqrt{16t^{2} /(4t^{2} +1)^{2}+(-8t^{2}/4t^{2} +1)^{2}}, \ 0 \][/tex]
Now, we have the unit tangent vector [tex]\(T(t)\)[/tex] and the unit normal vector[tex]\(N(t)\)[/tex]:
[tex]\[ \begin {array}{l} T(t)=\left \langle \frac{1} {\sqrt {4t^{2}+1}}, \ \frac {2t}{\sqrt {4t^{2}+1}}, \ 0 \right \rangle \\ \\ N(t )=[-4t/((4t^{2}+1)\sqrt{16t^{2} /(4t^{2} +1)^{2} + (-8t^{2}/4t^{2}+1) + 2)^{2 }), \ ( -8t^{2}/(4t^{2}+1)+2) /\sqrt{(4t^{2}+2)+ \left ((-8t^{2})^2 \right) }/(4t^{2}+1)), 0] \\ \][/tex]
