Problem 10 - Contemporary Precalculus

Question

Find a viewing window that shows a complete graph of the curve. $$x=t^{3}-3 t-8, \quad y=3 t^{2}-15, \quad-4 \leq t \leq 4$$

Step-by-Step Solution

Answer

Question: Determine the viewing window that shows a complete graph of the curve given by the parametric equations x(t) = t^3 - 3t - 8 and y(t) = 3t^2 - 15 for t in the interval [-4, 4]. Answer: The viewing window for the complete graph of the curve is given by the following boundaries: - x values range from -60 to 44 - y values range from -15 to 33

Step 1: Plug in the endpoints

Evaluate the parametric equations for x and y for the given interval's endpoints. For t=-4: $$x(-4)=(-4)^3 -3(-4)-8=-64+12-8=-60$$ $$y(-4)=3(-4)^2-15=3(16)-15=33$$ For t=4: $$x(4)=(4)^3 -3(4)-8=64-12-8=44$$ $$y(4)=3(4)^2-15=3(16)-15=33$$ Boundaries obtained so far: - x ranges from -60 to 44 - y is 33 for both endpoints

Step 2: Analyze critical points of x(t) and y(t)

Now we must find critical points of x(t) and y(t) by finding their respective derivatives and solving for t when sets equal to 0. Find the first derivative of x(t) and set it equal to zero: $$\begin{aligned}\frac{d}{dt}(t^3-3t-8) &= 0\\ 3t^2-3 &= 0\\ t^2 &= 1\\ t &= \pm 1\end{aligned}$$ Now plug in t = 1 and t = -1 to find their corresponding x and y values: For t = 1: $$x(1) = (1)^3 - 3(1) - 8 = -10$$ $$y(1) = 3(1)^2 - 15 = -12$$ For t = -1: $$x(-1) = (-1)^3 + 3(-1) - 8 = -10$$ $$y(-1) = 3(-1)^2 - 15 = -12$$ Find the first derivative of y(t) and set it equal to zero: $$\begin{aligned}\frac{d}{dt}(3t^2-15) &= 0\\ 6t &= 0\\ t &= 0\end{aligned}$$ Now plug in t = 0 to find the corresponding x and y values: For t = 0: $$x(0) = (0)^3 - 3(0) - 8 = -8$$ $$y(0) = 3(0)^2 - 15 = -15$$ Thus, the viewing window for the complete graph of the curve is given by the final boundaries: - x ranges from -60 to 44 - y ranges from -15 to 33

Key Concepts

Graphing Parametric Curves

Graphing parametric curves can seem tricky at first, but it's a powerful way to visualize relationships between two variables, often representing a curve in space. In parametric equations, the x and y coordinates are expressed as functions of a third variable, typically time, denoted by \( t \). In the exercise provided, the equations for \( x \) and \( y \) are given as functions of \( t \), allowing us to plot \( x \) and \( y \) values corresponding to various values of \( t \).

• To graph, we calculate pairs of \( (x, y) \) coordinates for specified \( t \) values, which define the curve.
• This curve will often illustrate how two measurements coexist over time or another index.

Fundamentally, the parametric equations found in the exercise trace a curve based on inputs from \( t \), specifically from \( -4 \) to \( 4 \). Reviewing the step-by-step solution, plugging these bounds into our equations gives us a complete trajectory of the curve.

Derivative of Parametric Functions

The derivative of parametric functions provides insights into the curve's slopes and helps identify critical points, which are locations where the function’s rate of change alters. In our case, we determined these derivatives to find where they equaled zero, suggesting either maxima, minima, or points of inflection.

• The derivative of each function is computed by differentiating the parametric equations with respect to \( t \).
• The zeroes of the derivatives indicate critical values, points where the slope of the curve is either zero or undefined.

This concept is essential for understanding where the graph might change direction or display specific features, helping us draw more accurate shapes of curves. In this exercise, derivatives help us ascertain certain \( t \) values, assisting in defining the shape of the parametric plot.

Critical Points

Critical points are significant in analyzing the behavior of parametric curves because they indicate where the curve’s direction changes. These points arise where the derivative either vanishes or becomes undefined. In the practice exercise, we found critical points for \( x(t) \) and \( y(t) \) by setting their respective derivatives equal to zero.

• For \( x(t) \), the critical points were located at \( t = \pm 1 \), indicating potential extremum or bend.
• For \( y(t) \), the critical point surfaced at \( t = 0 \).

Identifying these points allows for a more precise depiction of the curve, ensuring all inflection points and bends are noted. Acknowledging critical points leads to deeper understanding and more detailed graphing of the parametric curve.

Viewing Window for Graphing

Selecting an appropriate viewing window is vital for visualizing the complete graph of a parametric curve. This involves setting the right x and y boundaries that encapsulate all the important aspects of the curve. The solution given defines the optimal window by first calculating extremities of the equations based on edge \( t \) values and critical points.

• The determined range for \( x \) is from \( -60 \) to \( 44 \), providing a span to contain all variations of the curve.
• For \( y \), it extends from \( -15 \) to \( 33 \), ensuring ease in capturing its full oscillation.

By setting this window, students can ensure no critical component of the graph is missed, paving way for better analysis and comprehension of the curve's properties as delineated by the parametric equations.