Use the vector-valued function r(t) to find the principal unit normal vector N(t) using the...

Question:

Use the vector-valued function r(t) to find the principal unit normal vector N(t) using the alternative formula

{eq}N=\frac{(v\cdot v)a-(v\cdot a)v}{\left \| (v\cdot v)a-(v\cdot a)v \right \|}. {/eq}

{eq}r(t)=3ti+2t^2j {/eq}

N(t) = _____

Principal Unit Normal Vector:

Usually, the unit normal vector is found by dividing the differentiated unit tangent vector by its norm of derivative.

But, the alternative formula is given by using that, we have to find the unit normal vector.

Answer and Explanation:

Given vector-valued function is {eq}\displaystyle r(t)=3t \vec{i}+2t^2 \vec{j} {/eq} to find the principal unit normal vector.

{eq}\begin{align*} \displaystyle r(t) &=3t \vec{i}+2t^2 \vec{j} \\ \displaystyle \frac{dr}{dt} &=\frac{d}{dt}(3t) \vec{i}+\frac{d}{dt}(2t^2) \vec{j} \\ \displaystyle v(t) &= 3 \vec{i}+4t \vec{j} \\ \displaystyle \frac{dv}{dt} &=\frac{d}{dt}(3) \vec{i}+\frac{d}{dt}(4t) \vec{j} \\ \displaystyle a(t) &=0 \vec{i} + 4 \vec{j} \\ \displaystyle v \cdot v &=\left \langle 3, 4t \right \rangle \cdot \left \langle 3, 4t \right \rangle \\ \displaystyle &=(3)(3)+(4t)(4t) \\ \displaystyle v \cdot v &=16t^2+9 \\ \displaystyle v \cdot a &=\left \langle 3, 4t \right \rangle \cdot \left \langle 0, 4 \right \rangle \\ \displaystyle &=(3)(0)+(4t)(4) \\ \displaystyle v \cdot a &=16t \\ \displaystyle (v\cdot v)a &=\left( 16t^2+9 \right)\left \langle 0, 4 \right \rangle \\ \displaystyle &=\left \langle (0)(16t^2+9), (4)(16t^2+9) \right \rangle \\ \displaystyle (v\cdot v)a &=\left \langle 0, 64t^2+36 \right \rangle \\ \displaystyle (v\cdot a)v &=(16t)\left \langle 3, 4t \right \rangle \\ \displaystyle &=\left \langle (16t)(3), (16t)(4t) \right \rangle \\ \displaystyle (v\cdot a)v &=\left \langle 48t, 64t^2 \right \rangle \\ \displaystyle (v\cdot v)a-(v\cdot a)v &=\left \langle 0, 64t^2+36 \right \rangle - \left \langle 48t, 64t^2 \right \rangle \\ \displaystyle &=\left \langle 0-48t, 64t^2+36-64t^2 \right \rangle \\ \displaystyle (v\cdot v)a-(v\cdot a)v &=\left \langle -48t, 36 \right \rangle \\ \displaystyle \left \| (v\cdot v)a-(v\cdot a)v \right \| &=\sqrt{(-48t)^{2}+(36)^{2}} \\ \displaystyle \left \| (v\cdot v)a-(v\cdot a)v \right \| &=12\sqrt{16t^2+9} \\ \displaystyle N(t) &=\frac{(v\cdot v)a-(v\cdot a)v}{\left \| (v\cdot v)a-(v\cdot a)v \right \|} \\ \displaystyle N(t) &=\left \langle \frac{-48t}{12\sqrt{16t^2+9}}, \frac{36}{12\sqrt{16t^2+9}} \right \rangle \\ \displaystyle N(t) &=\left \langle -\frac{4t}{\sqrt{16t^2+9}}, \frac{3}{\sqrt{16t^2+9}} \right \rangle \end{align*} {/eq}

The principal unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ N(t) =\left \langle -\frac{4t}{\sqrt{16t^2+9}}, \frac{3}{\sqrt{16t^2+9}} \right \rangle }} {/eq}.


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