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

The values are \(a = -2\), \(b = 7\), and \(c = 0\).

Step 1: Apply Conditions to Find Equations

We first use the condition \(g(1) = 5\). Substitute \(t=1\) into \(g(t) = a t^2 + b t + c\), which gives the equation \(a(1)^2 + b(1) + c = 5\). Simplifying, we obtain Equation 1: \(a + b + c = 5\).

Step 2: Differentiate and Apply Derivative Condition 1

Calculate \(g'(t)\) by differentiating \(g(t)\): \(g'(t) = 2at + b\). Use the condition \(g'(1) = 3\). Substitute \(t = 1\) to get \(2a(1) + b = 3\), which simplifies to Equation 2: \(2a + b = 3\).

Step 3: Differentiate Again and Apply Derivative Condition 2

Calculate \(g''(t)\) by differentiating \(g'(t)\): \(g''(t) = 2a\). Use the condition \(g''(1) = -4\). Since \(g''(t)\) is constant, substitute to obtain Equation 3: \(2a = -4\).

Step 4: Solve for \(a\)

From Equation 3, \(2a = -4\), solve for \(a\) by dividing both sides by 2: \(a = -2\).

Step 5: Solve for \(b\)

Substitute \(a = -2\) in Equation 2: \(2(-2) + b = 3\). Simplify to determine \(b = 7\).

Step 6: Solve for \(c\)

Substitute \(a = -2\) and \(b = 7\) in Equation 1: \(-2 + 7 + c = 5\). Simplify to find \(c = 0\).

Key Concepts

Differentiation

Differentiation is an essential concept in calculus. It focuses on finding the derivative of a function. This process tells us how the value of the function changes as its input changes. In simple terms, it's a way to measure how "steep" something is.
In the context of our exercise, the function given is a quadratic function, notably:

• Quadratic functions have a general form of \(ax^2 + bx + c\).
• The derivative of a quadratic function \(at^2 + bt + c\) is \(2at + b\).
• Differentiation helps in finding the slope of the tangent line at any point \(t\) on the curve.

By using the given conditions and calculating the derivative, one can find specific values for the coefficients \(a\), \(b\), and \(c\). This step involves setting up equations by applying the known conditions about the derivatives and solving them accordingly.

Second Derivative

The second derivative is the derivative of the derivative. Mathematically, if the first derivative tells us the slope of the function, then the second derivative tells us how this slope is changing.
For a quadratic function like in this exercise:

• The first derivative is \(g'(t) = 2at + b\).
• The second derivative is \(g''(t) = 2a\).

The second derivative being constant, as in our case \(g''(t) = 2a\), indicates that the rate of change of the slope is uniform. It's akin to looking at acceleration in physics, where uniform acceleration corresponds to constant speed changes. In the exercise, the condition \(g''(1) = -4\) allows us to directly solve for \(a = -2\). Understanding the second derivative helps in determining concavity and inflection points as well.

System of Equations

A system of equations involves multiple equations that need to be solved simultaneously. In this scenario, we deal with three equations stemming from the conditions given in the problem and our differentiated functions.
Solving such a system requires:

• Step-by-step substitution or elimination methods.
• From the given equations, you can sequentially isolate and solve for variables \(a\), \(b\), and \(c\).

This technique is crucial in algebra and calculus for finding unique solutions to problems that involve multiple constraints or conditions.
Understanding how to navigate through a system of equations allows for precise solutions, especially when dealing with real-world problems modeled by polynomial or algebraic expressions.