Answer :
Certainly! Let's solve the problem step-by-step to find the unit tangent vector [tex]\( T(t) \)[/tex] and the unit normal vector [tex]\( N(t) \)[/tex] for the given vector function:
[tex]\[ r(t) = \langle 4t^2, \sin(t) - t \cos(t), \cos(t) + t \sin(t) \rangle, \quad t > 0. \][/tex]
### Step 1: Find the Derivative [tex]\( r'(t) \)[/tex]
First, we need to find the derivative of [tex]\( r(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ r'(t) = \langle \frac{d}{dt}(4t^2), \frac{d}{dt}(\sin(t) - t\cos(t)), \frac{d}{dt}(\cos(t) + t\sin(t)) \rangle. \][/tex]
Calculate each component:
[tex]\[ \frac{d}{dt}(4t^2) = 8t, \][/tex]
[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]
Therefore:
[tex]\[ r'(t) = \langle 8t, t\sin(t), t\cos(t) \rangle. \][/tex]
### Step 2: Find the Magnitude of [tex]\( r'(t) \)[/tex]
Next, we find the magnitude (norm) of [tex]\( r'(t) \)[/tex]:
[tex]\[ \| r'(t) \| = \sqrt{(8t)^2 + (t\sin(t))^2 + (t\cos(t))^2}. \][/tex]
Simplify inside the square root:
[tex]\[ \| r'(t) \| = \sqrt{64t^2 + t^2\sin^2(t) + t^2\cos^2(t)}. \][/tex]
Use the Pythagorean identity [tex]\(\sin^2(t) + \cos^2(t) = 1\)[/tex]:
[tex]\[ \| r'(t) \| = \sqrt{64t^2 + t^2}. \][/tex]
Combine like terms:
[tex]\[ \| r'(t) \| = \sqrt{65t^2} = t\sqrt{65}. \][/tex]
### Step 3: Find the Unit Tangent Vector [tex]\( T(t) \)[/tex]
The unit tangent vector [tex]\( T(t) \)[/tex] is given by:
[tex]\[ T(t) = \frac{r'(t)}{\| r'(t) \|}. \][/tex]
Substitute [tex]\( r'(t) \)[/tex] and [tex]\( \| r'(t) \| \)[/tex]:
[tex]\[ T(t) = \frac{\langle 8t, t\sin(t), t\cos(t) \rangle}{t\sqrt{65}}. \][/tex]
Simplify the components:
[tex]\[ T(t) = \langle \frac{8t}{t\sqrt{65}}, \frac{t\sin(t)}{t\sqrt{65}}, \frac{t\cos(t)}{t\sqrt{65}} \rangle = \langle \frac{8}{\sqrt{65}}, \frac{\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \rangle. \][/tex]
### Step 4: Find the Derivative of the Unit Tangent Vector [tex]\( T'(t) \)[/tex]
Find the derivative of [tex]\( T(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ T'(t) = \left\langle \frac{d}{dt} \left( \frac{8}{\sqrt{65}} \right), \frac{d}{dt} \left( \frac{\sin(t)}{\sqrt{65}} \right), \frac{d}{dt} \left( \frac{\cos(t)}{\sqrt{65}} \right) \right\rangle. \][/tex]
Since [tex]\(\frac{8}{\sqrt{65}}\)[/tex] is a constant:
[tex]\[ \frac{d}{dt} \left( \frac{8}{\sqrt{65}} \right) = 0, \][/tex]
[tex]\[ \frac{d}{dt} \left( \frac{\sin(t)}{\sqrt{65}} \right) = \frac{\cos(t)}{\sqrt{65}}, \][/tex]
[tex]\[ \frac{d}{dt} \left( \frac{\cos(t)}{\sqrt{65}} \right) = \frac{-\sin(t)}{\sqrt{65}}. \][/tex]
So,
