Jeramiah Campos

2022-06-22

Let $\gamma :[0,2\pi ]\to {\mathbb{R}}^{3}$ be

$\gamma (t)=(4t,\mathrm{cos}(3t),\mathrm{sin}(3t)).$

Justify that $\gamma $ is differentiable.

$\gamma (t)=(4t,\mathrm{cos}(3t),\mathrm{sin}(3t)).$

Justify that $\gamma $ is differentiable.

drumette824ed

Beginner2022-06-23Added 19 answers

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,

${f}_{1}(t)=4t\phantom{\rule{0ex}{0ex}}{f}_{2}(t)=\mathrm{cos}(3t)\phantom{\rule{0ex}{0ex}}{f}_{3}(t)=\mathrm{sin}(3t)$

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),

$\mathrm{\nabla}f=\left(\begin{array}{c}{\mathrm{\partial}}_{t}{f}_{1}\\ {\mathrm{\partial}}_{t}{f}_{2}\\ {\mathrm{\partial}}_{t}{f}_{3}\end{array}\right)=\left(\begin{array}{c}4\\ -3\mathrm{sin}(3t)\\ 3\mathrm{cos}(3t)\end{array}\right)$

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,

${f}_{1}(t)=4t\phantom{\rule{0ex}{0ex}}{f}_{2}(t)=\mathrm{cos}(3t)\phantom{\rule{0ex}{0ex}}{f}_{3}(t)=\mathrm{sin}(3t)$

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),

$\mathrm{\nabla}f=\left(\begin{array}{c}{\mathrm{\partial}}_{t}{f}_{1}\\ {\mathrm{\partial}}_{t}{f}_{2}\\ {\mathrm{\partial}}_{t}{f}_{3}\end{array}\right)=\left(\begin{array}{c}4\\ -3\mathrm{sin}(3t)\\ 3\mathrm{cos}(3t)\end{array}\right)$