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Exploring Parametric Curves: Finding the Area of an Inner Loop

Can you calculate the area enclosed by one of the inner loops formed by the curve x(t)=2cos(t)+cos(5t) and y(t)=2sin(t)+sin(5t)?

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Table of Contents

The Problem

From VOL. 55, NO. 22, MARCH 2024 edition of THE COLLEGE MATHEMATICS JOURNAL:

x(t)=2cos(t)+cos(5t)y(t)=2sin(t)+sin(5t).

Find the area enclosed by one of the four inner loops.
— Greg Dresden, Washington and Lee University, Lexington, VA

Breaking Down the Problem

This problem revolves around a parametric curve, with equations for x(t) and y(t) that describe a looping, symmetric structure. The challenge is to calculate the area enclosed by one of the four inner loops. As soon as I saw the parametric nature of the problem, I realized I would need to use calculus, specifically integrals, to compute the area. The parametric area formula was the right tool for this task.

Parametric Equations for the Curve

The parametric equations for the curve are:

x(t)=2cos(t)+cos(5t),y(t)=2sin(t)+sin(5t).

Desmos graph of the parametric curve

Notice that the point closest to the origin on the inner loop in the first quadrant is given by x(π4)=22=y(π4), which lies on the line y=x. Let the point P where the graph intersects itself in the first quadrant be given by t=α and t=β, where 0<α<π4<β<π2.

From the symmetry of the curve, α and β must be complementary, so that:

x(β)=2cos(β)+cos(5β)=2cos(π2α)+cos(5(π2α))=2sin(α)+cos(π25α)=2sin(α)+sin(5α)=y(α)

Since x(α)=x(β) and y(α)=y(β), then P must lie on the line y=x. This is clearly true, as shown below:

y=x going through parametric curve (sorry that P is maladjusted!)

Solving for α and β

Now, let’s examine the points of self-intersection we can apply the multiple-angle formulas to rewrite the parametric equations in terms of trigonometric identities:

cos(5t)=cost(112sin2t+16sin4t),sin(5t)=sint(520sin2t+16sin4t).

Thus, the parametric equations become:

x(t)=2cos(t)+cos(5t)=cost(312sin2t+16sin4t),y(t)=2sin(t)+sin(5t)=sint(720sin2t+16sin4t).sint(720sin2t+16sin4t)=cost(312sin2t+16sin4t)

Dividing both sides by cost, multiplying the right-hand side by sec4tsec4t, and simplifying:

sintcost=312sin2t+16sin4t720sin2t+16sin4tsec4tsec4ttant=3(1+tan2t)212tan2t(1+tan2t)+16tan4t7(1+tan2t)220tan2t(1+tan2t)+16tan4ttant=36tan2t+7tan4t76tan2t+3tan4t

To simplify further, let u=tant, leading to the following equation:

u(76u2+3u4)=36u2+7u4.

This expands to:

3u57u46u3+6u2+7u3=0,

which factors as:

(u1)(3u44u310u24u+3)=0.

Given that tant=u=1, we know t=π4 is a solution, corresponding to the least upper bound before α. The remaining real solutions for t yield values for α and β, which can be determined by solving polynomial:

3u44u310u24u+3=0.

Since 3u44u34u+3 is palindromic, we can divide by u2 to get:

3u24u101u+3u2=03(u2+1u2)4(u+1u)10=03((u+1u)22)4(u+1u)10=03(u+1u)24(u+1u)16=0

Using the quadratic formula, we find

u+1u=2±2133,

and

u2(2(1±13)3)u+1=0.

Since the discriminant Δ of this quadratic is 49(5±213), and 5213<0, the only real solutions are obtained for the positive case. Therefore:

u=1+13±5+2133

and the solutions for α and β are:

α=tan1(1+135+2133),β=tan1(1+13+5+2133).

For future reference, it’s important to notice that:

βα=tan1(25+21332)=tan1(5+2133)

and α+β=π2.

Finding the Area

To compute the area enclosed by one of the inner loops, we use the parametric area formula:

A=12Cxdyydx.

Substituting the parametric expressions for x(t) and y(t) and simplifying:

A=12αβ((2cost+cos5t)(2cost+5cos5t)(2sint+sin5t)(2sint5sin5t)) dt=12αβ(4cos2t+12costcos5t+5cos25t+4sin2t+12sintsin5t+5sin25t) dt=12αβ(9+12(costcos5t+sintsin5t)) dt=αβ(92+6cos4t) dtA=9t2+32sin4t]αβ=92(βα)+3sin2(βα)cos2(βα)

Using the values of α and β, we find:

A=92tan1(5+2133)+6sin(βα)cos(βα)cosπ=92tan1(5+2133)95+2137+13

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A=92tan1(5+2133)9(5+213)7+130.9108.

This problem elegantly ties together parametric equations, trigonometric identities, and calculus. Through careful manipulation and integration, we arrived at a precise answer for the area (as well as a numerical approximation).