## Q. 12.10

If a = 3ti − t²j and b = 2t²i + 3j, verify

(a) $\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \cdot \mathbf{b})=\mathbf{a} \cdot \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \cdot \mathbf{b}$               (b) $\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \times \mathbf{b})=\mathbf{a} \times \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \times \mathbf{b}$

## Verified Solution

(a)

$\begin{gathered}\mathbf{a} \cdot \mathbf{b}=\left(3 t \mathbf{i}-t^2 \mathbf{j}\right) \cdot\left(2 t^2 \mathbf{i}+3 \mathbf{j}\right)=6 t^3-3 t^2 \\\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \cdot \mathbf{b})=18 t^2-6 t\end{gathered}$

Also

$\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t}=3 \mathbf{i}-2 t \mathbf{j} \quad \frac{\mathrm{db}}{\mathrm{d} t}=4 t \mathbf{i}$

So,

\begin{aligned}\mathbf{a} \cdot \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\mathbf{b} \cdot \frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} & =\left(3 t \mathbf{i}-t^2 \mathbf{j}\right) \cdot(4 t \mathbf{i})+\left(2 t^2 \mathbf{i}+3 \mathbf{j}\right) \cdot(3 \mathbf{i}-2 t \mathbf{j}) \\& =12 t^2+6 t^2-6 t=18 t^2-6 t\end{aligned}

We have verified  $\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \cdot \mathbf{b})=\mathbf{a} \cdot \frac{\mathrm{d} \mathbf{d}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \cdot \mathbf{b}$

(b)

\begin{aligned}& \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\3 t & -t^2 & 0 \\2 t^2 & 3 & 0\end{array}\right| \\&=\left(9 t+2 t^4\right) \mathbf{k} \\& \frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \times \mathbf{b})=\left(9+8 t^3\right) \mathbf{k}\end{aligned}

Also,

\begin{aligned}\mathbf{a} \times \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t} & =\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\3 t & -t^2 & 0 \\4 t & 0 & 0\end{array}\right| \\& =4 t^3 \mathbf{k} \\\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \times \mathbf{b} & =\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\3 & -2 t & 0 \\2 t^2 & 3 & 0\end{array}\right| \\& =\left(9+4 t^3\right) \mathbf{k}\end{aligned}

and so

$\mathbf{a} \times \frac{\mathrm{d} \mathbf{b}}{\mathrm{d} t}+\frac{\mathrm{d} \mathbf{a}}{\mathrm{d} t} \times \mathbf{b}=4 t^3 \mathbf{k}+\left(9+4 t^3\right) \mathbf{k}=\left(9+8 t^3\right) \mathbf{k}=\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{a} \times \mathbf{b})$

as required.