Jeramiah Campos

2022-06-22

Let $\gamma :\left[0,2\pi \right]\to {\mathbb{R}}^{3}$ be
$\gamma \left(t\right)=\left(4t,\mathrm{cos}\left(3t\right),\mathrm{sin}\left(3t\right)\right).$
Justify that $\gamma$ is differentiable.

### Answer & Explanation

drumette824ed

Step 1
A vector valued function $f:{\mathbb{R}}^{n}\to {\mathbb{R}}^{m}$ is differentiable if its component functions ${f}_{i}:{\mathbb{R}}^{n}\to \mathbb{R}$ are differentiable. Here, $n=1$ and $m=3$ The component functions are,
${f}_{1}\left(t\right)=4t\phantom{\rule{0ex}{0ex}}{f}_{2}\left(t\right)=\mathrm{cos}\left(3t\right)\phantom{\rule{0ex}{0ex}}{f}_{3}\left(t\right)=\mathrm{sin}\left(3t\right)$
Hopefully it is clear that each of these component functions are differentiable. Then, we have that f is differentiable and its derivative is the Jacobian (matrix of partial derivatives),
$\mathrm{\nabla }f=\left(\begin{array}{c}{\mathrm{\partial }}_{t}{f}_{1}\\ {\mathrm{\partial }}_{t}{f}_{2}\\ {\mathrm{\partial }}_{t}{f}_{3}\end{array}\right)=\left(\begin{array}{c}4\\ -3\mathrm{sin}\left(3t\right)\\ 3\mathrm{cos}\left(3t\right)\end{array}\right)$

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