Problem 44 - Basic Technical Mathematics with Calculus

Question

$$ \text { Solve the given quadratic equations by factoring.}$$ The mass \(m\) (in \(\mathrm{Mg}\) ) of the fuel supply in the first-stage booster of a rocket is \(m=135-6 t-t^{2},\) where \(t\) is the time (in \(\mathrm{s}\) ) after launch. When does the booster run out of fuel?

Step-by-Step Solution

Answer

The booster runs out of fuel at 9 seconds.

Step 1: Understand the Problem

The given equation for mass \(m\) in terms of time \(t\) is \(m = 135 - 6t - t^2\). To find when the booster runs out of fuel, we need to determine when \(m = 0\). This requires solving the equation \(135 - 6t - t^2 = 0\).

Step 2: Rearrange the Equation

Start by rearranging the equation \(135 - 6t - t^2 = 0\) into standard quadratic form. This means writing it as \(-t^2 - 6t + 135 = 0\). Next, multiply through by \(-1\) to simplify the factoring process: \(t^2 + 6t - 135 = 0\).

Step 3: Factor the Quadratic Equation

We need to factor \(t^2 + 6t - 135\). Look for two numbers whose product is \(-135\) and sum is \(6\). The numbers \(15\) and \(-9\) work since \(15 \times (-9) = -135\) and \(15 + (-9) = 6\). Thus, the factored form is \((t + 15)(t - 9) = 0\).

Step 4: Solve for the Roots

Set each factor equal to zero: \(t + 15 = 0\) and \(t - 9 = 0\). Solving these gives \(t = -15\) and \(t = 9\).

Step 5: Interpret the Results

Since negative time does not make sense in this context, we discard \(t = -15\). Thus, the booster runs out of fuel at \(t = 9\) seconds.

Key Concepts

Factoring

Factoring is an essential skill when working with quadratic equations. It involves breaking down a quadratic expression into a product of simpler linear factors. This is useful in solving equations where setting the quadratic expression to zero allows us to find its roots.
To factor a quadratic equation like \( t^2 + 6t - 135 \), we look for two numbers that multiply to give the constant term (here, \(-135\)) and add up to the linear coefficient (here, \(6\)). In this particular case, the numbers \(15\) and \(-9\) fulfill these conditions. Thus, the quadratic can be expressed as \((t + 15)(t - 9) = 0\).
Factoring is not only crucial for solving equations but also simplifies expressions and can offer insights into the behavior of graphical functions. It breaks down complex expressions into simpler parts, making them easier to work with.

Roots of Quadratic Equation

The roots of a quadratic equation are the values of the variable that make the equation equal to zero. For the equation \(135 - 6t - t^2 = 0\), once factored into \((t + 15)(t - 9) = 0\), it becomes straightforward to find its roots.
Setting each factor equal to zero, we have:

• \( t + 15 = 0 \) which simplifies to \( t = -15 \)
• \( t - 9 = 0 \) which simplifies to \( t = 9 \)

These roots indicate where the parabola represented by the quadratic crosses the time axis on a graph. However, in practical scenarios such as this, where the negative root \( t = -15 \) is not feasible due to the nature of time, only the positive solution \( t = 9 \) is considered acceptable.
Understanding roots is vital, as they represent crucial turning points in many real-life applications, like determining when to expect zero fuel in a rocket booster.

Physics Applications

Quadratic equations frequently appear in physics, embedded in formulas and scenarios where variables change over time. For example, in this exercise, the equation models the fuel mass over time for a rocket booster.
The quadratic form \( m = 135 - 6t - t^2 \) describes how the mass diminishes as time passes. By setting this equation to zero, we can predict precisely when a key event occurs—such as the total depletion of fuel at \( t = 9 \) seconds.
Using quadratics in physics can help predict:

• Projectiles' paths, where the equation models the object's height over time.
• The behavior of accelerating vehicles, calculating distances or times based on speed changes.
• Rocket dynamics, as seen here, tracking fuel use for mission planning.

Recognizing these applications enriches our understanding of both mathematics and physics, illustrating how abstract equations can depict real-world phenomena. This makes the study of quadratic equations invaluable in comprehending and solving numerous scientific and engineering problems.