reparametrizing the curve in terms of arc length (KristaKingMath)

[@matj1833]

Your videos are excellent. When Sal's voice puts me to sleep I can throw you in the mix and liven back up. Keep up the great work!

[@accidentalfrenchfries]

I've been trying to understand my textbook's version of this for an hour. This saved me! I loved how you represented t as a function of s.

[@SneakyZeke]

sweet jesus thank you you much, multi variable calc exam in 2 hours

[@kathrynhund5925]

I read multiple explanations of this that left out pieces of the explanation and only made me more confuse than when I started. This was the first source to actually make it clear what I needed to do. Thank you for the helpful, informative video!

[@kristakingmath]

You're welcome, Kathryn, I'm so glad it all made sense! :)

[@observer4322]

That is so much easier that I thought. You are so good at hitting all the points, but keeping the video short. Perfect explanation, thanks.

[@kristakingmath]

Thank you so much, I'm so glad everything made sense! :)

[@asamonson6866]

Again, you are awesome. You do a great job not just explaining the process but the reason behind it.

[@kristakingmath]

Thanks Asa!

[@maria1100]

omg youre the BESTTTTTTTT TYSM

[@Zydico]

Thank you very much for this clear explanation. Explained much better than my textbook and professor's lecture combined with one example.

[@kristakingmath]

So glad it was helpful! :D

[@yj29885]

Wonderfully explained!

[@noelabitzen6580]

UR MY LIFE SAVER

[@ethanheikens5705]

Thank you! Wish my professor taught this as well as you

[@kristakingmath]

You're welcome, I'm glad it helped!

[@evrik78]

Outstanding: clear and easy to understand.

[@MiKe-y8e]

Thanks for your help may God bless you 🤝

[@romegulay2493]

Thanks! it really helps a lot for my finals tomorrow. GOD BLESS :D

[@kristakingmath]

+Rome Gulay I'm glad I could help. Good luck tomorrow!

[@parad0x928]

Thank you very much, Your explanation was so clear and easy to follow (plus the question you solved happens to be the same i was looking at in my textbook).

[@kristakingmath]

+alex Laleyan I'm so glad it helped!

[@vincentwolfgramm-russell7263]

Haha funny I was having trouble with this question. This question is from Stewarts Calculus: Early Transcendental!

[@rolandsgradovskis9793]

wow thnx. i liked this so much i watched it 3 times in a row ha

[@Neueregel]

Good example. But in order to find an integer reparametrization result , instead of Root(29), one has to find a *sum of three squares that form a perfects square*. The deravatives of 2t,3t,4t, obviously parametrize to 29, a non-perfect square. So, the whole solution for *integer* reparametrization for x,y,z < 30 are : 1 ^2 + 2 ^2 + 2 ^2 = 3 ^2 1 ^2 + 4 ^2 + 8 ^2 = 9 ^2 1 ^2 + 6 ^2 + 18 ^2 = 19 ^2 1 ^2 + 12 ^2 + 12 ^2 = 17 ^2 2 ^2 + 3 ^2 + 6 ^2 = 7 ^2 2 ^2 + 4 ^2 + 4 ^2 = 6 ^2 2 ^2 + 5 ^2 + 14 ^2 = 15 ^2 2 ^2 + 6 ^2 + 9 ^2 = 11 ^2 2 ^2 + 7 ^2 + 26 ^2 = 27 ^2 2 ^2 + 8 ^2 + 16 ^2 = 18 ^2 2 ^2 + 10 ^2 + 11 ^2 = 15 ^2 2 ^2 + 10 ^2 + 25 ^2 = 27 ^2 2 ^2 + 14 ^2 + 23 ^2 = 27 ^2 2 ^2 + 24 ^2 + 24 ^2 = 34 ^2 2 ^2 + 26 ^2 + 29 ^2 = 39 ^2 3 ^2 + 4 ^2 + 12 ^2 = 13 ^2 3 ^2 + 6 ^2 + 6 ^2 = 9 ^2 3 ^2 + 6 ^2 + 22 ^2 = 23 ^2 3 ^2 + 12 ^2 + 24 ^2 = 27 ^2 3 ^2 + 14 ^2 + 18 ^2 = 23 ^2 3 ^2 + 16 ^2 + 24 ^2 = 29 ^2 3 ^2 + 24 ^2 + 28 ^2 = 37 ^2 4 ^2 + 4 ^2 + 7 ^2 = 9 ^2 4 ^2 + 5 ^2 + 20 ^2 = 21 ^2 4 ^2 + 6 ^2 + 12 ^2 = 14 ^2 4 ^2 + 8 ^2 + 8 ^2 = 12 ^2 4 ^2 + 8 ^2 + 19 ^2 = 21 ^2 4 ^2 + 10 ^2 + 28 ^2 = 30 ^2 4 ^2 + 12 ^2 + 18 ^2 = 22 ^2 4 ^2 + 13 ^2 + 16 ^2 = 21 ^2 4 ^2 + 17 ^2 + 28 ^2 = 33 ^2 4 ^2 + 20 ^2 + 22 ^2 = 30 ^2 5 ^2 + 10 ^2 + 10 ^2 = 15 ^2 6 ^2 + 6 ^2 + 7 ^2 = 11 ^2 6 ^2 + 6 ^2 + 17 ^2 = 19 ^2 6 ^2 + 8 ^2 + 24 ^2 = 26 ^2 6 ^2 + 9 ^2 + 18 ^2 = 21 ^2 6 ^2 + 10 ^2 + 15 ^2 = 19 ^2 6 ^2 + 12 ^2 + 12 ^2 = 18 ^2 6 ^2 + 13 ^2 + 18 ^2 = 23 ^2 6 ^2 + 14 ^2 + 27 ^2 = 31 ^2 6 ^2 + 18 ^2 + 27 ^2 = 33 ^2 6 ^2 + 21 ^2 + 22 ^2 = 31 ^2 7 ^2 + 14 ^2 + 14 ^2 = 21 ^2 7 ^2 + 14 ^2 + 22 ^2 = 27 ^2 7 ^2 + 16 ^2 + 28 ^2 = 33 ^2 8 ^2 + 8 ^2 + 14 ^2 = 18 ^2 8 ^2 + 9 ^2 + 12 ^2 = 17 ^2 8 ^2 + 11 ^2 + 16 ^2 = 21 ^2 8 ^2 + 12 ^2 + 24 ^2 = 28 ^2 8 ^2 + 16 ^2 + 16 ^2 = 24 ^2 8 ^2 + 20 ^2 + 25 ^2 = 33 ^2 8 ^2 + 24 ^2 + 27 ^2 = 37 ^2 9 ^2 + 12 ^2 + 20 ^2 = 25 ^2 9 ^2 + 18 ^2 + 18 ^2 = 27 ^2 10 ^2 + 10 ^2 + 23 ^2 = 27 ^2 10 ^2 + 20 ^2 + 20 ^2 = 30 ^2 11 ^2 + 12 ^2 + 24 ^2 = 29 ^2 11 ^2 + 22 ^2 + 22 ^2 = 33 ^2 12 ^2 + 12 ^2 + 14 ^2 = 22 ^2 12 ^2 + 12 ^2 + 21 ^2 = 27 ^2 12 ^2 + 15 ^2 + 16 ^2 = 25 ^2 12 ^2 + 16 ^2 + 21 ^2 = 29 ^2 12 ^2 + 21 ^2 + 28 ^2 = 37 ^2 12 ^2 + 24 ^2 + 24 ^2 = 36 ^2 13 ^2 + 26 ^2 + 26 ^2 = 39 ^2 14 ^2 + 18 ^2 + 21 ^2 = 31 ^2 14 ^2 + 22 ^2 + 29 ^2 = 39 ^2 14 ^2 + 28 ^2 + 28 ^2 = 42 ^2 15 ^2 + 18 ^2 + 26 ^2 = 35 ^2 16 ^2 + 16 ^2 + 28 ^2 = 36 ^2 16 ^2 + 18 ^2 + 24 ^2 = 34 ^2 17 ^2 + 20 ^2 + 20 ^2 = 33 ^2 18 ^2 + 18 ^2 + 21 ^2 = 33 ^2 19 ^2 + 22 ^2 + 26 ^2 = 39 ^2 20 ^2 + 28 ^2 + 29 ^2 = 45 ^2 23 ^2 + 24 ^2 + 24 ^2 = 41 ^2 24 ^2 + 24 ^2 + 28 ^2 = 44 ^2 lol even Kronecker would like that

