February 2022 – Page 2

February 4, 2022October 24, 2022

202204021524 Example 1, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If $\mathbf{f}(\theta)=\cos\theta \,\hat{\mathbf{i}}+e^{2\theta}\,\hat{\mathbf{j}}$ , find $\mathbf{f}'(\theta)$ .

Setting.

If $\mathbf{f}(t)=f_1(t)\,\hat{\mathbf{i}}+f_2(t)\,\hat{\mathbf{j}}+f_3(t)\,\hat{\mathbf{k}}$ is a vector function and $f_1$ , $f_2$ , and $f_3$ are differentiable, then the derivative of $f(t)$ is defined as

$\displaystyle{\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}=\mathbf{f}'(t)=f_1'(t)\,\hat{\mathbf{i}}+f_2'(t)\,\hat{\mathbf{j}}+f_3'(t)\,\hat{\mathbf{k}}$ .

Solution.

From $\mathbf{f}(\theta)=\cos\theta \,\hat{\mathbf{i}}+e^{2\theta}\,\hat{\mathbf{j}}$ , identify that

$\begin{aligned} f_1(\theta ) & =\cos\theta \\ f_2(\theta ) & = e^{2\theta} \\ f_1'(\theta ) & = -\sin\theta \\ f_2'(\theta ) & = 2e^{2\theta }\\ \end{aligned}$

Thus, $\mathbf{f}'(\theta )= -\sin\theta\,\hat{\mathbf{i}}+2e^{2\theta}\,\hat{\mathbf{j}}$ .

February 4, 2022October 24, 2022

202202041514 Example 1, Chapter 1.2, Methods in Physics II (2015-2016 Lectures)

Find the Cartesian equation of the locus represented by the parametric equation:

$\mathbf{f}(t)=t\,\hat{\mathbf{i}}+e^t\,\hat{\mathbf{j}}$ .

Solution.

$\begin{aligned} x(t) & =t \\ y(t) & = e^t\\ \end{aligned}$

Hence, $y=e^x$ .

物理後生﹑掣數學之肘.
小子算數﹑指物理之月.

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