How do you find the principal unit normal vector to the curve at the specified value of the parameter \[r(t)=\cos (3t)i+2\sin (3t)j+k\] where \[t=\pi \].
Answer
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Hint: We are given an equation whose principal unit vector we have to find. Firstly, we will have to find the unit tangent vector, which is, \[T(t)=\dfrac{r'(t)}{\left\| r'(t) \right\|}\]. We will find the differentiation of the given equation with respect to ‘t’ and substitute accordingly in the formula of the unit tangent vector. Then, we will substitute the values of unit tangent vector in the formula for principal unit normal vector, which is, \[N(t)=\dfrac{T'(t)}{\left\| T'(t) \right\|}\]. We will then have the required principal unit normal vector.
Complete step-by-step solution:
According to the given question, we are given an equation whose principal unit normal vector we have to find out.
The principal unit normal vector can be expressed for a vector with a smooth curve as
\[N(t)=\dfrac{T'(t)}{\left\| T'(t) \right\|}\]
Here, \[T(t)=\dfrac{r'(t)}{\left\| r'(t) \right\|}\] is called the unit tangent vector.
So, we will begin with finding the unit tangent vector and then proceed on to the principal unit normal vector.
The given equation we have is,
\[r(t)=\cos (3t)i+2\sin (3t)j+k\]-----(1)
We will find the differentiation of the equation (1), we get,
\[r'(t)=\dfrac{d}{dt}\left( (\cos 3t)i+(2\sin 3t)j+k \right)\]
We get,
\[\Rightarrow r'(t)=\dfrac{d}{dt}\left( (\cos 3t)i \right)+\dfrac{d}{dt}\left( (2\sin 3t)j \right)+\dfrac{d}{dt}(k)\]
\[\Rightarrow r'(t)=-3(\sin 3t)i+6(\cos 3t)j\]-----(2)
Now, we will find the value of the mod of the equation (2),
We have,
\[\left\| r'(t) \right\|=\left\| -3(\sin 3t)i+6(\cos 3t)j \right\|\]
\[\Rightarrow \left\| r'(t) \right\|=\sqrt{{{\left( -3(\sin 3t) \right)}^{2}}+{{\left( 6(\cos 3t) \right)}^{2}}}\]
\[\Rightarrow \left\| r'(t) \right\|=\sqrt{9{{\sin }^{2}}3t+36{{\cos }^{2}}3t}\]
Taking the common terms out, we get,
\[\Rightarrow \left\| r'(t) \right\|=\sqrt{9\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]
\[\Rightarrow \left\| r'(t) \right\|=3\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]-----(3)
We know that the formula for the unit tangent vector is,
\[T(t)=\dfrac{r'(t)}{\left\| r'(t) \right\|}\]-----(4)
We will now substitute the values of equation (2) and equation (3) in equation (4), we get,
\[\Rightarrow T(t)=\dfrac{-3(\sin 3t)i+6(\cos 3t)j}{3\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]
\[\Rightarrow T(t)=\dfrac{3\left( -(\sin 3t)i+2(\cos 3t)j \right)}{3\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]
\[\Rightarrow T(t)=\dfrac{-(\sin 3t)i+2(\cos 3t)j}{\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]----(5)
We now have the expression for the unit tangent vector, we will now find the principal unit normal vector is, \[N(t)=\dfrac{T'(t)}{\left\| T'(t) \right\|}\]-----(6)
Now, we will differentiate the equation (5) with respect to ‘t’.
We will be using the quotient rule which is, \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{u'v-v'u}{{{v}^{2}}}\]. Applying this rule, we get,
\[T'(t)=\dfrac{\dfrac{d}{dt}\left( -(\sin 3t)i+2(\cos 3t)j \right)\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}{{{\left( \sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)} \right)}^{2}}}\] \[-\dfrac{\dfrac{d}{dt}\left( \sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)} \right)\left( -(\sin 3t)i+2(\cos 3t)j \right)}{{{\left( \sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)} \right)}^{2}}}\]
\[\Rightarrow T'(t)=\dfrac{\left( -3(\cos 3t)i-6(\sin 3t)j \right)\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]
\[-\dfrac{\dfrac{1}{2}{{\left( \left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right) \right)}^{-\dfrac{1}{2}}}\left( 6\sin 3t\cos 3t+8\cos 3t(-\sin 3t)3 \right)\left( -(\sin 3t)i+2(\cos 3t)j \right)}{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]
Solving the terms further, we have,
\[\Rightarrow T'(t)=\dfrac{2\left( -3(\cos 3t)i-6(\sin 3t)j \right)\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)+9\sin 6t\left( -(\sin 3t)i+2(\cos 3t)j \right)}{2\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]
We are given that \[t=\pi \], we get now,
\[\Rightarrow T'(\pi )=\dfrac{2\left( -3(\cos 3\pi )i-6(\sin 3\pi )j \right)\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)+9\sin 6\pi \left( -(\sin 3\pi )i+2(\cos 3\pi )j \right)}{2\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)\sqrt{\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)}}\]
\[\Rightarrow T'(\pi )=\dfrac{2\left( -3(\cos 3\pi )i-6(\sin 3\pi )j \right)\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)+9\sin 6\pi \left( -(\sin 3\pi )i+2(\cos 3\pi )j \right)}{2\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)\sqrt{\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)}}\]
We will be substituting the values of the trigonometric functions in the above expression, we get.
