Problem 5 - 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 \(t\). $$x=3 \sin ^{3} t, y=3 \cos ^{3} t ; t=\pi / 4$$

Step-by-Step Solution

Answer

The point is \( \left( \frac{3\sqrt{2}}{4}, \frac{3\sqrt{2}}{4} \right) \).

Step 1: Substitute parameter into the x-equation

We are given the equation for x: \( x = 3 \sin^3(t) \). Substitute \( t = \frac{\pi}{4} \) into this equation. So, we have: \( x = 3 \sin^3\left(\frac{\pi}{4}\right) \).

Step 2: Evaluate \( \sin(\pi/4) \)

The value of \( \sin\left(\frac{\pi}{4}\right) \) is \( \frac{\sqrt{2}}{2} \). Substitute this value back into the equation for x: \( x = 3 \left(\frac{\sqrt{2}}{2}\right)^3 \).

Step 3: Simplify the calculation for x

First, calculate \( \left(\frac{\sqrt{2}}{2}\right)^3 = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}\). Then, finish calculating x: \( x = 3 \times \frac{\sqrt{2}}{4} = \frac{3\sqrt{2}}{4} \).

Step 4: Substitute parameter into the y-equation

We are given the equation for y: \( y = 3 \cos^3(t) \). Substitute \( t = \frac{\pi}{4} \) into this equation. So, we have: \( y = 3 \cos^3\left(\frac{\pi}{4}\right) \).

Step 5: Evaluate \( \cos(\pi/4) \)

The value of \( \cos\left(\frac{\pi}{4}\right) \) is \( \frac{\sqrt{2}}{2} \). Substitute this value back into the equation for y: \( y = 3 \left(\frac{\sqrt{2}}{2}\right)^3 \).

Step 6: Simplify the calculation for y

First, calculate \( \left(\frac{\sqrt{2}}{2}\right)^3 = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}\). Then, finish calculating y: \( y = 3 \times \frac{\sqrt{2}}{4} = \frac{3\sqrt{2}}{4} \).

Step 7: Conclusion

The coordinates of the point on the curve when \( t = \frac{\pi}{4} \) are \( x = \frac{3\sqrt{2}}{4} \) and \( y = \frac{3\sqrt{2}}{4} \). Therefore, the point is \( \left( \frac{3\sqrt{2}}{4}, \frac{3\sqrt{2}}{4} \right) \).

Key Concepts

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in understanding parametric equations. These functions help describe how angles relate to ratios of sides in right triangles.
In this exercise, both the sine and cosine functions are raised to the power of three, which might look intimidating at first.
However, it simply means taking the standard trigonometric value for the angle and cubing it.Some important things to remember are:

• \( \sin(\pi/4) \) and \( \cos(\pi/4) \) both equal \( \frac{\sqrt{2}}{2} \). These values come from the unit circle, a circle with a radius of 1.


• The unit circle helps us easily find values of trigonometric functions for special angles like \( \pi/4 \), which is equivalent to 45 degrees.

Understanding these basic properties allows us to modify and work with trigonometric functions in parametric equations effectively.

Curve Coordinates

The coordinates of curves in parametric equations can be found by evaluating given expressions for \( x \) and \( y \). These expressions are functions of a parameter, often denoted by \( t \).
In our exercise, both \( x \) and \( y \) depend on the parameter \( t \). This means to figure out their values, we substitute the given \( t \) into the parametric equation.Understanding curve coordinates involves:

• Recognizing that \( x \) and \( y \) in parametric equations create points on a graph.
• Knowing each substitution creates a unique point depending on the given \( t \).


• The ultimate result of these substitutions and evaluations can be a specific curve on which all point coordinates lie.

In essence, you're building a set of related points, rather than independent \( x \), \( y \) pairs typically found in non-parametric equations.

Parameter Substitution

Parameter substitution is a crucial technique in solving parametric equations. Here, you replace the parameter \( t \) with a specific value to find corresponding \( x \) and \( y \) coordinates.
By substituting \( t = \pi/4 \), you can solve for both \( x \) and \( y \) in the given equations.How to effectively use parameter substitution:

• Insert the given \( t \) value directly into each parametric formula.


• Calculate the trigonometric functions with the substituted parameter. This often involves using known values from the unit circle.
• Simplify the algebraic expressions by completing any necessary calculations.

Each step ensures that you correctly find a single point on a curve that adheres to the terms of its parametric equations. This method transforms a potentially complex problem into a manageable set of tasks.