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Magazine Chapter 8: Problem 30
A curve is defined parametrically by the equations $$ \begin{aligned} &x=2 \cos \theta+\cos 2 \theta \\ &y=2 \sin \theta-\sin 2 \theta \end{aligned} $$ Draw a sketch of the curve for \(0 \leqslant \theta \leqslant 2 \pi\). Find the equation of the tangent to the curve at the point where \(\theta=\pi / 4\)
Short Answer
Expert verified
The tangent line at \( \theta = \frac{\pi}{4} \) is \( y - (\sqrt{2} - 1) = \frac{\sqrt{2}}{-\sqrt{2} - 2}(x - \sqrt{2}) \).
Step by step solution
01
Parametric Derivatives
To find the equation of the tangent, we need to compute the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). The derivative of \( x \) with respect to \( \theta \) is \( \frac{dx}{d\theta} = -2 \sin \theta - 2 \sin 2\theta \), and for \( y \) it is \( \frac{dy}{d\theta} = 2 \cos \theta - 2 \cos 2\theta \).
02
Slope of the Tangent Line
The slope of the tangent line is given by \( \frac{dy}{dx} \), which is the ratio \( \frac{dy/d\theta}{dx/d\theta} \). Substitute the derivatives: \( \frac{dy}{dx} = \frac{2 \cos \theta - 2 \cos 2\theta}{-2 \sin \theta - 2 \sin 2\theta} \).
03
Evaluate Derivatives at \( \theta = \frac{\pi}{4} \)
Substitute \( \theta = \frac{\pi}{4} \) into the derivatives to evaluate the slope at this point. Compute \( \frac{dy}{d\theta} = 2 \cos \frac{\pi}{4} - 2 \cos \frac{\pi}{2} = \sqrt{2} \) and \( \frac{dx}{d\theta} = -2 \sin \frac{\pi}{4} - 2 \sin \frac{\pi}{2} = -\sqrt{2} - 2 \).
04
Slope of the Tangent Line at \( \theta = \frac{\pi}{4} \)
Calculate the slope of the tangent by evaluating \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{4} \). This results in \( \frac{dy}{dx} = \frac{\sqrt{2}}{-\sqrt{2} - 2} \).
05
Determine the Cartesian Coordinates at \( \theta = \frac{\pi}{4} \)
Calculate the \( x \) and \( y \) coordinates. Substituting \( \theta = \frac{\pi}{4} \), \( x = 2 \cos \frac{\pi}{4} + \cos \frac{\pi}{2} = \sqrt{2} \) and \( y = 2 \sin \frac{\pi}{4} - \sin \frac{\pi}{2} = \sqrt{2} - 1 \).
06
Equation of the Tangent Line
Use point-slope form \( y - y_1 = m(x - x_1) \). Here, \( y_1 = \sqrt{2} - 1 \), \( x_1 = \sqrt{2} \), and \( m = \frac{\sqrt{2}}{-\sqrt{2} - 2} \). Substitute to get: \( y - (\sqrt{2} - 1) = \frac{\sqrt{2}}{-\sqrt{2} - 2}(x - \sqrt{2}) \).
07
Sketch the Curve
To sketch the curve, calculate the \( x \) and \( y \) values for key values of \( \theta \) (such as 0, \( \frac{\pi}{2} \), \( \pi \), \( \frac{3\pi}{2} \), and \( 2\pi \)) and plot the resulting points on the graph. Connect the points smoothly to form the curve.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
When dealing with parametric equations, derivatives play an essential role in understanding how a curve behaves. Derivatives essentially represent the rate of change. To derive the curve's tangent, we calculate two derivatives: \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). These derivatives focus on how changes in \( \theta \) affect \( x \) and \( y \) respectively.
- The Process: Derive \( x = 2 \cos \theta + \cos 2\theta \), and obtain \( \frac{dx}{d\theta} = -2 \sin \theta - 2 \sin 2\theta \).
- For \( y = 2 \sin \theta - \sin 2\theta \), the derivative is \( \frac{dy}{d\theta} = 2 \cos \theta - 2 \cos 2\theta \).
Tangent Line
The tangent line to a curve at a specific point provides a linear approximation to the curve at that point. This tangent can help us understand the direction the curve is heading at the given point. To find a tangent line for a parametric curve, we need to calculate its slope and use a standard point-slope form.
- Slope: The slope of the tangent is the ratio \( \frac{dy}{dx} \), where \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
- Point Slope Form: With \( y-y_1 = m(x-x_1) \), and by knowing our point and the slope, we can construct the tangent line.
Sketching Curves
Sketching a curve that is defined by parametric equations involves calculating and plotting key points that represent the curve鈥檚 path. This visual representation aids in understanding the geometric behavior of the curve,
- Key Values: Begin by evaluating the parametric equations at turning points such as \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \).
- Smooth Connection: Plot these points on the coordinate plane, then smoothly connect them.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, are pivotal in expressing parametric equations. These functions naturally allow the modeling of periodic patterns and behaviors, which are typical in geometry involving circles and waves.
- Why Trig Functions: Functions like \( \cos \theta \) and \( \sin \theta \) inherently cycle through peaks and troughs, displaying regular patterns that are well-suited for parametric descriptions.
- Application: In our exercise, the curve鈥檚 equations, \( x = 2\cos\theta + \cos 2\theta \) and \( y = 2\sin \theta - \sin 2 \theta \), use trigonometric terms to encode complex motion paths.
