﻿ Mathematics Help Station - Page 8

1. Originally Posted by Blizzard Yup, they both come from that.
Oh, thanks. We just did hyperbolic functions in school, so I'm glad I could actually use some of it xD
I hear integrating hyperbolic functions is probably the hardest topic we learn at my level (upper sixth - thats the British equivalent of senior at high school. Going to uni next year!). Just looking at all of those standard forms....  Reply With Quote

2. So I'm doing an arc length problem (parametric) that is
X=e^(t)cost
Y=e^(t)sint
My teacher gave the solution of 2^(1/2)(e^(pi)-1)
BUT he's old and his answers are wrong sometimes, and i got [2^(1/2)/2][e^(2pi)-1]
So basically it comes down to if i did my algebra right, so my question is if i squared the following terms correctly (the derivatives of the x & y)
[(e^(t))(cost-sint)]^2 = e^(2t)(cost^(2)-2sintcost+sint^(2)
And the second is very similar but i don't feel like typing it.
Basically i forget algebra basics sometimes, but that e gets raised to the 2t, when squaring, right? The 2 in the denominator on my solution's outside term comes from simplifying the two squared derivatives and adding them, then factoring.
Sorry if my notation is confusing, in hindsight a picture would have been better  Reply With Quote

3. Noooo fucking clue, forgot that shit. @Stein  Reply With Quote

4. @Super Neko yes that basic algebra is of course correct. Perhaps your mistake is elsewhere.

Usually these are solved with euler's and some calculations so maybe you have made a careless miscalculation somewhere.  Reply With Quote

5. Originally Posted by Super Neko So I'm doing an arc length problem (parametric) that is
X=e^(t)cost
Y=e^(t)sint
My teacher gave the solution of 2^(1/2)(e^(pi)-1)
BUT he's old and his answers are wrong sometimes, and i got [2^(1/2)/2][e^(2pi)-1]
So basically it comes down to if i did my algebra right, so my question is if i squared the following terms correctly (the derivatives of the x & y)
[(e^(t))(cost-sint)]^2 = e^(2t)(cost^(2)-2sintcost+sint^(2)
And the second is very similar but i don't feel like typing it.
Basically i forget algebra basics sometimes, but that e gets raised to the 2t, when squaring, right? The 2 in the denominator on my solution's outside term comes from simplifying the two squared derivatives and adding them, then factoring.
Sorry if my notation is confusing, in hindsight a picture would have been better
Ya, a picture would have been way clearer. The difference betwee your answer and his seems to be whether or not you're multiplying by root 2(his answer) or dividing by root 2 (your answer).

I know that you can take this: e^(2t)(cost^(2)-2sintcost+sint^(2) (what you had above)

and make this: e^(2t)(1+sin(2t)). You can then distribute through the parentheses

cos^2 + sin^2 = 1, and 2sintcost = sin2t. Pretty sure those are just basic trig identities. Does that help you figure out if you're right or the teacher is right? Originally Posted by Stein @Super Neko yes that basic algebra is of course correct. Perhaps your mistake is elsewhere.

Usually these are solved with euler's and some calculations so maybe you have made a careless miscalculation somewhere.
Love euler. His calculus stuff was boring to me, but I really liked his stuff on graph theory and topology. Ever heard of the 7 bridges of Konigsberg problem?  Reply With Quote

6. Originally Posted by ShichiDrewKai Love euler. His calculus stuff was boring to me, but I really liked his stuff on graph theory and topology. Ever heard of the 7 bridges of Konigsberg problem?
Yeap does ring a bell from graphs iirc.  Reply With Quote

7. Originally Posted by Stein Yeap does ring a bell from graphs iirc. you math?  Reply With Quote

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•