Question

Find the tangential and normal components of acceleration for the position functions in Exercises 18–22

\(r\left ( t \right )=\left<t,t^{2} \right>\)

Short answer

The tangential and normal components of acceleration for the position functions are \(a_{T}=\frac{4t}{\sqrt{1+4t^{2}}}\) and \(a_{N}=\frac{2}{\sqrt{1+4t^{2}}}\).

Step-by-step solution

Step 1. Find the tangential component of acceleration 

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,

\(r\left ( t \right )=\left<t,t^{2} \right>\)

\(v(t)=r^{\prime}\left ( t \right )=\left<1,2t \right>\)

\(a(t)=r^{\prime \prime} \left ( t \right )=\left<0,2 \right>\)

\(\left\|v \right\|=\left\| \left<1,2t \right>\right\|\)

\(\left\| v\right\|=\sqrt{1+4t^{2}}\)

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

\(v\cdot a=\left<1,2t \right> \cdot \left<0,2 \right>\)

\(v\cdot a=1(0)+2t(2)\)

\(v\cdot a=4t\)

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

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

\(a_{T}=\frac{4t}{\sqrt{1+4t^{2}}}\)

Step 2. Find the normal component of acceleration 

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

So,

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

\(v\times a=\left|\begin{array}{ccc}i & j & k \\1 & 2 t & 0 \\0 & 2 & 0\end{array}\right|\)

\(v\times a=i(0)-j(0)+k(2)\)

\(v\times a=\left<0,0,2 \right>\)

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

\(\left\|v \right\|=\left\| \left<1,2t \right>\right\|\)

\(\left\| v\right\|=\sqrt{1+4t^{2}}\)

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

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

\(a_{N}=\frac{2}{\sqrt{1+4t^{2}}}\)