Question

Find the value(s) of \(t\) where the curve defined by the parametric equations is not smooth. \(x=2 \cos t-\cos (2 t), \quad y=2 \sin t-\sin (2 t)\)

Short answer

The curve is not smooth at \(t = n\pi\), where \(n\) is an integer.

Step-by-step solution

Understand the Problem

To determine where the curve defined by the given parametric equations is not smooth, we need to find where the curve's derivative does not exist or where it has a discontinuity. This often occurs when the derivatives of both parametric equations equal zero simultaneously.

Find the Derivatives

Calculate the derivatives of the parametric equations with respect to the parameter \(t\):\[ \frac{dx}{dt} = -2\sin(t) + 2\sin(2t) \]\[ \frac{dy}{dt} = 2\cos(t) - 2\cos(2t) \]

Set the Derivatives Equal to Zero

Determine where both derivatives are zero at the same time:1. \(-2\sin(t) + 2\sin(2t) = 0\)2. \(2\cos(t) - 2\cos(2t) = 0\)Solve these equations to find common solutions where both derivatives are zero simultaneously.

Solve \(-2\sin(t) + 2\sin(2t) = 0\)

Use the identity \(\sin(2t) = 2\sin(t)\cos(t)\) to rewrite the equation:\(-2\sin(t) + 4\sin(t)\cos(t) = 0\)Factor out \(2\sin(t)\):\[ 2\sin(t)(2\cos(t) - 1) = 0 \]This gives us the solutions \(\sin(t) = 0\) or \(2\cos(t) - 1 = 0\).

Solve for \(\sin(t) = 0\)

The solutions to \(\sin(t) = 0\) are:\[ t = n\pi \quad \text{where} \; n \; \text{is an integer.} \]

Solve for \(2\cos(t) - 1 = 0\)

Setting \(2\cos(t) - 1 = 0\) yields:\[ \cos(t) = \frac{1}{2} \]The solutions are: \[ t = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad t = -\frac{\pi}{3} + 2n\pi \quad \text{where} \; n \; \text{is an integer.} \]

Solve \(2\cos(t) - 2\cos(2t) = 0\)

Rewrite using \(\cos(2t) = 2\cos^2(t) - 1\):\[ 2\cos(t) - 2(2\cos^2(t) - 1) = 0 \]Simplify:\[ 2\cos(t) - 4\cos^2(t) + 2 = 0 \]\[ 4\cos^2(t) - 2\cos(t) - 2 = 0 \]Divide by 2:\[ 2\cos^2(t) - \cos(t) - 1 = 0 \]This is a quadratic in \(\cos(t)\). Solve using the quadratic formula.

Quadratic Formula Solution

Use the quadratic formula \(\cos(t) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2, b = -1, c = -1\):\[ \cos(t) = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]\[ \cos(t) = \frac{1 \pm \sqrt{1 + 8}}{4} \]\[ \cos(t) = \frac{1 \pm 3}{4} \]This gives \(\cos(t) = 1\) or \(\cos(t) = -\frac{1}{2}\).

Solve for \(\cos(t) = 1\)

The solution to \(\cos(t) = 1\) is:\[ t = 2n\pi \quad \text{where} \; n \; \text{is an integer.} \]

Solve for \(\cos(t) = -\frac{1}{2}\)

The solutions are:\[ t = \frac{2\pi}{3} + 2n\pi \quad \text{and} \quad t = -\frac{2\pi}{3} + 2n\pi \quad \text{where} \; n \; \text{is an integer.} \]

Determine Common Solutions

From previous steps, identify common solutions where both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are zero. These occur when:- \(t = n\pi \)- Combine with conditions from quadratic solutions: - \(t = 2n\pi \) is already included. - Solutions for \(t = \frac{\pi}{3} + 2n\pi\) and \(t = -\frac{\pi}{3} + 2n\pi\) should be verified if they match with \(n\pi\).

Unified Solution

The curve is not smooth where the parameter \(t\) makes both derivatives simultaneously zero, which commonly occurs when \(t = n\pi\), where \(n\) is an integer.

Key concepts

Curve Smoothness

When we talk about the smoothness of a curve defined by parametric equations, we are interested in points where the curve might not flow without interruption. A smooth curve has no sharp corners or cusps. For a curve to be smooth at a certain point, the derivatives of its parametric equations must be defined and not vanish at the same time.

• A sharp corner or cusp occurs on a parametric curve if both the derivatives of the parametric functions ( \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) ) are zero simultaneously.
• Why? Because this indicates a sudden directional change which causes a lack of smoothness.

Hence, identifying these points is key to discussing the curve's smoothness.

Derivatives

Derivatives represent the rates at which things change. For a parametric equation, the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) indicate the rate of change of \(x\) and \(y\) with respect to the parameter \(t\). These derivatives help us to find the slope of the tangent to the curve at any point.

• To find where a parametric curve might not be smooth, we solve for where both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are zero simultaneously.
• This often requires solving trigonometric equations and systems of equations as illustrated in the steps.

Understanding and computing these derivatives becomes crucial to analyze the behavior of the curve.

Trigonometric Identities

Trigonometric identities are invaluable tools when working with parametric equations that involve trigonometric functions like sine and cosine. These identities allow us to simplify and solve equations, aiding in our understanding of the curve's behavior.

Common Identities

• \( \sin(2t) = 2\sin(t)\cos(t) \)
• \( \cos(2t) = 2\cos^2(t) - 1 \)

These identities were crucial in simplifying the equations \(-2\sin(t) + 2\sin(2t) = 0\) and \(2\cos(t) - 2\cos(2t) = 0\).By substituting these identities, as shown in the solution steps, we convert the given problem into simpler mathematical forms, making it easier to solve and analyze.

Quadratic Equations

Quadratic equations often appear in parametric equations when you use trigonometric identities or when factoring. These look like \(ax^2 + bx + c = 0\) and can be solved using the quadratic formula.

The Quadratic Formula

Given \(ax^2 + bx + c = 0\), use:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our context, using trigonometric identities, we stumbled upon a quadratic equation \(2\cos^2(t) - \cos(t) - 1 = 0\), leading us to solutions for the trigonometric values.

• These solutions help isolate values of \(t\) that affect the smoothness of the parametric curve.
• In parametric contexts, solving a quadratic can often reveal critical points where the behavior of the curve changes.