We know that, \[\sin 3\pi =0\], \[\cos 3\pi =-1\] and \[\sin 6\pi =0\], we get,
\[\Rightarrow T'(\pi )=\dfrac{2\left( -3(-1)i-6(0)j \right)\left( 0+4{{(-1)}^{2}} \right)+9(0)\left( -(0)i+2(-1)j \right)}{2\left( 0+4{{(-1)}^{2}} \right)\sqrt{\left( 0+4{{(-1)}^{2}} \right)}}\]
\[\Rightarrow T'(\pi )=\dfrac{2\left( 3i \right)\left( 4 \right)}{2\left( 4 \right)\sqrt{\left( 4 \right)}}=\dfrac{3i}{2}\]-----(7)
We will now find the mod of the equation (7), we get,
\[\left\| T'(t) \right\|=\left\| T'(\pi ) \right\|=\left\| \dfrac{3i}{2} \right\|\]
\[\Rightarrow \left\| T'(\pi ) \right\|=\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}}=\dfrac{3}{2}\]----(8)
Now, we will substitute equations (7) and (8) in equation (6), we get,
\[N(t)=N(\pi )=\dfrac{T'(\pi )}{\left\| T'(\pi ) \right\|}\]
\[\Rightarrow N(\pi )=\dfrac{\dfrac{3i}{2}}{\dfrac{3}{2}}=i\]
Therefore, the principal unit normal vector of the given equation is \[N(\pi )=i\].
Note: The calculation of the unit tangent should be done step wise and carefully so that the principal unit normal vector can be computed correctly. The differentiation of the required equations should be done carefully so that mistakes can be avoided. Also, the value of \[t=\pi \] should be put only after solving the equations to some extent, else the calculation of values will get complicated and can lead to wrong answers.
Complete step-by-step solution:
According to the given question, we are given an equation whose principal unit normal vector we have to find out.
The principal unit normal vector can be expressed for a vector with a smooth curve as
\[N(t)=\dfrac{T'(t)}{\left\| T'(t) \right\|}\]
Here, \[T(t)=\dfrac{r'(t)}{\left\| r'(t) \right\|}\] is called the unit tangent vector.
So, we will begin with finding the unit tangent vector and then proceed on to the principal unit normal vector.
The given equation we have is,
\[r(t)=\cos (3t)i+2\sin (3t)j+k\]-----(1)
We will find the differentiation of the equation (1), we get,
\[r'(t)=\dfrac{d}{dt}\left( (\cos 3t)i+(2\sin 3t)j+k \right)\]
We get,
\[\Rightarrow r'(t)=\dfrac{d}{dt}\left( (\cos 3t)i \right)+\dfrac{d}{dt}\left( (2\sin 3t)j \right)+\dfrac{d}{dt}(k)\]
\[\Rightarrow r'(t)=-3(\sin 3t)i+6(\cos 3t)j\]-----(2)
Now, we will find the value of the mod of the equation (2),
We have,
\[\left\| r'(t) \right\|=\left\| -3(\sin 3t)i+6(\cos 3t)j \right\|\]
\[\Rightarrow \left\| r'(t) \right\|=\sqrt{{{\left( -3(\sin 3t) \right)}^{2}}+{{\left( 6(\cos 3t) \right)}^{2}}}\]
\[\Rightarrow \left\| r'(t) \right\|=\sqrt{9{{\sin }^{2}}3t+36{{\cos }^{2}}3t}\]
Taking the common terms out, we get,
\[\Rightarrow \left\| r'(t) \right\|=\sqrt{9\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]
\[\Rightarrow \left\| r'(t) \right\|=3\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]-----(3)
We know that the formula for the unit tangent vector is,
\[T(t)=\dfrac{r'(t)}{\left\| r'(t) \right\|}\]-----(4)
We will now substitute the values of equation (2) and equation (3) in equation (4), we get,
\[\Rightarrow T(t)=\dfrac{-3(\sin 3t)i+6(\cos 3t)j}{3\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]
\[\Rightarrow T(t)=\dfrac{3\left( -(\sin 3t)i+2(\cos 3t)j \right)}{3\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]
\[\Rightarrow T(t)=\dfrac{-(\sin 3t)i+2(\cos 3t)j}{\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]----(5)
We now have the expression for the unit tangent vector, we will now find the principal unit normal vector is, \[N(t)=\dfrac{T'(t)}{\left\| T'(t) \right\|}\]-----(6)
Now, we will differentiate the equation (5) with respect to ‘t’.
