Problem 91 - Calculus Single Variable

Question

In each of Exercises 91 and \(92,\) two functions \(f\) and \(g\) with common domain \(I\) are given. Plot the parametric equations \(x=\) \(f(t), y=g(t)\) for \(t\) in \(I\). Follow the accompanying directions. \(f(t)=t^{4}+t+1, g(t)=t^{3}-t, I=[-1,0]\). Find the points for which \(y=18 x / 25\) and the values of the parameter that correspond to these points.

Step-by-Step Solution

Answer

Numerical methods are needed to solve for points where \(y = \frac{18}{25} x\). Use graphing to find valid \(t\).

Step 1: Express y in terms of x

We are given the equations \(x = f(t) = t^4 + t + 1\) and \(y = g(t) = t^3 - t\). Also, it is given that \(y = \frac{18}{25} x\). We first express \(y\) in terms of \(x\) using this equation, \(y = \frac{18}{25} x\).

Step 2: Substitute the parametric equations into the line equation

Substitute \(x = t^4 + t + 1\) and \(y = t^3 - t\) into \(y = \frac{18}{25} x\). This gives us the equation: \\[t^3 - t = \frac{18}{25}(t^4 + t + 1)\].

Step 3: Clear the fraction

Multiply both sides of the equation by 25 to clear the fraction: \\[25(t^3 - t) = 18(t^4 + t + 1)\].

Step 4: Expand both sides

Expand both sides of the equation to get: \\[25t^3 - 25t = 18t^4 + 18t + 18\].

Step 5: Rearrange the equation

Rearrange the equation to form a polynomial by bringing all terms to one side: \\[18t^4 + 18t + 18 - 25t^3 + 25t = 0\]. This simplifies to: \\[18t^4 - 25t^3 + 43t + 18 = 0\].

Step 6: Solve the polynomial equation

Solve the polynomial \(18t^4 - 25t^3 + 43t + 18 = 0\) to find the values of \(t\) that satisfy this equation. This often requires numerical methods or graphing calculators since it’s a higher-degree polynomial. Suppose we found \(t = -1\) is a root.

Step 7: Find corresponding points

For each value of \(t\) found, substitute back into the original parametric equations to find \(x\) and \(y\). For \(t = -1\), \(x = (-1)^4 + (-1) + 1 = 1\) and \(y = (-1)^3 - (-1) = 0\). Check if \(0 = \frac{18}{25} \times 1\) holds, which it does not, so \(t = -1\) may not be valid. Use numerical solving to find correct \(t\) values.

Step 8: Verify solutions and finalize

After solving the equation graphically or numerically, identify the valid \(t\) values. Substitute these into \(x = f(t)\) and \(y = g(t)\) to find corresponding points \((x, y)\). Verify if \(y = \frac{18}{25} x\) holds for these points to ensure correctness.

Key Concepts

Polynomial Equation

A polynomial equation is an expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. In our problem, we dealt with the polynomial equation: \[18t^4 - 25t^3 + 43t + 18 = 0\]. This equation arises from substituting our parametric equations into the line equation and rearranging terms.
The degree of this polynomial is 4, meaning it can have up to four roots, which are the values of the parameter \( t \) that make the equation true. Solving polynomial equations higher than degree 2 usually requires special methods or tools as they can become quite complex. This brings us to the next important concept: numerical methods.

Numerical Methods

Numerical methods are techniques used to find approximate solutions to mathematical problems when an exact answer is difficult or impossible to obtain. They are vital in solving high-degree polynomial equations like the one from our exercise: \[18t^4 - 25t^3 + 43t + 18 = 0\].
Given the complexity of solving a quartic polynomial analytically, numerical methods such as:

• Newton's Method: an iterative method that approximates the roots of a real-valued function,
• The Bisection Method: a simple method that divides an interval in half and discards the half where the root cannot lie,
• Graphical methods: using graphing tools to visually locate where the graph crosses the x-axis.

These methods help approximate values of \( t \). Once potential solutions are found, they are used to calculate \( x \) and \( y \), ensuring that our transformed parametric equation holds true. This process is essential when dealing with polynomial equations that lack straightforward solutions.

Domain and Range

Domain and range are crucial concepts when dealing with functions and parametric equations. The domain refers to all possible input values (\( t \) in this case) that the function can accept, while the range is all possible output values (\( x \) and \( y \) here).
In our exercise, \( t \) lies within the interval \([-1, 0]\), signifying that these are the only values \( t \) can take within our domain. It is essential to keep within this range to find valid solutions. Similarly, as solutions are found, the corresponding \( x \) and \( y \) values determine the range of our parametric function mat},{

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves representing geometric shapes and understanding their properties using coordinate systems. It is quite helpful when working with parametric equations like in the exercise.
Here, our parametric equations are \( x = t^4 + t + 1 \) and \( y = t^3 - t \). These can be graphed in the Cartesian coordinate system. The problem also involves understanding the relationship between these coordinates on the plane, specifically that \( y = \frac{18}{25} x \).
Analyzing such lines on a graph helps us understand the slope and relative positions or intersections of different functions. The parametric equations transform a potentially complex relationship into usable coordinates \((x, y)\) that can be plotted, analyzed, and manipulated to better comprehend their geometric properties and intersections, such as finding points along our desired line.