Problem 76 - AP Calculus with 8 Practice Tests 15th

Question

The equation of the curve whose slope at point \((x y)\) is \(x^{2}-2\) and which contains the point (1,-3) is (A) \(y=\frac{1}{3} x^{3}-2 x\) (B) \(y=2 x-1\) (C) \(y=\frac{1}{3} x^{3}-\frac{10}{3}\) (D) \(y=\frac{1}{3} x^{3}-2 x-\frac{4}{3}\)

Step-by-Step Solution

Answer

The equation is (D) \(y=\frac{1}{3} x^{3}-2 x-\frac{4}{3}\).

Step 1: Understand the Problem

We are given the slope of a curve at any point \( (x, y) \) as \( x^2 - 2 \) and need to find the equation of the curve that passes through the point \( (1, -3) \).

Step 2: Set Up the Differential Equation

The slope of the curve is given by the derivative \( \frac{dy}{dx} = x^2 - 2 \). To find the curve, we need to integrate this expression.

Step 3: Integrate to Find the General Solution

Integrate \( \frac{dy}{dx} = x^2 - 2 \) with respect to \( x \). The integral of \( x^2 \) is \( \frac{1}{3}x^3 \) and the integral of \( -2 \) is \( -2x \), so we get the general solution: \[ y = \frac{1}{3}x^3 - 2x + C. \]

Step 4: Solve for the Constant C

Use the point \( (1, -3) \) to solve for \( C \). Substitute \( x = 1 \) and \( y = -3 \) into the equation: \[ -3 = \frac{1}{3}(1)^3 - 2(1) + C. \] Simplifying, this becomes \( -3 = \frac{1}{3} - 2 + C \).

Step 5: Simplify and Solve for C

Combine the numbers: \( \frac{1}{3} - 2 = -\frac{5}{3} \). This gives the equation: \[ -3 = -\frac{5}{3} + C. \] Solve for \( C \): \[ C = -3 + \frac{5}{3} = -\frac{9}{3} + \frac{5}{3} = -\frac{4}{3}. \]

Step 6: Write the Specific Solution

Substitute the value of \( C \) back into the general solution: \[ y = \frac{1}{3} x^3 - 2x - \frac{4}{3}. \] This matches option (D).

Key Concepts

Differential Equations

Differential equations are equations that relate a function with its derivatives. In this context, the derivative represents the slope, or rate of change, of the function. A differential equation can describe how a curve behaves by specifying the curve's slope at various points.
For example, in our exercise, the differential equation is given by \( \frac{dy}{dx} = x^2 - 2 \). This tells us that at any point \((x, y)\), the slope of the curve is \(x^2 - 2\). This is an ordinary differential equation (ODE) because it involves functions of only one variable, \(x\), and its derivatives.
To find the equation of the curve, we solve the differential equation. Solving typically involves integrating the differential equation to find a general solution, followed by applying any given initial conditions to find a specific solution.

Integration

Integration is essentially the reverse process of differentiation. It involves finding a function given its derivative. In calculus, integration is used to solve differential equations by finding antiderivatives.
In our scenario, we are integrating the expression \( \frac{dy}{dx} = x^2 - 2 \) with respect to \(x\). The integral of \(x^2\) is \(\frac{1}{3}x^3\), and the integral of \(-2\) is \(-2x\).
This results in a general solution:

• \( y = \frac{1}{3}x^3 - 2x + C \)

The constant \(C\) represents an unknown that arises because indefinite integration can result in multiple functions with the same derivative.
Finding the value of \(C\) specifically requires additional information, typically provided by initial conditions.

Curve Slope

The slope of a curve at a particular point indicates how steep the curve is at that point. It's essentially the derivative of the curve's function with respect to \(x\).
The given slope of the curve in the problem is \(x^2 - 2\). This means at each point \(x\), the slope of the curve is determined by plugging the \(x\)-value into the expression \(x^2 - 2\). For example, at \(x = 1\), the slope of the curve would be \(1^2 - 2 = -1\).
Understanding the slope helps us visualize or predict how the curve behaves: whether it ascends, descends, or is flat.
This knowledge is foundational for sketching the curve and calculating areas under the curve when using integrals.

Initial Conditions

Initial conditions are specific values that provide necessary information to determine the particular solution of a differential equation. They help to find the exact value of the constant \(C\) in the general solution.
In this exercise, the initial condition is the point \((1, -3)\) on the curve. By substituting the \(x\)-value of 1 and the \(y\)-value of -3 into the general solution, we can solve for \(C\).
This process goes as follows:

• Substitute \(x = 1\) and \(y = -3\)
• Simplify the resulting equation to find \(C\)
• Use this \(C\) value to write the specific solution

In our solution, solving the equation \(-3 = \frac{1}{3}(1)^3 - 2(1) + C\) gives \(C = -\frac{4}{3}\).
Using this, we obtain the specific equation of the curve: \( y = \frac{1}{3} x^3 - 2x - \frac{4}{3} \). These conditions ensure the curve passes through the given point and accurately reflects the curve's behavior as defined by the differential equation.