Binormal Vectors - Calculus 3

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Question

Find the binormal vector of $v(t)=sin(t)\hat{i}+cos(t)\hat{j}$ .

Answer

To find the binormal vector, you must first find the unit tangent vector, then the unit normal vector.

The equation for the unit tangent vector, , is

$T(t)=\frac{r(t)}{\left | r(t)\right |}$

where is the vector and $\left | r(t)\right |$ is the magnitude of the vector.

The equation for the unit normal vector,, is

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}$

where is the derivative of the unit tangent vector and $\left | T'(t)\right |$ is the magnitude of the derivative of the unit vector.

The binormal vector is the cross product of unit tangent and unit normal vectors, or

$B(t)=T(t)\times N(t)$

For this problem

$v(t)=sin(t)\hat{i}+cos(t)\hat{j}$

$\left | v(t)\right |=\sqrt{\left (sin(t) \right )^{2}+\left (cos(t) \right )^{2}}=\sqrt{1}=1$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{sin(t)\hat{i}+cos(t)\hat{j}}{1}=sin(t)\hat{i}+cos(t)\hat{j}$

$T'(t)=\frac{d}{dt}\left (sin(t)\hat{i}+cos(t)\hat{j} \right )=cos(t)\hat{i}-sin(t)\hat{j}$

$\left | T'(t)\right |=\sqrt{\left (cos(t) \right )^{2}+\left (sin(t) \right )^{2}}=\sqrt{1}=1$

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}=\frac{cos(t)\hat{i}-sin(t)\hat{j}}{1}=cos(t)\hat{i}-sin(t)\hat{j}$

$B(t)=T(t)\times N(t)=\begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \ sin(t)& cos(t)& 0\ cos(t)&-sin(t) & 0 \end{vmatrix}$

$B(t)=\hat{i}\left [ cos(t)0+sin(t)0\right ] -\hat{j}\left [ sin(t)0-cos(t)0\right ] +\hat{k}\left [ sin(t)(-sin(t))-cos(t)cos(t)\right ]$

$B(t)=0\hat{i}-0\hat{j} -\hat{k}\left [ sin^{2}(t)+cos^{2}(t)\right ]=-\hat{k}$

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Question

Find the binormal vector of $v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$ .

Answer

To find the binormal vector, you must first find the unit tangent vector, then the unit normal vector.

The equation for the unit tangent vector, , is

$T(t)=\frac{r(t)}{\left | r(t)\right |}$

where is the vector and $\left | r(t)\right |$ is the magnitude of the vector.

The equation for the unit normal vector,, is

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}$

where is the derivative of the unit tangent vector and $\left | T'(t)\right |$ is the magnitude of the derivative of the unit vector.

For this problem

$v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$

$\left | v(t)\right |=\sqrt{\left (5cos(t) \right )^{2}+\left (5sin(t) \right )^{2}+(2)^{2}}=\sqrt{25+4}=\sqrt{29}$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}}{\sqrt{29}}$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k}$

$T'(t)=\frac{d}{dt}\left (\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k} \right )$

$T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}+0\hat{k}$

$T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}$

$\left | T'(t)\right |=\sqrt{\left (\frac{-5}{\sqrt{29}}sin(t) \right )^{2}+\left (\frac{5}{\sqrt{29}}cos(t) \right )^{2}}$

$\left | T'(t)\right |=\sqrt{\frac{25}{29}sin^2(t)+\frac{25}{29}cos^2(t)}=\sqrt{\frac{25}{29}}=\frac{5}{\sqrt{29}}$

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}=\frac{\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}}{\frac{5}{\sqrt{29}}}=-sin(t)\hat{i}+cos(t)\hat{j}$

$B(t)=T(t)\times N(t)=\begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \ \frac{5}{\sqrt{29}}cos(t)& \frac{5}{\sqrt{29}}sin(t)& \frac{2}{\sqrt{29}}\ -sin(t)& cos(t) & 0 \end{vmatrix}$

$B(t)=\hat{i}\left [ \frac{5}{\sqrt{29}}sin(t)0-cos(t)\frac{2}{\sqrt{29}}\right ] -\hat{j}\left [ \frac{5}{\sqrt{29}}cos(t)0+sin(t)\frac{2}{\sqrt{29}}\right ] +\hat{k}\left [ \frac{5}{\sqrt{29}}cos(t)cos(t)+\frac{5}{\sqrt{29}}sin(t)(sin(t))\right ]$

$B(t)=\hat{i}\left [ -\frac{2}{\sqrt{29}}cos(t)\right ] -\hat{j}\left [ \frac{2}{\sqrt{29}}sin(t)\right ] +\hat{k}\left [ \frac{5}{\sqrt{29}}cos^{2}(t)+\frac{5}{\sqrt{29}}sin^{2}(t)\right ]$

$B(t)=-\frac{2}{\sqrt{29}}cos(t)\hat{i} -\frac{2}{\sqrt{29}}sin(t)\hat{j} + \frac{5}{\sqrt{29}}\hat{k}$

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Question

Find the binormal vector for:

$\mathbf{r}(t)=[2\sin(t), t, 2\cos(t)]$

Answer

The binormal vector is defined as

$\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)$ $\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)$

Where T(t) (the tangent vector) and N(t) (the normal vector) are:

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\frac{1}{\sqrt{5}}[2\cos (t), 1, -2\sin (t)]$

and

$\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=[-\sin (t), 0, -\cos (t)]$

The binormal vector is:

$\mathbf{B}(t)=\left [ \frac{-\cos(t)}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{\sin(t)}{\sqrt{5}}\right ]$

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