[tex]\[ T'(t) = \left\langle 0, \frac{\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle. \][/tex]
### Step 5: Find the Magnitude of [tex]\( T'(t) \)[/tex]
Calculate the magnitude of [tex]\( T'(t) \)[/tex]:
[tex]\[ \| T'(t) \| = \sqrt{0^2 + \left( \frac{\cos(t)}{\sqrt{65}} \right)^2 + \left( \frac{-\sin(t)}{\sqrt{65}} \right)^2}. \][/tex]
Simplify inside the square root:
[tex]\[ \| T'(t) \| = \sqrt{0 + \frac{\cos^2(t)}{65} + \frac{\sin^2(t)}{65}}. \][/tex]
Use the Pythagorean identity again:
[tex]\[ \| T'(t) \| = \sqrt{\frac{\cos^2(t) + \sin^2(t)}{65}} = \sqrt{\frac{1}{65}} = \frac{1}{\sqrt{65}}. \][/tex]
### Step 6: Find the Unit Normal Vector [tex]\( N(t) \)[/tex]
The unit normal vector [tex]\( N(t) \)[/tex] is given by:
[tex]\[ N(t) = \frac{T'(t)}{\| T'(t) \|}. \][/tex]
Substitute [tex]\( T'(t) \)[/tex] and [tex]\( \| T'(t) \| \)[/tex]:
[tex]\[ N(t) = \frac{\left\langle 0, \frac{\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle}{\frac{1}{\sqrt{65}}}. \][/tex]
Simplify:
[tex]\[ N(t) = \left\langle 0 \cdot \sqrt{65}, \frac{\cos(t)}{\sqrt{65}} \cdot \sqrt{65}, \frac{-\sin(t)}{\sqrt{65}} \cdot \sqrt{65} \right\rangle = \langle 0, \cos(t), -\sin(t) \rangle. \][/tex]
### Summary
To summarize, we have:
[tex]\[ T(t) = \langle \frac{8}{\sqrt{65}}, \frac{\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \rangle, \][/tex]
[tex]\[ N(t) = \langle 0, \cos(t), -\sin(t) \rangle. \][/tex]
[tex]\[ r(t) = \langle 4t^2, \sin(t) - t \cos(t), \cos(t) + t \sin(t) \rangle, \quad t > 0. \][/tex]
### Step 1: Find the Derivative [tex]\( r'(t) \)[/tex]
First, we need to find the derivative of [tex]\( r(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ r'(t) = \langle \frac{d}{dt}(4t^2), \frac{d}{dt}(\sin(t) - t\cos(t)), \frac{d}{dt}(\cos(t) + t\sin(t)) \rangle. \][/tex]
Calculate each component:
[tex]\[ \frac{d}{dt}(4t^2) = 8t, \][/tex]
[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]
Therefore:
[tex]\[ r'(t) = \langle 8t, t\sin(t), t\cos(t) \rangle. \][/tex]
### Step 2: Find the Magnitude of [tex]\( r'(t) \)[/tex]
Next, we find the magnitude (norm) of [tex]\( r'(t) \)[/tex]:
[tex]\[ \| r'(t) \| = \sqrt{(8t)^2 + (t\sin(t))^2 + (t\cos(t))^2}. \][/tex]
Simplify inside the square root:
[tex]\[ \| r'(t) \| = \sqrt{64t^2 + t^2\sin^2(t) + t^2\cos^2(t)}. \][/tex]
Use the Pythagorean identity [tex]\(\sin^2(t) + \cos^2(t) = 1\)[/tex]:
[tex]\[ \| r'(t) \| = \sqrt{64t^2 + t^2}. \][/tex]
Combine like terms:
[tex]\[ \| r'(t) \| = \sqrt{65t^2} = t\sqrt{65}. \][/tex]
### Step 3: Find the Unit Tangent Vector [tex]\( T(t) \)[/tex]
The unit tangent vector [tex]\( T(t) \)[/tex] is given by:
