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r/HomeworkHelp•
Airade

[University? Calculus] Integration help

Sorry I'm not sure what the grade level is. I have the two parametric equations:

x(θ)=399.567cosθ+cos(13.369θ)

and

y(θ)=399.567sinθ+sin(13.369θ).

I then need to use the arc length formula for parametric equations, L=a∫b √[ (x')^2 + (y')^2 ] dt, and I can't figure out for the life of me how to integrate! I substituted 399.567 and 13.369 with variables to make it a little neater/easier. I can easily differentiate the two parametric equations, and square the two, but now I still need to take the square root of it and integrate, which is where I am stuck. I am using the point 0 for the lower limit and pi/2 for the upper limit.

Any and all help would be greatly appreciated.

Edit: In case I formatted the formula wrong, this is the equation I need to use:http://calculusapplets.com/assets/images/Eqn220.gif

Posted Jan 14, 2016, 3:41 AM UTC · edited

18
showmesomereddit•

What did you get for x' and y'?

Don't forget that acos^2 (x) + asin^2 (x) = a(cos^2 (x) + sin^2 (x)) = a.

2
AiradeOP•

I got -399.567sinθ-13.369sin13.369θ for x' and 339.567cosθ+13.369cos13.369θ for y'. I'm sorry, but I fail to see how that trig identity would help, since both coefficients are different and I have cos13.369θ, not a regular cosθ.

2
showmesomereddit•

Your differentiations look good.

The above trig identity is good for any x, even x=13.369θ. As long as cos and sin have the same input, the identity holds.

What do you get for your integrand after you simplify?

1
AiradeOP•

Thanks. I set a=339.567 and b=13.369

L=0∫(pi/2) √[a^(2)sin^(2)θ+2absinθsinbθ+b^(2)sin^(2)bθ+a^(2)cos^(2)θ+2abcosθcosbθ+b^(2)cos^(2)bθ]

I basically expanded it all, and I'm stuck at this point.

2
showmesomereddit•

Great. So, I get

(copied from above)

a^2 sin^2 θ+2absinθsinbθ+b^2 sin^2 bθ+a^2 cos^2 θ+2abcosθcosbθ + b^2 cos^2 bθ

(moving terms around)

a^2 sin^2 θ +a^2 cos^2 θ + b^2 sin^2 bθ + b^2 cos^2 bθ + 2abcosθcosbθ + 2absinθsinbθ

(using trig identity)

a^2 + b^2 + 2abcosθcosbθ + 2absinθsinbθ

(factoring)

a^2 + b^2 + 2ab(cosθcosbθ + sinθsinbθ)

So, this is our 'simplified' integrand. Have you seen any techniques for integrating something of this form (keeping in mind it is under a square root)

Edit: Actually, it looks like another set of trig identities will help. Applying a product of sines and a product of cosines will cancel things nicely.

1