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Chapter 3: Problem 9

Evaluate \(\gamma\) if a. \(s(t)=5 t^{2}+\gamma \quad\) and \(\quad s(0)=15\) b. \(s(t)=-8 t^{2}+12 t+\gamma\) and \(\quad s(0)=11\) c. \(s(t)=-8 t^{2}+12 t+\gamma \quad\) and \(\quad s(1)=11\) d. \(s(t)=-\frac{849}{2} t^{2}+126 t+\gamma\) and \(s(0.232)=245.9\)

Short Answer

Expert verified
a. \(\gamma = 15\), b. \(\gamma = 11\), c. \(\gamma = 7\), d. \(\gamma \approx 239.52\)

Step by step solution

01

Solve for \( \gamma \) in part (a)

For the first equation, we substitute \( t = 0 \) into the equation \( s(t)=5t^2+\gamma \) and use the given condition \( s(0)=15 \).- Substitute \( t = 0 \): \[ s(0) = 5(0)^2 + \gamma = 15 \]- This simplifies to: \[ \gamma = 15 \]
02

Solve for \( \gamma \) in part (b)

For the second equation, substitute \( t = 0 \) into \( s(t) = -8t^2 + 12t + \gamma \) with the given \( s(0) = 11 \).- Substitute \( t = 0 \): \[ s(0) = -8(0)^2 + 12(0) + \gamma = 11 \]- This simplifies to: \[ \gamma = 11 \]
03

Solve for \( \gamma \) in part (c)

For the third equation, substitute \( t = 1 \) into \( s(t) = -8t^2 + 12t + \gamma \) with the given \( s(1) = 11 \).- Substitute \( t = 1 \): \[ s(1) = -8(1)^2 + 12(1) + \gamma = 11 \]- Simplify the equation: \[ -8 + 12 + \gamma = 11 \] \[ 4 + \gamma = 11 \]- Therefore, \( \gamma \) is: \[ \gamma = 11 - 4 = 7 \]
04

Solve for \( \gamma \) in part (d)

For the fourth equation, substitute \( t = 0.232 \) into \( s(t) = -\frac{849}{2}t^2 + 126t + \gamma \) with \( s(0.232) = 245.9 \).- Substitute \( t = 0.232 \) into the equation: \[ s(0.232) = -\frac{849}{2}(0.232)^2 + 126(0.232) + \gamma = 245.9 \]- Calculate \( -\frac{849}{2}(0.232)^2 \): \[ -\frac{849}{2} \times 0.053824 = -22.852824 \]- Calculate \( 126 \times 0.232 \): \[ 29.232 \]- Substitute the calculated values in: \[ -22.852824 + 29.232 + \gamma = 245.9 \]- Simplify: \[ 6.379176 + \gamma = 245.9 \]- Solve for \( \gamma \): \[ \gamma = 245.9 - 6.379176 = 239.520824 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Evaluating Constants
When working with algebraic expressions like those in the exercise above, evaluating constants often involves finding a missing number in an equation. Here, the constant is represented by \( \gamma \), and to find its value, we substitute the given condition into the function.
For example, in part (a), the equation is \( s(t)=5t^2+\gamma \) with the condition \( s(0)=15 \). By substituting \( t=0 \) into the function, we find \( \gamma \) by solving the simplified equation \( \gamma = 15 \).
In each part, the approach is similar:
  • Identify the function and the given condition.
  • Substitute the value of \( t \) provided in the condition into the function.
  • Simplify and solve for \( \gamma \).
Evaluating constants is essential for understanding the role of fixed values in algebraic expressions. It allows us to tailor functions to specific conditions and solve various mathematical problems.
Solving Equations
Solving equations is a critical skill in algebra that involves finding the variable that makes an equation true. In the given exercise, solving equations entails substituting the value of \( t \) into the function to find the unknown constant \( \gamma \).
Let's look at solving for \( \gamma \) in part (c):
  • Start with the function \( s(t)=-8t^2+12t+\gamma \) and the condition \( s(1)=11 \).
  • Substitute \( t=1 \): \( s(1) = -8(1)^2 + 12(1) + \gamma = 11 \).
  • Simplify: \( -8 + 12 + \gamma = 11 \), which gives \( 4 + \gamma = 11 \).
  • Isolate \( \gamma \): \( \gamma = 11 - 4 = 7 \).
Each step is about performing operations that simplify the equation until the unknown becomes isolated and calculable. Understanding each step aids in demystifying the process and equips you for more complex equations.
Function Values
Function values are the outputs you get when a specific input is substituted into a function. In this exercise, determining the value of \( \gamma \) relies heavily on interpreting function values.
Take part (d): You have \( s(t) = -\frac{849}{2}t^2 + 126t + \gamma \) with \( s(0.232)=245.9 \). The goal is to find the corresponding function value when \( t = 0.232 \).
This involves several steps:
  • Substitute \( t=0.232 \) into the equation.
  • Calculate: \(-\frac{849}{2}(0.232)^2 = -22.852824 \) and \( 126 \times 0.232 = 29.232 \).
  • Combine these with \( \gamma \) into the equation so that \( -22.852824 + 29.232 + \gamma = 245.9 \).
  • Finally, solve: \( \gamma = 245.9 - 6.379176 \), yielding \( \gamma = 239.520824 \).
Being comfortable with the calculation of function values lets you determine necessary constants and adjust equations to meet specific outcomes, reinforcing your algebraic fluency.

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Most popular questions from this chapter

If \(b\) approaches 3 a. \(b\) approaches _________. b. \(2 \div b\) approaches _________. c. \(\pi\) approaches _________. d. \(\frac{2}{\sqrt{b}+\sqrt{3}}\) approaches _________. e. \(b^{3}+b^{2}+b\) approaches _________. f. \(\frac{b}{1+b}\) approaches _________. g. \(2^{b}\) approaches _________. h. \(\log _{3} b\) approaches _________.

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