Question

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and normal components of acceleration .r(t) = cos 2t i + sin 2t j + k.

Answer 1

The curvature is
`\kappa=1`

The tangential component of acceleration is
`a_{\boldsymbol{T}}=0`

The normal component of acceleration is
`a_{\boldsymbol{N}}=1 (2)^2=4`

Explanation:

To find the curvature of the path we are going to use this formula:

`\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}`

where

`\boldsymbol{T}}` is the unit tangent vector.

`(ds)/(dt)=|| \boldsymbol{r}'(t)}||` is the speed of the object

We need to find
`\boldsymbol{r}'(t)`, we know that
`\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}` so

`\boldsymbol{r}'(t)=(d)/(dt)\left(cos\left(2t\right)\right)\:\boldsymbol{i}+(d)/(dt)\left(sin\left(2t\right)\right)\:\boldsymbol{j}+(d)/(dt)\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}`

Next , we find the magnitude of derivative of the position vector

`|| \boldsymbol{r}'(t)}||=√((-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2) \\|| \boldsymbol{r}'(t)}||=√(2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right))\\|| \boldsymbol{r}'(t)}||=√(4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right))\\|| \boldsymbol{r}'(t)}||=√(4)√(\sin ^2\left(2t\right)+\cos ^2\left(2t\right))\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2√(1)=2`

The unit tangent vector is defined by

`\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}`

`\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)`

We need to find the derivative of unit tangent vector

`\boldsymbol{T}'=(d)/(dt)(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})`

And the magnitude of the derivative of unit tangent vector is

`||\boldsymbol{T}'||=2√(\cos ^2\left(x\right)+\sin ^2\left(x\right)) =2`

The curvature is

`\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=(2)/(2) =1`

The tangential component of acceleration is given by the formula

`a_{\boldsymbol{T}}=(d^2s)/(dt^2)`

We know that
`(ds)/(dt)=|| \boldsymbol{r}'(t)}||` and
`||\boldsymbol{r}'(t)}||=2`

`(d)/(dt)\left(2\right)\: = 0` so

`a_{\boldsymbol{T}}=0`

The normal component of acceleration is given by the formula

`a_{\boldsymbol{N}}=\kappa ((ds)/(dt))^2`

We know that
`\kappa=1` and
`(ds)/(dt)=2` so

`a_{\boldsymbol{N}}=1 (2)^2=4`