[@sleverlight]

Wow thanks. I didnt know it was this easy the book was confusing

[@debendragurung3033]

ok , but what if we end up with polynomial expression under integral. eg X(t)= 2t Y(t) = 1+3t^2 Z(t) = 2+ 4t^3

[@mimimoon1827]

wow thank you so much!

[@dwang2805]

I encountered the exact problem and didn't know how to solve, came here, and found out that it is the exact problem, and even numbers match perfectly....

[@kristakingmath]

I'm glad the video helped! :)

[@junebug9892]

Thank you for your help! The explanation was very helpful : )

[@kristakingmath]

+June Bug You're so welcome! I'm glad I could help!

[@Sameergupta1508_]

omg....your voice is so sweet !!

[@wtfchazpwnt]

Very helpful video

[@EricIskandarZulkarnain]

thank it help alot i have final in 2 days time. but one quick question what if the end part of the question is change to with respect to arc length measured from (1,-1,2) in the direction of increasing t ? thank you !

[@heyitsrin6556]

Thank you!

[@kristakingmath]

You're welcome, Rin! :D

[@stephaniebeagle2030]

This was very helpful! Thank you :)

[@kristakingmath]

Glad it could help! :)

[@jaredwright3342]

I thought you couldn't integrate a function with its parameter as a bound? i.e. how can t be the upper bound and be part of the function at the same time?

[@vinaydeshmukh8446]

When you take the integral, doesn't there need to be a + C(vector) that you must calculate for using rvetor=0?

[@cassied9327]

This is a definite integral so we there will not be a +C :)

[@wasimakram-zf2xb]

Thank u ......may Allah's bless u

[@eswyatt]

Some people substitute a dummy variable like "u" when they take the integral. I suspect it's less confusing to evaluate u by plugging in t than plugging in t or t. But the u really threw me the first time I saw it.

[@kibrika]

I'm confused as to why YT search for Krista's channel of "Chebyshev" gave out this video. I was hoping to get a nice example of how to use Chebyshev substitution. Not that this is not interesting, just unrelated.

[@navyaaaa01]

What if the limits are not specified???

[@theodorearmah3370]

Determine where on the curve given by r ⃗(t)=〈t^2,2t^3,1-t^3 〉 we are after travelling a distance of 20 units. Can someone help with the above question?

[@BrooksHu]

tysm :)

[@tristan3577]

What of my integral is a constant?

[@kristakingmath]

Remember that the integral (with respect to x) of 3 is 3x, the integral of -7 is -7x, and the integral of k is kx. So integrating a constant just means you need to attach the variable to that constant.

[@alemyatimothy3498]

❤

[@turkeydota2]

Goat

[@MinhTri-io9om]

Jesus blesses for u

[@trevor3905]

thank you!

[@kristakingmath]

You're welcome, Trevor! :)