Find the unit tangent T and the principal normal vector N to the curve at t=2. The acceleration...

Question:

Find the unit tangent T and the principal normal vector N to the curve at t=2.

(4) The acceleration of a particle is a(t)=(1,2,2t). If the particle starts at the point (0,1,-3) with an initial velocity of (2,-1,2).

a. Find the position r(t) of the particle at t=2.

b. Find the unit tangent T and the principal normal vector N to the curve at t=2

c. Find the components of the acceleration on T and N

d. What is the curvature of the trajectory of the particle at t=2

e. Find the bi-normal vector B at t=2

Finding Position Vector and Unit Vectors:

The given acceleration vector is integrated to get the velocity vector, then the velocity vector is integrated to get the position vector. Indicated unit vectors are obtained by dividing the differentiated vectors by their norms.

Answer and Explanation:

(4) Given acceleration of a particle is {eq}\displaystyle a(t)=(1,2,2t) {/eq} with an initial position vector {eq}\displaystyle (0,1,-3) {/eq} and an initial velocity of {eq}\displaystyle (2,-1,2) {/eq}:

a. Finding the position r(t) of the particle at {eq}\displaystyle t=2 {/eq}:

{eq}\begin{align*} \displaystyle \int a(t) dt &= \int \left \langle 1, 2, 2t \right \rangle dt \\ \displaystyle &=\left \langle \int 1 dt, \int 2 dt, \int 2t dt \right \rangle \\ \displaystyle \int a(t) dt &=\left \langle t+C_{1}, 2t+C_{2}, t^2+C_{3} \right \rangle \\ \displaystyle C &=v(0) \\ \displaystyle v(t) &=\left \langle t+2, 2t-1, t^2+2 \right \rangle \\ \\ \displaystyle \int v(t) dt &=\left \langle \int t+2 dt, \int 2t-1 dt, \int t^2+2 dt \right \rangle \\ \displaystyle \int v(t) dt &=\left \langle \frac{t^2}{2}+2t+C_{1}, t^2-t+C_{2}, \frac{t^3}{3}+2t+C_{3} \right \rangle \\ \displaystyle C &=r(0) \\ \displaystyle r(t) &=\left \langle \frac{t^2}{2}+2t+0, t^2-t+1, \frac{t^3}{3}+2t-3 \right \rangle \\ \displaystyle r(t) &=\left \langle \frac{t^2}{2}+2t, t^2-t+1, \frac{t^3}{3}+2t-3 \right \rangle \\ \displaystyle r(2) &=\left \langle \frac{(2)^2}{2}+2(2), (2)^2-2+1, \frac{(2)^3}{3}+2(2)-3 \right \rangle \\ \displaystyle r(2) &=\left \langle 6, 3, \frac{11}{3} \right \rangle \end{align*} {/eq}

The position of the particle is at the parameter value is {eq}\ \displaystyle \mathbf{\color{blue}{ r(2)=\left \langle 6, 3, \frac{11}{3} \right \rangle }} {/eq}.


b. Finding the unit tangent T and the principal normal vector N to the curve at {eq}\displaystyle t=2 {/eq}:

{eq}\begin{align*} \displaystyle \text{Unit tangent vector}: \\ \displaystyle {r}'(t) &=\left \langle t+2, 2t-1, t^2+2 \right \rangle \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{t^4+9t^2+9} \\ \displaystyle T(t) &=\frac{{r}'(t)}{\left \| {r}'(t) \right \|} \\ \displaystyle T(t) &=\left \langle \frac{t+2}{\sqrt{t^4+9t^2+9}}, \frac{2t-1}{\sqrt{t^4+9t^2+9}}, \frac{t^2+2}{\sqrt{t^4+9t^2+9}} \right \rangle \\ \displaystyle T(2) &=\left \langle \frac{2+2}{\sqrt{(2)^4+9(2)^2+9}}, \frac{2(2)-1}{\sqrt{(2)^4+9(2)^2+9}}, \frac{(2)^2+2}{\sqrt{(2)^4+9(2)^2+9}} \right \rangle \\ \displaystyle T(2) &=\left \langle \frac{4}{\sqrt{61}}, \frac{3}{\sqrt{61}}, \frac{6}{\sqrt{61}} \right \rangle \\ \\ \displaystyle \text{Unit normal vector}: \\ \displaystyle T(t) &=\left \langle \frac{t+2}{\sqrt{t^4+9t^2+9}}, \frac{2t-1}{\sqrt{t^4+9t^2+9}}, \frac{t^2+2}{\sqrt{t^4+9t^2+9}} \right \rangle \\ \displaystyle \frac{dT}{dt} &=\frac{d}{dt}\left \langle \frac{t+2}{\sqrt{t^4+9t^2+9}}, \frac{2t-1}{\sqrt{t^4+9t^2+9}}, \frac{t^2+2}{\sqrt{t^4+9t^2+9}} \right \rangle \\ \displaystyle {T}'(t) &=\left \langle \frac{-t^4-4t^3-18t+9}{\left(t^4+9t^2+9\right)^{\frac{3}{2}}}, \frac{-2t^4+2t^3+9t+18}{\left(t^4+9t^2+9\right)^{\frac{3}{2}}}, \frac{5t^3}{\left(t^4+9t^2+9\right)^{\frac{3}{2}}} \right \rangle \\ \displaystyle {T}'(2) &=\left \langle \frac{-(2)^4-4(2)^3-18(2)+9}{\left((2)^4+9(2)^2+9\right)^{\frac{3}{2}}}, \frac{-2(2)^4+2(2)^3+9(2)+18}{\left((2)^4+9(2)^2+9\right)^{\frac{3}{2}}}, \frac{5(2)^3}{\left((2)^4+9(2)^2+9\right)^{\frac{3}{2}}} \right \rangle \\ \displaystyle {T}'(2) &=\left \langle -\frac{75}{61^{\frac{3}{2}}}, \frac{20}{61^{\frac{3}{2}}}, \frac{40}{61^{\frac{3}{2}}} \right \rangle \\ \displaystyle \left \| {T}'(2) \right \| &=\sqrt{\left(-\frac{75}{61^{\frac{3}{2}}}\right)^2+\left(\frac{20}{61^{\frac{3}{2}}}\right)^2+\left(\frac{40}{61^{\frac{3}{2}}}\right)^2} \\ \displaystyle \left \| {T}'(2) \right \| &=\frac{5\sqrt{5}}{61} \\ \displaystyle N(t) &=\frac{{T}'(t)}{\left \| {T}'(t) \right \|} \\ \displaystyle N(2) &=\left \langle -\frac{\frac{75}{61^{\frac{3}{2}}}}{\frac{5\sqrt{5}}{61}}, \frac{\frac{20}{61^{\frac{3}{2}}}}{\frac{5\sqrt{5}}{61}}, \frac{\frac{40}{61^{\frac{3}{2}}}}{\frac{5\sqrt{5}}{61}} \right \rangle \\ \displaystyle N(2) &=\left \langle -\frac{3\sqrt{5}}{\sqrt{61}}, \frac{4}{\sqrt{305}}, \frac{8}{\sqrt{305}} \right \rangle \end{align*} {/eq}

The unit tangent vector is {eq}\ \displaystyle \mathbf{\color{blue}{ T(2)=\left \langle \frac{4}{\sqrt{61}}, \frac{3}{\sqrt{61}}, \frac{6}{\sqrt{61}} \right \rangle }} {/eq} and the principal unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ N(2)=\left \langle -\frac{3\sqrt{5}}{\sqrt{61}}, \frac{4}{\sqrt{305}}, \frac{8}{\sqrt{305}} \right \rangle }} {/eq}.


c. Finding the components of the acceleration on T and N:

{eq}\begin{align*} \displaystyle \text{Tangential component}: \\ \displaystyle {r}'(t) &=\left \langle t+2, 2t-1, t^2+2 \right \rangle \\ \displaystyle {r}''(t) &=\left \langle 1, 2, 2t \right \rangle \\ \displaystyle {r}'(t) \cdot {r}''(t) &=\left \langle t+2, 2t-1, t^2+2 \right \rangle \cdot \left \langle 1, 2, 2t \right \rangle \\ \displaystyle &=(t+2)(1)+(2t-1)(2)+(t^2+2)(2t) \\ \displaystyle {r}'(t) \cdot {r}''(t) &= 2t^3+9t \\ \displaystyle a_{T} &=\frac{{\vec{r}}'(t)\cdot {\vec{r}}''(t)}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle a_{T} &=\frac{2t^3+9t}{\sqrt{t^4+9t^2+9}} \\ \\ \displaystyle \text{Normal component}: \\ \displaystyle {r}'(t) \times {r}''(t) &=\begin{vmatrix} i & j & k\\ t+2 & 2t-1 & t^2+2\\ 1 & 2 & 2t \end{vmatrix} \\ \displaystyle &=((2t)(2t-1)-(2)(t^2+2))\vec{i}-((2t)(t+2)-(1)(t^2+2))\vec{j}+((2)(t+2)-(1)(2t-1))\vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=(2t^2-2t-4)\vec{i} +(-t^2-4t+2)\vec{j}+ 5 \vec{k} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{\left(2t^2-2t-4\right)^2+\left(-t^2-4t+2\right)^2+\left(5\right)^2} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{5t^4+45} \\ \displaystyle a_{N} &=\frac{\left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle a_{N} &=\frac{\sqrt{5t^4+45}}{\sqrt{t^4+9t^2+9}} \end{align*} {/eq}

Tangential component of acceleration vector is {eq}\ \displaystyle \mathbf{\color{blue}{ a_{T} =\frac{2t^3+9t}{\sqrt{t^4+9t^2+9}} }} {/eq} and normal component of acceleration vector is {eq}\ \displaystyle \mathbf{\color{blue}{ a_{N} =\frac{\sqrt{5t^4+45}}{\sqrt{t^4+9t^2+9}} }} {/eq}.


d. Finding the curvature of the trajectory of the particle at {eq}\displaystyle t=2 {/eq}:

{eq}\begin{align*} \displaystyle \kappa &=\frac{ \left \| {\vec{T}}'(t) \right \|}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle \kappa &=\frac{ \left \| {\vec{T}}'(2) \right \|}{\left \| {\vec{r}}'(2) \right \|} \\ \displaystyle \kappa &=\frac{\frac{5\sqrt{5}}{61}}{\sqrt{61}} \\ \displaystyle \kappa &=\frac{5\sqrt{5}}{61\sqrt{61}} \end{align*} {/eq}

The curvature of the particle is {eq}\ \displaystyle \mathbf{\color{blue}{ \kappa =\frac{5\sqrt{5}}{61\sqrt{61}} }} {/eq}.


e. Finding the bi-normal vector B at {eq}\displaystyle t=2 {/eq}:

{eq}\begin{align*} \displaystyle \vec{B}(t) &=\vec{T}(t)\times \vec{N}(t) \\ \displaystyle \vec{B}(2) &=\vec{T}(2)\times \vec{N}(2) \\ \displaystyle &=\begin{vmatrix} i & j & k\\ \frac{4}{\sqrt{61}} & \frac{3}{\sqrt{61}} & \frac{6}{\sqrt{61}} \\ -\frac{3\sqrt{5}}{\sqrt{61}} & \frac{4}{\sqrt{305}} & \frac{8}{\sqrt{305}} \end{vmatrix} \\ \displaystyle &=\left( \left( \frac{8}{\sqrt{305}} \right)\left( \frac{3}{\sqrt{61}} \right)-\left( \frac{4}{\sqrt{305}} \right)\left( \frac{6}{\sqrt{61}} \right) \right) \vec{i} - \left( \left( \frac{8}{\sqrt{305}} \right)\left( \frac{4}{\sqrt{61}} \right)-\left( -\frac{3\sqrt{5}}{\sqrt{61}} \right)\left( \frac{6}{\sqrt{61}} \right) \right) \vec{j} + \left( \left( \frac{4}{\sqrt{305}} \right)\left( \frac{4}{\sqrt{61}} \right)-\left( -\frac{3\sqrt{5}}{\sqrt{61}} \right)\left( \frac{3}{\sqrt{61}} \right) \right)\vec{k} \\ \displaystyle \vec{B}(2) &=0 \vec{i} -\frac{2}{\sqrt{5}} \vec{j} + \frac{1}{\sqrt{5}} \vec{k} \end{align*} {/eq}

The bi-normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{B}(2) = -\frac{2}{\sqrt{5}} \vec{j} + \frac{1}{\sqrt{5}} \vec{k} }} {/eq}.


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