Problem 4 - Precalculus

Question

You are given the parametric equations of a curve and a value for the parameter \(t\). Find the coordinates of the point on the curve corresponding to the given value of \(t\). $$x=4 \cos 2 t, y=6 \sin 2 t ; t=\pi / 3$$

Step-by-Step Solution

Answer

The coordinates are \((-2, 3\sqrt{3})\).

Step 1: Substitute the given t-value into the equations

Since we need to find the coordinates of the point when \( t = \frac{\pi}{3} \), we'll substitute \( t = \frac{\pi}{3} \) into the parametric equations:\ \( x = 4 \cos (2t) \) and \( y = 6 \sin (2t) \).

Step 2: Calculate x-coordinate

Substitute \( t = \frac{\pi}{3} \) into the expression for \( x \):\[\begin{align*}x &= 4 \cos(2 \times \frac{\pi}{3}) \&= 4 \cos(\frac{2\pi}{3})\end{align*}\]Now, calculate \( \cos(\frac{2\pi}{3}) \). The angle \( \frac{2\pi}{3} \) is in the second quadrant, and \( \cos(\frac{2\pi}{3}) = -\frac{1}{2} \). Substitute this value:\[ x = 4 \times -\frac{1}{2} = -2 \]

Step 3: Calculate y-coordinate

Substitute \( t = \frac{\pi}{3} \) into the expression for \( y \):\[\begin{align*}y &= 6 \sin(2 \times \frac{\pi}{3}) \&= 6 \sin(\frac{2\pi}{3})\end{align*}\]Now, calculate \( \sin(\frac{2\pi}{3}) \). The angle \( \frac{2\pi}{3} \) is in the second quadrant, and \( \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} \). Substitute this value:\[ y = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \]

Step 4: Combine x and y to find the coordinates

The \(x\)-coordinate is \(-2\), and the \(y\)-coordinate is \(3\sqrt{3}\). Therefore, the coordinates of the point on the curve corresponding to \( t = \frac{\pi}{3} \) are \((-2, 3\sqrt{3})\).

Key Concepts

Trigonometric Functions

Trigonometric functions, primarily sine and cosine, are mathematical functions that relate the angles of a triangle to the lengths of its sides. In the context of parametric equations, these functions help us determine the coordinates of points on a curve. Here, the parametric equations given are

• \(x = 4\cos(2t)\)
• \(y = 6\sin(2t)\)

When \(t\) is substituted, these trigonometric functions give specific values, resulting in exact coordinates.
Let's take an example: If \(t = \frac{\pi}{3}\), the function \(\cos(2t)\) calculates to \(\cos(\frac{2\pi}{3})\), while \(\sin(2t)\) calculates to \(\sin(\frac{2\pi}{3})\). The way sine and cosine function depend on the quadrant determines the sign and value of the coordinates.
These functions encapsulate periodic behavior, which is essential in understanding the curves they describe. This is why they are fundamental in calculus, physics, and engineering. Knowing how to substitute and compute these values is key to grasping more advanced concepts in math.

Coordinate Calculation

Calculating coordinates using parametric equations involves evaluating trigonometric functions at specific parameter values, like \(t = \frac{\pi}{3}\).
Parametric equations, such as \(x = 4 \cos 2t\) and \(y = 6 \sin 2t\), represent a curve on the plane.
By substituting \(t\) into these equations, we derive specific \(x\) and \(y\) values. This substitution translates the abstract mathematical expressions into concrete points.

• For the \(x\)-coordinate, substitute \(t = \frac{\pi}{3}\): \( x = 4 \cos\left(\frac{2\pi}{3}\right)\). The cosine of \(\frac{2\pi}{3}\), which is in the second quadrant, is \(-\frac{1}{2}\). Thus, \(x = 4 \times -\frac{1}{2} = -2\).
• For the \(y\)-coordinate, substitute \(t = \frac{\pi}{3}\): \( y = 6 \sin\left(\frac{2\pi}{3}\right)\). The sine of \(\frac{2\pi}{3}\) is \(\frac{\sqrt{3}}{2}\), leading to \(y = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\).

These calculations effectively "translate" the parameter \(t\) into coordinates that represent a real point on the curve. Understanding this process is crucial for analyzing more complex curves and systems in mathematics and applied sciences.

Second Quadrant Angles

Understanding the behavior of angles in different quadrants is key to handling trigonometric functions correctly. Each quadrant in the Cartesian plane dictates the sign and sometimes the value of trigonometric functions.
For instance, the angle \(\frac{2\pi}{3}\) lies in the second quadrant.

• In this quadrant, the cosine of any angle is negative, while the sine is positive. This is fundamental to determining the correct sign for the x- and y-coordinates in parametric equations.
• Therefore, when calculating \(\cos\left(\frac{2\pi}{3}\right)\), we get \(-\frac{1}{2}\). This aligns with standard trigonometric rules.
• Similarly, \(\sin\left(\frac{2\pi}{3}\right)\) is \(\frac{\sqrt{3}}{2}\), a positive value. Recognizing the sign and quadrant of an angle ensures accurate coordinate outcomes.

Quadrant analysis is a habit to develop for dealing with more complex problems involving trigonometry. It ensures precision when predicting the position of points on curves or evaluating various periodic functions.