Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The set of parabolas passing through \((0,3)\) with a slope of 2 at \((0,3)\) Strategize. In the example above we used \(y-y_{1}=m\left(x-x_{1}\right)\) as the equation of our line, not \(y=m x+b .\) Similarly, when working with parabolas, different forms of the equation for a parabola are best suited to different situations. In Problems 17 through 21, nd the equation of the parabola with the speci cations given.

Step-by-Step Solution

Answer

The equation of the parabola passing through the point (0,3) with a slope of 2 at that point is \(y = x^2 + 2x + 3\).

Step 1- Assign generic form to the parabola and define given point and slope

Let us use the generic form of a parabola which is given as \(p(x)= ax^2 + bx + c\). The problem gives us that the parabola passes through the point (0,3) which gives us \(p(0)=3\). Moreover, the derivative at point x=0 is equal to 2 which gives us \(p'(0) = 2\).

Step 2- Calculate derivative of the parabola

First, we need to find the derivative of the generic parabola \(p(x)= ax^2 + bx + c\). Using the power rule, the derivative of the parabola is given by \(p'(x)= 2ax + b\).

Step 3- Substitute for the derivative at x=0

Substituting x = 0 in \(p'(x)\) equation should result into the given slope of the parabola, i.e.,2. We get \(p'(0) = 2a*0 + b = b\). Since we know \(p'(0) = 2\), we can equate this to find b=2.

Step 4- Substitute for parabola equation at x=0

Substituting x = 0 in \(p(x)\) equation should give us the given y-coordinate of the point, i.e., 3. We get \(p(0) = a*0^2 +b*0 + c = c\). Since we know \(p(0) = 3\), we can equate these to find that c=3.

Step 5- Formulate final equation of the parabola

Since we know b and c now, and the equation of the parabola is not given explicitly, we assume \(a=1\). Hence, the final equation of the parabola is given by \(y = x^2 + 2x + 3\).

Key Concepts

Understanding the Derivative of a Parabola

A parabola is a type of curve that can be defined by a quadratic function, often written in the form \( y = ax^2 + bx + c \). The derivative of a parabola is essential because it helps us understand the behavior of the curve at various points. In simple terms, the derivative tells us how the slope of the curve changes as we move along it. To find the derivative of a parabola, we apply the power rule from calculus. Given our generic parabola \( p(x) = ax^2 + bx + c \), the derivative \( p'(x) \) is calculated as \( 2ax + b \). This derivative represents the rate of change of the parabola's slope at any given point on the curve.

Exploring the Slope of a Curve

The slope of a curve is a fundamental concept that helps us describe how steep the curve is at any particular point. When we talk about the slope of a parabola, we're referring to how the curve rises or falls as we move along the x-axis. The derivative of a parabola, such as \( p'(x) = 2ax + b \), directly represents this slope at any specific point. For instance, if we have a slope of 2 at \( x = 0 \), this means that for every unit we move horizontally to the right, the curve moves up by two units. This information is critical when we want to understand the shape and direction in which the parabola is pointing at any given coordinate.

Identifying Points on a Parabola

Points on a parabola are essentially coordinates \( (x, y) \) that satisfy the parabola's equation, such as \( y = ax^2 + bx + c \). In many problems, like the one you're working on, specific points are given that need to lie on the parabola. These points are crucial because they provide constraints that can help determine the specific coefficients of the parabola equation. For example, if we know a parabola passes through the point \((0, 3)\) and has a slope of 2 at \( x = 0 \), we can use these two pieces of information to solve for any unknown coefficients and write the precise equation of the parabola. This ensures that the derived parabola truly follows the given path and gradient at the designated point.