# calculus

posted by .

Find complete length of curve r=a sin^3(theta/3).
I have gone thus- (theta written as t)
r^2= a^2 sin^6 t/3 and
(dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3)
s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt
=a Int Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int Sqrt[sin^4(t/3)dt
=a Int sin^2(t/3)dt
=a[1/2(t/3)-1/2{3sin(2t/3)}] from t=0 to 2pi
Or, s=a.pi/3-3a[sin(4pi/3)]/4
=a.pi/3+3a[sin(pi/3)]/4
=a.pi/3+3/4*rt3/2=a.pi/3+3rt3/8.
The answer in the book is 3a.pi/2. Where have I gone wrong?

• calculus -

r = a sin^3(θ/3)

r' = a sin^2(θ/3) cos(θ/3)

Sketch the curve. sin^2(θ/3) has a period of 3π.

The complete length of r must be measured over one whole period; thus integrate from 0 to 3π, (not 2π).

Arc length in polar coordinates:
s = ∫{0↔3π} √(r^2 + r'^2) dθ

s = ∫{0↔3π} √(a^2 sin^6(θ/3) + a^2 sin^4(θ/3)cos^2(θ/3)) dθ

s = a ∫{0↔3π} sin^2(θ/3) √(sin^2(θ/3) + cos^2(θ/3)) dθ

s = a ∫{0↔3π} sin^2(θ/3) dθ

Use sin^2(A) = (1-cos(2A))/2

s = (a/2) ∫{0↔3π} 1 - cos(2θ/3) dθ

s = (a/2) [θ - (3/2) sin(2θ/3)]{θ=0↔3π}

s = (a/2) (3π - (3/2) sin(4π))

s = 3aπ/2

---
Error 1: The period over which you integrate

Error 2: The integration of sin^2(θ/3).

• calculus -

Thanks a lot for guiding. Kindly let me know if it should be a routine procedure to plot the complete curve everytime or can it be conveniently worked out analytically?

I used this formula for integrating Int sin^2x dx=1/2*(x-1/2*sin2x)which is a std formula. Why the result differs?

• calculus -

Yes, the general formula is.
∫sin^2(x) dx = x/2 - sin(2x)/4 +C

However, a more general formula is:
∫sin^2(ax) dx = x/2 - sin(2ax)/4a +C

Proof:
Let I = ∫ sin^2(ax) dx

Substitute u = ax, and dx = (1/a)du
I = (1/a)∫sin^2(u) du

I = (1/a)(u/2 - sin(2u)/4) +C

Substitute back, u = ax
I = (1/a)(ax/2 - sin(2ax)/4) +C

Simplify and conclude:
.: ∫ sin^2(ax) dx = x/2 - sin(2ax)/4 +C

Q.E.D.

Basically; when you substitute for x, you must remember to also substitute for dx.

• calculus -

Mr. Graham,
Thank you very much for elaborate explanation. Kindly also let me know if it is ok to find at what values of theta the r becomes 0 and take that as period of curve? If so, plotting the complete curve every time may not be required.

• calculus -

For a closed curve, the complete length is the distance along the curve from a starting point, and to the starting point again. That is, measure the length of a complete circuit.

When using polar coordinates, the domain over which you need to integrate to measure the complete length is called the period. The period being the rotation you need to traverse to trace the curve.

This is often, but not necessarily 2π.

• calculus -

Thank you very much.

## Similar Questions

1. ### tigonometry

expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) …
2. ### Trig

Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos …

4. ### Calculus

Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
5. ### Mathematics - Trigonometric Identities

Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y …
6. ### Trig(3 all with work)

1)Find the exact value of cos 105 by using a half-angle formula. A)sqrt 2 - sqrt 3 /2 B)-sqrt 2 - sqrt 3 /2 C)-sqrt 2 + sqrt 3 /2 D)sqrt 2 + sqrt 3 /2 cos 105 cos 105 = cos 210/2 sqrt 1 + 210/2 sqrt 1 + sqrt 3/2 /2 sqrt 2 + sqrt 3/2 …
7. ### TRIG!

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …
8. ### Precalculs

I have no idea how to do these type of problems. -------Problem-------- Solve each equation on the interval 0 less than or equal to theta less than 2 pi 42. SQRT(3) sin theta + cos theta = 1 ---------------------- There is an example …
9. ### Trigonometry

Solve the equation for solutions in the interval 0<=theta<2pi Problem 1. 3cot^2-4csc=1 My attempt: 3(cos^2/sin^2)-4/sin=1 3(cos^2/sin^2) - 4sin/sin^2 = 1 3cos^2 -4sin =sin^2 3cos^2-(1-cos^2) =4sin 4cos^2 -1 =4sin Cos^2 - sin=1/4 …
10. ### Mathematics-Integration

Question: For any positive integer n,show that integrate.[(sin x)^2n dx ] from o - π/2 = [(2n)!*π]/[(2)^(2n+1)*(n!)^2 ] What I thought: Let I =int.[(sinx)^2n dx] And again I= int.[ (sin(π/2-x))^2n dx] = int.[ (cos)^2n dx] 2I= int.[(sin …

More Similar Questions