[tex]\[ T(t) = \frac{r'(t)}{\| r'(t) \|}. \][/tex]
Substitute [tex]\( r'(t) \)[/tex] and [tex]\( \| r'(t) \| \)[/tex]:
[tex]\[ T(t) = \frac{\langle 8t, t\sin(t), t\cos(t) \rangle}{t\sqrt{65}}. \][/tex]
Simplify the components:
[tex]\[ T(t) = \langle \frac{8t}{t\sqrt{65}}, \frac{t\sin(t)}{t\sqrt{65}}, \frac{t\cos(t)}{t\sqrt{65}} \rangle = \langle \frac{8}{\sqrt{65}}, \frac{\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \rangle. \][/tex]
### Step 4: Find the Derivative of the Unit Tangent Vector [tex]\( T'(t) \)[/tex]
Find the derivative of [tex]\( T(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ T'(t) = \left\langle \frac{d}{dt} \left( \frac{8}{\sqrt{65}} \right), \frac{d}{dt} \left( \frac{\sin(t)}{\sqrt{65}} \right), \frac{d}{dt} \left( \frac{\cos(t)}{\sqrt{65}} \right) \right\rangle. \][/tex]
Since [tex]\(\frac{8}{\sqrt{65}}\)[/tex] is a constant:
[tex]\[ \frac{d}{dt} \left( \frac{8}{\sqrt{65}} \right) = 0, \][/tex]
[tex]\[ \frac{d}{dt} \left( \frac{\sin(t)}{\sqrt{65}} \right) = \frac{\cos(t)}{\sqrt{65}}, \][/tex]
[tex]\[ \frac{d}{dt} \left( \frac{\cos(t)}{\sqrt{65}} \right) = \frac{-\sin(t)}{\sqrt{65}}. \][/tex]
So,
[tex]\[ T'(t) = \left\langle 0, \frac{\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle. \][/tex]
### Step 5: Find the Magnitude of [tex]\( T'(t) \)[/tex]
Calculate the magnitude of [tex]\( T'(t) \)[/tex]:
[tex]\[ \| T'(t) \| = \sqrt{0^2 + \left( \frac{\cos(t)}{\sqrt{65}} \right)^2 + \left( \frac{-\sin(t)}{\sqrt{65}} \right)^2}. \][/tex]
Simplify inside the square root:
[tex]\[ \| T'(t) \| = \sqrt{0 + \frac{\cos^2(t)}{65} + \frac{\sin^2(t)}{65}}. \][/tex]
Use the Pythagorean identity again:
[tex]\[ \| T'(t) \| = \sqrt{\frac{\cos^2(t) + \sin^2(t)}{65}} = \sqrt{\frac{1}{65}} = \frac{1}{\sqrt{65}}. \][/tex]
### Step 6: Find the Unit Normal Vector [tex]\( N(t) \)[/tex]
The unit normal vector [tex]\( N(t) \)[/tex] is given by:
[tex]\[ N(t) = \frac{T'(t)}{\| T'(t) \|}. \][/tex]
Substitute [tex]\( T'(t) \)[/tex] and [tex]\( \| T'(t) \| \)[/tex]:
[tex]\[ N(t) = \frac{\left\langle 0, \frac{\cos(t)}{\sqrt{65}}, \frac{-\sin(t)}{\sqrt{65}} \right\rangle}{\frac{1}{\sqrt{65}}}. \][/tex]
Simplify:
[tex]\[ N(t) = \left\langle 0 \cdot \sqrt{65}, \frac{\cos(t)}{\sqrt{65}} \cdot \sqrt{65}, \frac{-\sin(t)}{\sqrt{65}} \cdot \sqrt{65} \right\rangle = \langle 0, \cos(t), -\sin(t) \rangle. \][/tex]
### Summary
To summarize, we have:
[tex]\[ T(t) = \langle \frac{8}{\sqrt{65}}, \frac{\sin(t)}{\sqrt{65}}, \frac{\cos(t)}{\sqrt{65}} \rangle, \][/tex]
[tex]\[ N(t) = \langle 0, \cos(t), -\sin(t) \rangle. \][/tex]
