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202204021539 Example 3, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If \mathbf{f}(t)=t\,\hat{\mathbf{i}}+t^3\,\hat{\mathbf{j}}, \mathbf{g}(t)=\cos t\,\hat{\mathbf{i}}+\sin t\,\hat{\mathbf{j}}, and \mathbf{v}=2\,\hat{\mathbf{i}}-3\,\hat{\mathbf{j}}. Calculate

(a) (\mathbf{f}+\mathbf{g})';

(b) (\mathbf{v}\cdot\mathbf{f})';

(c) (\mathbf{f}\cdot\mathbf{g})'.

Settings. (Some properties of vector differentiation)

i. If \mathbf{f} and \mathbf{g} are differentiable vector functions,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}+\mathbf{g})=\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}}

ii. If \mathbf{f} is a differentiable vector function and \alpha a constant scalar,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\alpha\mathbf{f})=\alpha\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}

iii. If \mathbf{f}(t) is a vector function, \mathbf{v} a constant vector, and \mathbf{v}\cdot\mathbf{f} differentiable,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{v}\cdot\mathbf{f})=\mathbf{v}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}

iv. If h(t) is a scalar function, \mathbf{f}(t) a vector function and h\mathbf{f} differentiable,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(h\mathbf{f})=h\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\mathbf{f}\frac{\mathrm{d}h}{\mathrm{d}t}}

v. If \mathbf{f} and \mathbf{g} are vector functions and \mathbf{f}\cdot\mathbf{g} differentiable,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}\cdot\mathbf{g})=\mathbf{f}\cdot\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}+\mathbf{g}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}

vi. If \mathbf{f} and \mathbf{g} are vector functions and \mathbf{f}\times\mathbf{g} differentiable,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\mathbf{f}\times\mathbf{g})=\displaystyle{\bigg( \frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}\times\mathbf{g}\bigg) + \bigg( \mathbf{f}\times \frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}\bigg)}.

Solution.

(a)

\begin{aligned} \mathbf{f}+\mathbf{g} & = (t+\cos t)\,\hat{\mathbf{i}}+(t^3+\sin t)\,\hat{\mathbf{j}} \\ (\mathbf{f}+\mathbf{g})' & = (t+\cos t)'\,\hat{\mathbf{i}}+(t^3+\sin t)'\,\hat{\mathbf{j}} \\ & = (1-\sin t)\,\hat{\mathbf{i}}+(3t^2+\cos t)\,\hat{\mathbf{j}} \\ \end{aligned}

(b)

\begin{aligned} \mathbf{v}\cdot\mathbf{f} & = (2,-3)\cdot (t,t^3) \\ & = 2t-3t^3 \\ (\mathbf{v}\cdot\mathbf{f})' & = (2t-3t^3)' \\ & = 2-9t^2 \\ \end{aligned}

(c)

\begin{aligned} \mathbf{f}\cdot\mathbf{g} & = (t,t^3) \cdot (\cos t,\sin t) \\ & = t\cos t+t^3\sin t \\ (\mathbf{f}\cdot\mathbf{g})' & = \cos t-t\sin t+t^3\cos t + 3t^2\sin t\\ \end{aligned}