We will be using the quotient rule which is, \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{u'v-v'u}{{{v}^{2}}}\]. Applying this rule, we get,
\[T'(t)=\dfrac{\dfrac{d}{dt}\left( -(\sin 3t)i+2(\cos 3t)j \right)\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}{{{\left( \sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)} \right)}^{2}}}\] \[-\dfrac{\dfrac{d}{dt}\left( \sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)} \right)\left( -(\sin 3t)i+2(\cos 3t)j \right)}{{{\left( \sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)} \right)}^{2}}}\]
\[\Rightarrow T'(t)=\dfrac{\left( -3(\cos 3t)i-6(\sin 3t)j \right)\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]
\[-\dfrac{\dfrac{1}{2}{{\left( \left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right) \right)}^{-\dfrac{1}{2}}}\left( 6\sin 3t\cos 3t+8\cos 3t(-\sin 3t)3 \right)\left( -(\sin 3t)i+2(\cos 3t)j \right)}{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}\]
Solving the terms further, we have,
\[\Rightarrow T'(t)=\dfrac{2\left( -3(\cos 3t)i-6(\sin 3t)j \right)\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)+9\sin 6t\left( -(\sin 3t)i+2(\cos 3t)j \right)}{2\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)\sqrt{\left( {{\sin }^{2}}3t+4{{\cos }^{2}}3t \right)}}\]
We are given that \[t=\pi \], we get now,
\[\Rightarrow T'(\pi )=\dfrac{2\left( -3(\cos 3\pi )i-6(\sin 3\pi )j \right)\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)+9\sin 6\pi \left( -(\sin 3\pi )i+2(\cos 3\pi )j \right)}{2\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)\sqrt{\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)}}\]
\[\Rightarrow T'(\pi )=\dfrac{2\left( -3(\cos 3\pi )i-6(\sin 3\pi )j \right)\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)+9\sin 6\pi \left( -(\sin 3\pi )i+2(\cos 3\pi )j \right)}{2\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)\sqrt{\left( {{\sin }^{2}}3\pi +4{{\cos }^{2}}3\pi \right)}}\]
We will be substituting the values of the trigonometric functions in the above expression, we get.
We know that, \[\sin 3\pi =0\], \[\cos 3\pi =-1\] and \[\sin 6\pi =0\], we get,
\[\Rightarrow T'(\pi )=\dfrac{2\left( -3(-1)i-6(0)j \right)\left( 0+4{{(-1)}^{2}} \right)+9(0)\left( -(0)i+2(-1)j \right)}{2\left( 0+4{{(-1)}^{2}} \right)\sqrt{\left( 0+4{{(-1)}^{2}} \right)}}\]
\[\Rightarrow T'(\pi )=\dfrac{2\left( 3i \right)\left( 4 \right)}{2\left( 4 \right)\sqrt{\left( 4 \right)}}=\dfrac{3i}{2}\]-----(7)
We will now find the mod of the equation (7), we get,
\[\left\| T'(t) \right\|=\left\| T'(\pi ) \right\|=\left\| \dfrac{3i}{2} \right\|\]
\[\Rightarrow \left\| T'(\pi ) \right\|=\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}}=\dfrac{3}{2}\]----(8)
Now, we will substitute equations (7) and (8) in equation (6), we get,
\[N(t)=N(\pi )=\dfrac{T'(\pi )}{\left\| T'(\pi ) \right\|}\]
\[\Rightarrow N(\pi )=\dfrac{\dfrac{3i}{2}}{\dfrac{3}{2}}=i\]
Therefore, the principal unit normal vector of the given equation is \[N(\pi )=i\].
Note: The calculation of the unit tangent should be done step wise and carefully so that the principal unit normal vector can be computed correctly. The differentiation of the required equations should be done carefully so that mistakes can be avoided. Also, the value of \[t=\pi \] should be put only after solving the equations to some extent, else the calculation of values will get complicated and can lead to wrong answers.
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