Problem 43 - Thomas Calculus

Question

The curve with parametric equations $$ x=(1+2 \sin \theta) \cos \theta, \quad y=(1+2 \sin \theta) \sin \theta $$ is called a limacon and is shown in the accompanying figure. Find the points \((x, y)\) and the slopes of the tangent lines at these points for $$ \text { a. }\theta=0 . \quad \text { b. } \theta=\pi / 2, \quad \text { c. } \theta=4 \pi / 3 $$

Step-by-Step Solution

Answer

The points and slopes are: a. (1, 0), slope=1/2; b. (0, 3), slope=0; c. ((√3-1)/2, -3/2+√3/2), slope=(1-√3-3)/(3√3/4+1/2).

Step 1: Parameter Substitution for θ=0

First, we substitute \( \theta = 0 \) into the parametric equations to find the coordinates:\[ x = (1 + 2\sin(0))\cos(0) = 1 \times 1 = 1 \] \[ y = (1 + 2\sin(0))\sin(0) = 0 \] Thus, the point is \((x, y) = (1, 0)\).

Step 2: Parameter Substitution for θ=π/2

Next, substitute \( \theta = \frac{\pi}{2} \) into the parametric equations:\[ x = (1 + 2\sin(\frac{\pi}{2}))\cos(\frac{\pi}{2}) = 3 \times 0 = 0 \] \[ y = (1 + 2\sin(\frac{\pi}{2}))\sin(\frac{\pi}{2}) = 3 \times 1 = 3 \] Therefore, the point is \((x, y) = (0, 3)\).

Step 3: Parameter Substitution for θ=4π/3

Substitute \( \theta = \frac{4\pi}{3} \):\[ x = (1 + 2\sin(\frac{4\pi}{3}))\cos(\frac{4\pi}{3}) \] \( \sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2} \) and \( \cos(\frac{4\pi}{3}) = -\frac{1}{2} \). Calculate \( x \):\[ x = \left(1 + 2\left(-\frac{\sqrt{3}}{2}\right)\right)\left(-\frac{1}{2}\right) = \left(1 - \sqrt{3}\right)\left(-\frac{1}{2}\right) = \frac{\sqrt{3} - 1}{2} \] \[ y = (1 + 2\sin(\frac{4\pi}{3}))\sin(\frac{4\pi}{3}) = (1 - \sqrt{3}) \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}(\sqrt{3} - 1)}{2} = -\frac{3}{2} + \frac{\sqrt{3}}{2} \] Thus, the point is \((x, y) = \left(\frac{\sqrt{3} - 1}{2}, -\frac{3}{2} + \frac{\sqrt{3}}{2}\right)\).

Step 4: Find the Slope (dy/dx) for θ=0

To find the slope of the tangent line, compute \( \frac{dy}{dx} \) using \( \frac{dy/d\theta}{dx/d\theta} \). First, calculate the derivatives:\[ \frac{dx}{d\theta} = \frac{d}{d\theta}((1 + 2\sin\theta)\cos\theta) = -\sin\theta(1 + 2\sin\theta) + 2\cos^2\theta \] \[ \frac{dy}{d\theta} = \frac{d}{d\theta}((1 + 2\sin\theta)\sin\theta) = \cos\theta(1 + 2\sin\theta) + 2\sin^2\theta \] For \( \theta = 0 \), evaluate:\[ \frac{dx}{d\theta} = 2 \] \[ \frac{dy}{d\theta} = 1 \] Thus, \( \frac{dy}{dx} = \frac{1}{2} \).

Step 5: Find the Slope (dy/dx) for θ=π/2

Evaluate \( \frac{dy}{dx} \) for \( \theta = \frac{\pi}{2} \):\[ \frac{dx}{d\theta} = -3 \] \[ \frac{dy}{d\theta} = 0 \] Thus, \( \frac{dy}{dx} = 0 \).

Step 6: Find the Slope (dy/dx) for θ=4π/3

For \( \theta = \frac{4\pi}{3} \):\[ \sin\theta = -\frac{\sqrt{3}}{2}, \cos\theta = -\frac{1}{2} \] Evaluate derivatives:\[ \frac{dx}{d\theta} = -(-\frac{\sqrt{3}}{2})(1 - \sqrt{3}) + 2(-\frac{1}{2})^2 \] \[ \frac{dx}{d\theta} = \frac{3\sqrt{3}}{4} + \frac{1}{2} \] \[ \frac{dy}{d\theta} = (-\frac{1}{2})(1 - \sqrt{3}) + 2(-\frac{\sqrt{3}}{2})^2 \] \[ \frac{dy}{d\theta} = \frac{1-\sqrt{3}}{2} - \frac{3}{2}\] Therefore, \( \frac{dy}{dx} = \frac{(1-\sqrt{3}) - 3}{\frac{3\sqrt{3}}{4} + \frac{1}{2}} \). After simplification, the slope can be further evaluated numerically if necessary.

Key Concepts

Derivative of a Parametric Function

Parametric equations express curves in the plane using a third parameter, usually denoted as \( \theta \). Instead of \( y \) being defined directly in terms of \( x \), both \( x \) and \( y \) are defined via this parameter. To find the derivative of a parametric function, we use the chain rule. The derivative of \( y \) with respect to \( x \) in terms of \( \theta \) is given by \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] This formula allows us to find the slope of the tangent line to the curve at a given point. Calculating \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) separately involves using basic rules of differentiation, such as the product and chain rules. This step converts the parametric form to the more traditional derivative form, showing the rate of change of \( y \) with respect to \( x \). This is crucial in determining whether a tangent is horizontal, vertical, or at some other angle.
Understanding this process is essential in analyzing how a curve changes as we progress along it.

Slope of Tangent Line

The slope of a tangent line to a parametric curve at a specific point indicates the direction the curve is heading as it passes through that point. Tangent lines are straight lines that just touch the curve, meaning they have the same instantaneous direction as the curve does at that moment.
In parametric equations, the slope of the tangent line, represented as \( \frac{dy}{dx} \) using the formula mentioned earlier, can describe steepness and direction.

• If \( \frac{dy}{dx} = 0 \), the tangent line is horizontal, implying the curve is flat at that point.
• If \( \frac{dx}{d\theta} = 0 \), but \( \frac{dy}{d\theta} eq 0 \), this means the tangent is vertical.
• A positive slope means the curve is increasing, while a negative slope indicates the curve is decreasing.

Understanding how to find and interpret the slope of tangent lines is pivotal in mathematical analysis and in physical applications where directions of motion are described using curves.

Limaçon Curve

A limaçon is a type of polar curve that can exhibit a variety of shapes, from a simple loop to a cardioid-like form, all depending on its parameters. The name "limaçon" is derived from the resemblance some forms have to a snail shell. The parametric equations for a limaçon can be given as \[x = (1 + 2 \sin \theta) \cos \theta, \quad y = (1 + 2 \sin \theta) \sin \theta\] The factor \(1 + 2 \sin \theta\) modifies the radius as \(\theta\) changes.
Limaçon curves are fascinating for their symmetry and variety:

• They can take several shapes, including those with loops, dimples, or mere convex cuts.
• The polar form gives rise to unique points and features that can be explored using the parametric derivatives for thorough understanding.
• Limaçon curves can often be seen as special cases in physics or engineering problems involving rotational dynamics or motion along curved paths.

Comprehending limeçon curves involves not just mathematical computation, but spatial imagination to see where and how these curves fit into various scientific and practical contexts.