Respuesta :
Answer:
The curvature is [tex]\kappa=1[/tex]
The tangential component of acceleration is [tex]a_{\boldsymbol{T}}=0[/tex]
The normal component of acceleration is [tex]a_{\boldsymbol{N}}=1 (2)^2=4[/tex]
Step-by-step explanation:
To find the curvature of the path we are going to use this formula:
[tex]\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}[/tex]
where
[tex]\boldsymbol{T}}[/tex] is the unit tangent vector.
[tex]\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||[/tex] is the speed of the object
We need to find [tex]\boldsymbol{r}'(t)[/tex], we know that [tex]\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}[/tex] so
[tex]\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}[/tex]
Next , we find the magnitude of derivative of the position vector
[tex]|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\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\sqrt{1}=2[/tex]
The unit tangent vector is defined by
[tex]\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}[/tex]
[tex]\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)[/tex]
We need to find the derivative of unit tangent vector
[tex]\boldsymbol{T}'=\frac{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})[/tex]
And the magnitude of the derivative of unit tangent vector is
[tex]||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2[/tex]
The curvature is
[tex]\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1[/tex]
The tangential component of acceleration is given by the formula
[tex]a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}[/tex]
We know that [tex]\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||[/tex] and [tex] ||\boldsymbol{r}'(t)}||=2[/tex]
[tex]\frac{d}{dt}\left(2\right)\: = 0[/tex] so
[tex]a_{\boldsymbol{T}}=0[/tex]
The normal component of acceleration is given by the formula
[tex]a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2[/tex]
We know that [tex]\kappa=1[/tex] and [tex]\frac{ds}{dt}=2[/tex] so
[tex]a_{\boldsymbol{N}}=1 (2)^2=4[/tex]
