Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the plane curve \(\mathbf{r}(t)\) $$ \mathbf{r}(t)=a \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j}, \quad t=0 $$

Short answer

\r{T}(0) = -i, \r{N}(0) = -j, a_T(0) = 0, a_N(0) = -aω^2

Step-by-step solution

Calculate the first and second derivative of r(t)

We first find the velocity vector \r{v}(t), which is the first derivative of \r{r}(t): \r{v}(t) = -aω sin(ωt)i + bω cos(ωt)j. And then we find the acceleration vector \r{a}(t), which is the second derivative of \r{r}(t): \r{a}(t) = -aω^2 cos(ωt)i - bω^2 sin(ωt)j

Find the unit tangent vector T(t)

The unit tangent vector \r{T}(t) is the velocity vector normalized. We find: \r{T}(t) = -sin(ωt)i + cos(ωt)j

Find the unit normal vector N(t)

The unit normal vector \r{N}(t) is the derivative of the unit tangent vector normalized. We find: \r{N}(t) = -cos(ωt)i - sin(ωt)j

Find the tangential and normal components of acceleration

We find the tangential component of acceleration (a_T) by projecting the acceleration vector onto the unit tangent vector: a_T = \r{a}(t) . \r{T}(t). We find a_T = 0. We find the normal component of acceleration (a_N) by projecting the acceleration vector onto the unit normal vector: a_N = \r{a}(t) . \r{N}(t). We find a_N = -aω^2

Evaluate at t = 0

We now evaluate these quantities at the given time t = 0. We find \r{T}(0) = -i, \r{N}(0) = -j, a_T(0) = 0, a_N(0) = -aω^2