Regular Curves

[brisk goblet]

1. Do not ping the Moderators, unless someone is breaking the rules.
2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
3. Do not ping other members randomly for help.
4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
5. Wait patiently for a helper to come along.
6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[cursive fulcrum]

it's clear that $\gamma'(t)\neq 0$ for $t\neq0$

[jolly jetty]

Omegabet_

[cursive fulcrum]

You can directly compute that $\gamma'(0)=(1,0,0)$ using the usual definition

[jolly jetty]

Omegabet_

[grim jackal]

would that be using the limits from either side?

[cursive fulcrum]

yes

[grim jackal]

but how doi reconcile that with the actual value at 0

[cursive fulcrum]

wdym

[grim jackal]

wait I think I get it

[cursive fulcrum]

the fact it's 0 is irrelevant (bar making the function an actual curve lol)

[grim jackal]

fair

[cursive fulcrum]

just like in calc1

or calc3

[grim jackal]

the derivative isn't rly defined well for the t = 0 because it's just a point

not a function

[cursive fulcrum]

the curve is well defined on R

[grim jackal]

so I have to use either side of it

I get it

thanks for the help

[cursive fulcrum]

yes, you're just computing $\lim_{h\to 0}\frac{1}{h}\gamma(h)$

[jolly jetty]

Omegabet_

[cursive fulcrum]

which follows from the fact $\lim_{h\to 0}h^2\exp(-1/h^2)=0$

[jolly jetty]

Omegabet_

[grim jackal]

yeah thanks for the help

[opal cove]

@grim jackal

:HelpIcon:| Help Reminder

Hello .blinch, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.