MHB How do we prove the rule for differentiable curves in 3D?

[mathmari]

Hey!

How could we prove the following rule for differentiable curves in $\mathbb{R}^3$ ?? (Wondering)

$$\frac{d}{dt}[\overrightarrow{\sigma}(t)\times \overrightarrow{\rho}(t)]=\frac{d\overrightarrow{\sigma}}{dt}\times \overrightarrow{\rho}(t)+\overrightarrow{\sigma}(t)\times \frac{d\overrightarrow{\rho}}{dt}$$

 

[chisigma]

Hey!

How could we prove the following rule for differentiable curves in $\mathbb{R}^3$ ?? (Wondering)

$$\frac{d}{dt}[\overrightarrow{\sigma}(t)\times \overrightarrow{\rho}(t)]=\frac{d\overrightarrow{\sigma}}{dt}\times \overrightarrow{\rho}(t)+\overrightarrow{\sigma}(t)\times \frac{d\overrightarrow{\rho}}{dt}$$

Is...

$\displaystyle \overrightarrow{\sigma}(t)\times \overrightarrow{\rho}(t) = \sum_{i=1}^{3} \sigma_{i} (t)\ \rho_{i} (t)$

... and applying the product rule You esasily arrive to the result...

Kind regards

$\chi$ $\sigma$

 

[mathmari]

$\displaystyle \overrightarrow{\sigma}(t)\times \overrightarrow{\rho}(t) = \sum_{i=1}^{3} \sigma_{i} (t)\ \rho_{i} (t)$

Why does this formula stand?? (Wondering)

 

[Opalg]

Hey!

How could we prove the following rule for differentiable curves in $\mathbb{R}^3$ ?? (Wondering)

$$\frac{d}{dt}[\overrightarrow{\sigma}(t)\times \overrightarrow{\rho}(t)]=\frac{d\overrightarrow{\sigma}}{dt}\times \overrightarrow{\rho}(t)+\overrightarrow{\sigma}(t)\times \frac{d\overrightarrow{\rho}}{dt}$$

Two different proofs are given here.

 

[mathmari]

Two different proofs are given here.

To use the product rule do we have to write the cross product as followed??

$\overrightarrow{\sigma}(t)=(x_1(t), x_2(t), x_3)), \overrightarrow{\rho}(t)=(y_1(t), y_2(t), y_3(t))$

$$\overrightarrow{\sigma}(t) \times \overrightarrow{\rho}(t)=\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
x_1(t) & x_2(t) & x_3(t)\\
y_1(t) & y_2(t) & y_3(t)
\end{vmatrix} \\ =(x_2(t)y_3(t)-x_3(t)y_2(t))\overrightarrow{i}+(x_3(t)y_1(t)-x_1(t)y_3(t))\overrightarrow{j}+(x_1(t)y_2(t)-x_2(t)y_1(t))\overrightarrow{k}$$

 

[I like Serena]

To use the product rule do we have to write the cross product as followed??

$\overrightarrow{\sigma}(t)=(x_1(t), x_2(t), x_3)), \overrightarrow{\rho}(t)=(y_1(t), y_2(t), y_3(t))$

$$\overrightarrow{\sigma}(t) \times \overrightarrow{\rho}(t)=\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
x_1(t) & x_2(t) & x_3(t)\\
y_1(t) & y_2(t) & y_3(t)
\end{vmatrix} \\ =(x_2(t)y_3(t)-x_3(t)y_2(t))\overrightarrow{i}+(x_3(t)y_1(t)-x_1(t)y_3(t))\overrightarrow{j}+(x_1(t)y_2(t)-x_2(t)y_1(t))\overrightarrow{k}$$

Yep. (Nod)

 

[mathmari]

I have done the following:

$$\frac{d}{dt}[\overrightarrow{\sigma}(t)\times\overrightarrow{\rho}(t)] \\ =\frac{d}{dt}\left [(x_2(t)y_3(t)-x_3(t)y_2(t))\overrightarrow{i}+(x_3(t)y_1(t)-x_1(t)y_3(t))\overrightarrow{j}+(x_1(t)y_2(t)-x_2(t)y_1(t))\overrightarrow{k}\right ] \\ =(x_2'(t)y_3(t)+x_2(t)y_3'(t)-x_3'(t)y_2(t)-x_3(t)y_2'(t))\overrightarrow{i}+(x_3'(t)y_1(t)+x_3(t)y_1'(t)-x_1'(t)y_3(t)-x_1(t)y_3'(t))\overrightarrow{j}+(x_1'(t)y_2(t)+x_1(t)y_2'(t)-x_2'(t)y_1(t)-x_2(t)y_1'(t))\overrightarrow{k} \\ =\left ( (x_2'(t)y_3(t)-x_3'(t)y_2(t))\overrightarrow{i}+(x_3'(t)y_1(t)-x_1'(t)y_3(t))\overrightarrow{j}+(x_1'(t)y_2(t)-x_2'(t)y_1(t))\overrightarrow{k}\right ) +\\ \left ( (x_2(t)y_3'(t)-x_3(t)y_2'(t))\overrightarrow{i}+(x_3(t)y_1'(t)-x_1(t)y_3'(t))\overrightarrow{j}+(x_1(t)y_2'(t)-x_2(t)y_1'(t))\overrightarrow{k}\right ) \\ =\frac{d\overrightarrow{\sigma}(t)}{dt} \times \overrightarrow{\rho}(t)+\overrightarrow{\sigma}(t)\times \frac{\overrightarrow{\rho}(t)}{dt}$$ Is this correct?? (Wondering)

Could I improve something?? (Wondering)

At the equation of the lines $5-7$

$$\left ( (x_2'(t)y_3(t)-x_3'(t)y_2(t))\overrightarrow{i}+(x_3'(t)y_1(t)-x_1'(t)y_3(t))\overrightarrow{j}+(x_1'(t)y_2(t)-x_2'(t)y_1(t))\overrightarrow{k}\right ) +\\ \left ( (x_2(t)y_3'(t)-x_3(t)y_2'(t))\overrightarrow{i}+(x_3(t)y_1'(t)-x_1(t)y_3'(t))\overrightarrow{j}+(x_1(t)y_2'(t)-x_2(t)y_1'(t))\overrightarrow{k}\right ) \\ =\frac{d\overrightarrow{\sigma}(t)}{dt} \times \overrightarrow{\rho}(t)+\overrightarrow{\sigma}(t)\times \frac{\overrightarrow{\rho}(t)}{dt}$$

do I have to say something/explain it?? (Wondering)

 

[I like Serena]

It looks fine to me. (Smile)

No need for improvements or further explanations. (Wasntme)

 

[mathmari]

It looks fine to me. (Smile)

No need for improvements or further explanations. (Wasntme)

Nice... Thank you! (Star)