91Ó°ÊÓ

To find the tangential component of acceleration, we will use the formula \(a_{T}=\frac{v\cdot a}{\left\|v \right\|}\).

Now, if we differentiate \(r(t)\) we get \(r^{\prime}\left ( t \right )=v\left ( t \right )\) and \(r^{\prime \prime}\left ( t \right )=a\left ( t \right )\).

So,

\(r\left ( t \right )=\left<cos t, sin t, t \right>\)

\(v(t)=r^{\prime}\left ( t \right )=\left<- sint, cost, 1 \right>\)

\(a(t)=r^{\prime \prime} \left ( t \right )=\left<- cost, -sint, 0 \right>\)

\(\left\|v \right\|=\left\| \left<- sint, cost, 1 \right>\right\|\)

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

\(\left\| v\right\|=\sqrt{2}\)

\(v\cdot a=v\left ( t \right )\cdot a\left ( t \right )\)

\(v\cdot a=\left<- sint, cost, 1 \right> \cdot \left<- cost, -sint, 0 \right>\)

\(v\cdot a=sintcost-costsint\)

\(v\cdot a=0\)

Now, put all the above values we get in the formula

\(a_{T}=\frac{v\cdot a}{\left\|v \right\|}\)

\(a_{T}=\frac{0}{\sqrt{2}}\)

\(a_{T}=0\)

To find the normal component of acceleration, we will use the formula \(a_{N}=\frac{\left\|v\times a \right\|}{\left\|v \right\|}\).

So,

\(v\times a=v(t)\times a(t)\)

\(v\times a=\left|\begin{array}{ccc}i & j & k \\-sin t & cost & 1 \\-cost & -sint & 0\end{array}\right|\)

\(v\times a=i(sint)-j(cost)+k(sin^{2}t+cos^{2}t)\)

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

\(\left\|v\times a \right\|=\sqrt{2}\)

\(\left\| v\right\|=\sqrt{2}\)

Now, put all the above values we get in the formula,

\(a_{N}=\frac{\left\|v\times a \right\|}{\left\|v \right\|}\)

\(a_{N}=\frac{\sqrt{2}}{\sqrt{2}}\)

\(a_{N}=1\)