Math Problem Statement

Which of the following is the length of the circumference of the astroid defined for a > 0 by x = a cos^3 t, y = a sin^3 t, 0 ≤ t ≤ 2π?

Solution

To solve this problem, we need to find the length of the curve defined parametrically by the astroid equations:

x=acos3t,y=asin3t,0t2πx = a \cos^3 t, \quad y = a \sin^3 t, \quad 0 \leq t \leq 2\pi

Step-by-Step Solution

  1. Calculate dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}:

    Given: x=acos3tdxdt=3acos2t(sint)=3acos2tsintx = a \cos^3 t \quad \Rightarrow \quad \frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t

    And: y=asin3tdydt=3asin2tcosty = a \sin^3 t \quad \Rightarrow \quad \frac{dy}{dt} = 3a \sin^2 t \cos t

  2. Compute (dxdt)2+(dydt)2\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2:

    We need to evaluate: (dxdt)2+(dydt)2\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2

    Substituting the derivatives: (3acos2tsint)2+(3asin2tcost)2\left( -3a \cos^2 t \sin t \right)^2 + \left( 3a \sin^2 t \cos t \right)^2

    Expanding each term: =9a2cos4tsin2t+9a2sin4tcos2t= 9a^2 \cos^4 t \sin^2 t + 9a^2 \sin^4 t \cos^2 t

    Factor out 9a2cos2tsin2t9a^2 \cos^2 t \sin^2 t: =9a2cos2tsin2t(cos2t+sin2t)= 9a^2 \cos^2 t \sin^2 t (\cos^2 t + \sin^2 t)

    Since cos2t+sin2t=1\cos^2 t + \sin^2 t = 1: =9a2cos2tsin2t= 9a^2 \cos^2 t \sin^2 t

  3. Set Up the Integral for Arc Length:

    The arc length ss of a parametric curve from t=0t = 0 to t=2πt = 2\pi is given by: s=02π(dxdt)2+(dydt)2dts = \int_0^{2\pi} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt

    Substitute (dxdt)2+(dydt)2=9a2cos2tsin2t\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = 9a^2 \cos^2 t \sin^2 t: s=02π9a2cos2tsin2tdts = \int_0^{2\pi} \sqrt{9a^2 \cos^2 t \sin^2 t} \, dt

    Simplify the square root: s=02π3acostsintdts = \int_0^{2\pi} 3a |\cos t \sin t| \, dt

    Since costsint=12sin2t|\cos t \sin t| = \frac{1}{2} |\sin 2t|, we get: s=02π3a2sin2tdts = \int_0^{2\pi} \frac{3a}{2} |\sin 2t| \, dt

  4. Evaluate the Integral:

    By symmetry and periodicity, the integral can be evaluated over one period [0,2π][0, 2\pi], leading to a result of 6a6a.

Final Answer

The length of the circumference of the astroid is: 6a\boxed{6a}

Would you like further details on any step, or have any questions?

Here are five related questions for further exploration:

  1. How would the arc length formula change if the parametric equations were different?
  2. What is the role of symmetry in simplifying integrals in parametric curves?
  3. How would we compute the area enclosed by the astroid?
  4. Can the arc length integral be evaluated without breaking it down into trigonometric identities?
  5. How does the arc length calculation differ for curves defined implicitly?

Tip: Always check for symmetry in the parametric equations when calculating integrals, as it can simplify the problem significantly.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Arc Length
Astroid Curve

Formulas

Arc Length Formula for Parametric Curves: s = ∫ (sqrt((dx/dt)^2 + (dy/dt)^2)) dt
Astroid Equations: x = a cos^3 t, y = a sin^3 t

Theorems

Pythagorean Identity
Integral Calculus for Arc Length

Suitable Grade Level

Grades 11-12