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?

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

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Error 1: The period over which you integrate

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

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?

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

Use your standard formula
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.

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.

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.

Thank you very much.

To find the complete length of the curve, you correctly started by expressing the equation of the curve in polar coordinates: r = a sin^3(t/3), where t is the parameter and a is a constant.

Next, you correctly squared the equation and differentiated both sides with respect to t to find (dr/dt)^2. So far, so good.

However, when you integrated the expression to find the arc length (s), there was an error in your calculation. Let's go through the steps once again, starting from where you made the mistake.

You correctly wrote the expression for s as:
s = ∫ √[a^2 sin^6(t/3) + a^2 sin^4(t/3)cos^2(t/3)] dt.

Now, let's simplify this expression. Notice that both terms under the square root have a factor of sin^4(t/3). We can factor out sin^4(t/3) and simplify the expression.

s = a ∫ sin^2(t/3) √[sin^2(t/3) + cos^2(t/3)] dt.

Remember that sin^2(t/3) + cos^2(t/3) = 1, so we can simplify the expression further:

s = a ∫ sin^2(t/3) dt.

Now, let's integrate this expression. Recall that the integral of sin^2(t) dt can be found using the double angle formula for cosine: ∫sin^2(t) dt = (1/2)(t - sin(t)cos(t)).

Applying this formula to our integral, we have:

s = a [ (1/2)(t/3 - sin(t/3)cos(t/3)) ] from t = 0 to 2π.

Evaluating the integral at t = 2π and t = 0, we get:

s = a [ (1/2)(2π/3 - sin(2π/3)cos(2π/3)) - (1/2)(0 - sin(0)cos(0)) ].

Now, sin(2π/3) = sin(120 degrees) = √3/2, and cos(2π/3) = cos(120 degrees) = -1/2. Additionally, sin(0) = 0 and cos(0) = 1.

Substituting these values, we get:

s = a [ (1/2)(2π/3 - (√3/2)(-1/2)) - (1/2)(0 - 0) ].

Which simplifies to:

s = a [ (1/2)(2π/3 + √3/4) ].

Now, let's simplify the expression inside the brackets:

(1/2)(2π/3 + √3/4)
= (π/3) + (√3/8).

Therefore, the correct expression for the complete length of the curve is:

s = a (π/3 + √3/8).

This matches the expression you obtained in your calculation. However, the book's answer of 3aπ/2 is incorrect.

Please make sure to double-check the answer in the book or consult with your instructor or another reliable source for clarification.