Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the \(x y\)-equation of the curve through \((1,2)\) whose slope at any point is three times the square of its \(y\)-coordinate.

Step-by-Step Solution

Answer

The equation of the curve is \( y = \frac{1}{\left( \frac{7}{2} - 3x \right)} \).

Step 1: Understand the Problem

We are given that the slope of the curve at any point is three times the square of its y-coordinate. This means that the derivative of y with respect to x, \( \frac{dy}{dx} \), is equal to \( 3y^2 \). We also know that the curve passes through the point (1, 2).

Step 2: Set Up the Differential Equation

The problem can be translated into a differential equation: \( \frac{dy}{dx} = 3y^2 \). This means that the rate of change of y with respect to x is a function of y.

Step 3: Solve the Differential Equation

To solve \( \frac{dy}{dx} = 3y^2 \), we separate variables by dividing both sides by \( y^2 \) and multiplying by \( dx \):\[ \frac{1}{y^2} dy = 3 dx \]Integrate both sides:\[ \int \frac{1}{y^2} dy = \int 3 dx \]which gives\[ -\frac{1}{y} = 3x + C \]

Step 4: Solve for C Using the Initial Condition

We use the point \((1, 2)\) to determine C. Substitute \( x = 1 \) and \( y = 2 \) into \(-\frac{1}{y} = 3x + C\):\[ -\frac{1}{2} = 3(1) + C \]Solving for C, we get:\[ C = -\frac{1}{2} - 3 = -\frac{7}{2} \]

Step 5: Formulate the General Solution

Substitute the value of C back into the equation:\[ -\frac{1}{y} = 3x - \frac{7}{2} \]To express y in terms of x, solve for y:\[ \frac{1}{y} = \frac{7}{2} - 3x \]\[ y = \frac{1}{\left( \frac{7}{2} - 3x \right)} \]

Step 6: Finalize the Solution

Thus, the \(xy\)-equation of the curve is:\[ y = \frac{1}{\left( \frac{7}{2} - 3x \right)} \]

Key Concepts

Initial Conditions

Initial conditions are crucial when working with differential equations as they allow us to determine a particular solution from the family of solutions. Imagine you have a whole forest of possible paths that satisfy a differential equation. The initial condition helps you pick exactly one path, or solution, among all these possibilities.

In our original exercise, the curve we are looking for passes through the point \(1, 2\). This is our initial condition and it's essential for pinpointing the exact curve we're trying to describe. Without this information, there would be countless curves that could potentially satisfy the differential equation \(\frac{dy}{dx} = 3y^2\).

• Initial conditions specify a particular solution among an infinite number of them.
• They are provided as specific coordinates or sometimes as values of the function and its derivatives.

By incorporating this specific point, we solve for the constant of integration, helping us formulate the function describing the precise path our curve takes.

Separable Equations

Separable equations are a type of differential equation that can be transformed into a form where the variables can be divided onto separate sides of the equation, making them easier to solve. The goal is to rewrite the equation so that each side of the equation contains only one variable.

In our exercise, we started with the differential equation \(\frac{dy}{dx} = 3y^2\). We can see that the rate of change of \(y\) depends on \(y\) itself. This makes the equation separable, allowing us to isolate \(y\) and \(x\) on different sides:

• Move all \(y\)-related terms to one side and \(x\)-related terms to the other: \(\frac{1}{y^2} dy = 3 dx\).
• This rearrangement facilitates integration, as each integral involves only one variable.

Once separated, the equation resolves into two simpler integrals that can be dealt with independently, paving the way to a solution.

Integration

Integration is the process of finding the function whose derivative is the given function. It is a fundamental concept in calculus that essentially reverses differentiation.

In solving the differential equation \(\frac{dy}{dx} = 3y^2\), we use integration after separating the variables to find the antiderivatives of both sides of the equation:

• The left integral, \(\int \frac{1}{y^2} dy\), simplifies to \(-\frac{1}{y}\).
• The right integral, \(\int 3 dx\), results in \(3x\) and adds a constant \(C\) due to the indefinite nature of integration.

Thus, integration helps us transition from the rate of change back to the original relationship between \(x\) and \(y\), leading to a general solution.

Calculus

Calculus is the branch of mathematics that analyzes change and motion using derivatives and integrals. It's the backbone of solving differential equations like the one in our exercise: \(\frac{dy}{dx} = 3y^2\). Calculus allows us to model how something changes—its derivative—and to solve for the original function from that rate of change using integration.

Derivatives give us insights into instantaneous rate changes and tangents. For example, in this exercise, the derivative provided the slope of a curve at any given point. Using calculus, we managed to separate variables, integrate them, and apply initial conditions to uncover the specific curve function.

• It provides powerful techniques for evaluating limits and handling continuous change.
• Calculus is widely applied in physics, engineering, economics, statistics, and beyond.

It empowers us to solve complex problems involving varying conditions and find concrete solutions like the \(xy\)-equation of the curve in our exercise.