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how did this happen?


PlanetCpp

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hey guys, like i've said before i really don't like math..well i do when i understand it. when something goes right it feels the same as when i program i write works. but this is crazy

HOW DID IT DO THIS???

im bad with simplifying radicals, really bad. thanks for any help.

calc.gif

btw i understand that cos²x+sin²x = 1

i dont see how it was split, the way im looking at ti looks wierd. how can i hust take a t out of (t + ***)² its not the same as

(t(****))², it needs to be expanded right?:(:confused: :confused: :confused:

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Most of this is just a test of your algebra (simplifications) and the very basics of your knowledge of trig identities....

 

 

 

int(sqrt[(t*cos(t) + sin(t))^2 + (cos(t) - t*sin(t))^2]) dt

 

 

 

We first expand the binomials, therefore we have

 

(a+b)^2 = (a^2 + 2ab + b^2); (a-b)^2 = (a^2 - 2ab + b^2)

 

int(sqrt[[t^2*cos^2(t) + 2t*sin(t)cos(t) + sin^2(t)] + [cos^2(t) - 2t*sin(t)cos(t) + t^2*sin^2(t)]]) dt

 

 

 

2t*sin(t)cos(t) - 2t*sin(t)cos(t) = 0, therefore we have:

 

int(sqrt[t^2*cos^2(t) + [sin^2(t) + cos^2(t)] + t^2*sin^2(t)]) dt

 

 

 

Indeed, since since sin^2(t) + cos^2(t) = 1, we have:

 

int(sqrt[1 + t^2*cos^2(t) + t^2*sin^2(t)]) dt

 

 

 

Factoring inside the radical, we obtain:

 

int(sqrt[1 + t^2(cos^2(t) + sin^2(t))]) dt

 

 

Once again, since sin^2(t) + cos^2(t) = 1, we have:

 

int(sqrt[1 + t^2]) dt

 

 

 

Anyway, I get sqrt(t + (t^3)/3) + C (hmm, i think?) for my integral in case you want to check your answer.

 

Note: Scratch that, this involves hyperbolic functions which I have no clue about. Sorry.

 

Hope that helps!

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i expanded it like you did but when i saw all that i was like :eek:

i thought it did something really wrong. i see what you did.

my problem is that im not confident in math and need to just take things in small steps to eliminate the hard looking problems.

btw i dont have to integrate it, this is a parametric length problem that just needed to be simplified, it said not to do anything but set it up. thank you!

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it is, thats what he said

sin²t + cos²t = 1

the t might have made it look different, the problems ive been doing have been in respect to t, not x, because this is parametric equations, not based on cartesian coordinates (x,y) but the coordinates based on time. sad part is i understand all the new stuff we are learning but its the old stuff that still confuses me, things from years ago. so when an integral has simple simplification (simple to you), i go blank. sux but ;o\ what ya gonna do

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Unforunately, you're probably learning the material too fast to retain any of it for long-term memory. All I can recommend in that case is that you practice a lot.

 

Btw, I can't wait for this forum to grow :)

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I forgot to comment on sin^2(x) + cos^2(x) = 1. It is merely the Pythagorean Theorem on a unit circle. I'll explain why this is so:

 

The sine function, sin(x) (read sine of x) is defined to give the ratio of the side opposite of an angle on a right triangle to its hypotenuse.

 

The cosine function, cos(x) is defined to give the ratio of the side adjacent (the side that is next to the angle but is not the hypotenuse) of an angle on a right triangle to its hypotenuse.

 

sine = side opposite / hypotenuse

cosine = side adjacent / hypotenuse

 

You need to remember that a ratio can be given in many ways: One half can be written as 5/10, 30/60, 3/6, or 0.5/1 among an infinite amount of other ways.

 

Now, recall sin(30 degrees) = 1/2. Which is saying side opposite / hypotenuse is 1/2, which therefore means:

 

side opposite / hypotenuse = 0.5/1

 

Since all ratios can be written in the form of x/1 (meaning x can be reduced to satisify this property), the hypotenuse can always "be 1".

 

Hence when using the pythagorean theorem you're always going to get 1 when evaluating sin^2(x) + cos^2(x).

 

 

Hopefully, that made some sense :) In any case, I felt like trying to help you all understand this identity to see it is not magic.

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