Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(g(t)=a t^{2}+b t+c\) and \(g(1)=5\), \(g^{\prime}(1)=3\), and \(g^{\prime \prime}(1)=-4 .\) Find \(a, b\), and \(c .\)

Step-by-Step Solution

Answer

\( a = -2, b = 7, c = 0 \)

Step 1: Set up the function equation

The function is given by \( g(t) = a t^2 + b t + c \). We know that \( g(1) = 5 \). Substitute \( t = 1 \) into the function equation:\[ a(1)^2 + b(1) + c = 5 \]Simplify to obtain:\[ a + b + c = 5 \] This is our first equation.

Step 2: Differentiate g(t) and use g'(1)

Differentiate \( g(t) \) with respect to \( t \) to find \( g'(t) \):\[ g'(t) = 2at + b \]We are given \( g'(1) = 3 \). Substitute \( t = 1 \) into the derivative:\[ 2a(1) + b = 3 \]Simplify to get:\[ 2a + b = 3 \]This is our second equation.

Step 3: Differentiate g'(t) to find g''(t)

Find the second derivative \( g''(t) \):\[ g''(t) = 2a \]We are given that \( g''(1) = -4 \), therefore:\[ 2a = -4 \]Solve this equation to find \( a \):\[ a = -2 \]

Step 4: Substitute a back into the previous equations

Substitute \( a = -2 \) into the second equation from Step 2:\[ 2(-2) + b = 3 \]\[ -4 + b = 3 \]Solve for \( b \):\[ b = 7 \]

Step 5: Solve for c using all values

Substitute \( a = -2 \) and \( b = 7 \) into the first equation from Step 1:\[ -2 + 7 + c = 5 \]\[ 5 + c = 5 \]Simplify to find \( c \):\[ c = 0 \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function is changing at any given point. When you differentiate a function, you are essentially computing its derivative. In the context of quadratic functions, such as \(g(t) = a t^2 + b t + c\), the process involves calculating the slope of the tangent line at any point along the curve of the function. This helps in understanding the behavior of the function and finding critical points, like minima or maxima.

In the exercise, differentiation is performed to find the first derivative \(g'(t) = 2at + b\). This derivative tells us how the function \(g(t)\) changes with small changes in \(t\), allowing us to find points where the function increases or decreases. By setting \(g'(1) = 3\), the exercise gives us the condition to solve for unknown coefficients in the quadratic equation.

This calculated derivative shows that the rate of change of \(g(t)\) is influenced by both \(a\) and \(b\), given the linear relationship in \(g'(t)\). Differentiating functions like this is crucial for solving problems related to motion, optimization, and understanding functional changes.

System of Equations

A system of equations is a set of equations with multiple variables that you solve simultaneously. When solving a system, the solutions will satisfy all equations in the system at once. Such systems frequently arise in real-world applications where multiple conditions must be met.

In the exercise, we have three conditions:

• \(a + b + c = 5\)
• \(2a + b = 3\)
• \(2a = -4\)

These form a system of equations involving the variables \(a\), \(b\), and \(c\). Solving this system involves substituting known values back into the equations, often using the results from differentiation.

First, solve the simplest equation for one variable. For example, from \(2a = -4\), we get \(a = -2\). Then use this value to solve for other variables, such as \(b\) and \(c\), by fitting them back into the remaining equations. This systematic approach ensures that all the conditions are met, providing a complete solution to the problem.

Second Derivative

The second derivative of a function is the derivative of the function's first derivative. It provides information about the curvature or the concavity of the original function. The second derivative tells us how the rate of change of a function is changing.

In the exercise, the second derivative is calculated as \(g''(t) = 2a\). For quadratic functions, the second derivative is a constant. If the second derivative is positive, the function is "concave up," indicating a local minimum. If it's negative, as in \(g''(1) = -4\), the function is "concave down," indicating a local maximum.

This understanding of concavity is crucial for analyzing the behavior of the function over intervals and determining points of inflection. Knowing \(g''(t)\), particularly when combined with \(g'(t)\), provides a fuller picture of how the function behaves, assisting in problems that involve optimization or predicting movement directions.