Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 41-42: As a particle moves along a straight path, its speed \(v\) (in \(\mathrm{cm} / \mathrm{sec}\) ) at time \(t\) (in seconds) is given by the equation. For what subintervals of the given time interval \([a, b]\) will its speed be at least \(k \mathrm{~cm} / \mathrm{sec}\) ? $$ v=t^{3}-3 t^{2}-4 t+20 ; \quad[0,5] ; \quad k=8 $$

Step-by-Step Solution

Answer

The speed is at least 8 cm/sec for all \( t \) in the interval [0, 5].

Step 1: Define the Speed Equation

The speed of the particle is given by the equation \( v = t^3 - 3t^2 - 4t + 20 \). We'll find when this speed is at least 8 cm/sec.

Step 2: Set the Inequality

To find when the speed is at least 8 cm/sec, set the equation \( v = t^3 - 3t^2 - 4t + 20 \) greater than or equal to 8, resulting in the inequality: \( t^3 - 3t^2 - 4t + 20 \geq 8 \).

Step 3: Simplify the Inequality

Subtract 8 from both sides to simplify: \( t^3 - 3t^2 - 4t + 12 \geq 0 \). Now we need to solve for \( t \) in the interval \([0, 5]\).

Step 4: Identify Critical Points

To find where the inequality holds, find the critical points by solving \( t^3 - 3t^2 - 4t + 12 = 0 \). Attempt factoring or use numerical methods like the Rational Root Theorem.

Step 5: Factor the Polynomial

After trying possible rational roots, we find that \( t = 2 \) is a root. Thus, testing gives us factors as \((t-2)(t^2-t-6)=0\). Further factor the quadratic to get \((t-2)((t-3)(t+2))=0\).

Step 6: Determine Condition Intervals

The roots are \( t = 2 \), \( t = 3 \), and \( t = -2 \). Since \( t \) is positive and within \([0, 5]\), evaluate the intervals \( [0, 2] \), \( [2, 3] \), and \( [3, 5] \) to see where the inequality holds.

Step 7: Test Subintervals

Choose test points from each subinterval: For \([0, 2]\), choose \( t = 1 \); for \([2, 3]\), choose \( t = 2.5 \); for \([3, 5]\), choose \( t = 4 \). Evaluate the inequality \( t^3 - 3t^2 - 4t + 12 \geq 0 \) at these points.

Step 8: Verify Solutions

Calculating each, we find: \((1) = 6 \geq 0\), \((2.5) = 3.125 \geq 0\), \((4) = 12 \geq 0\). Thus, the speed is at least 8 cm/sec across all subintervals \([0, 5]\).

Key Concepts

Critical Points

In the context of polynomial inequalities, critical points are values of the variable where the polynomial is equal to zero. These points are crucial because they help determine where the inequality changes from true to false or vice versa. For example, in the problem where the speed of a moving particle is given by the polynomial equation \(v = t^3 - 3t^2 - 4t + 12\), finding the critical points involves solving the equation \(t^3 - 3t^2 - 4t + 12 = 0\).

Finding these critical points helps to divide the real number line into different intervals, which we need to evaluate separately. When you solve this equation for \(t\), each solution is a critical point. These points are the boundaries of our intervals.

• Consider using the Rational Root Theorem to find possible rational solutions for the polynomial.
• Use polynomial factoring if a root is found, to break down the polynomial into smaller degrees easier to solve parts.

In our exercise, the critical points found are \(t = 2\), \(t = 3\), and \(t = -2\). These points assist in interval testing to evaluate when the speed is within the desired range.

Interval Testing

Interval testing follows identifying critical points and involves checking true or false for the inequality within parts of the domain. After finding critical points that make the inequality equal to zero, we have different intervals to test the inequality's validity.

For each interval, choose a test point— thensign value within that interval—and substitute it into the inequality \(t^3 - 3t^2 - 4t + 12 \geq 0\). For our example, test the subintervals: \([0, 2]\), \([2, 3]\), and \([3, 5]\).

• If the test point satisfies the inequality, then every value in that interval satisfies the inequality.
• If it does not, then the inequality fails in that interval.

By testing in this way, you can easily determine which intervals meet the condition of the polynomial being greater than or equal to zero.

Factoring Polynomials

Factoring polynomials is a powerful technique to simplify computations and solve polynomial equations. The process of factoring involves expressing the polynomial as a product of its linear factors or sometimes irreducible higher degree polynomials.

In the exercise given, after identifying that \(t = 2\) is a root, you can express the polynomial \(t^3 - 3t^2 - 4t + 12\) in a factored form \((t-2)(t^2-t-6)=0\). Then, further factoring \((t^2-t-6)\) leads to \((t-3)(t+2)\), resulting in the complete factorization \((t-2)(t-3)(t+2) = 0\).

• Start with possible rational roots, which could be factors of the constant term over factors of the leading coefficient.
• Confirm each potential root by substituting it into the polynomial to see if it results in zero.

Once factored, solving the polynomial is greatly simplified to setting each factor to zero and finding the roots. This method is not only efficient but also provides insight into the behavior of polynomial expressions.