Finding \( \alpha - \beta \) using Tangent to a Curve

The problem provides the equation of a curve \( y^2 = \alpha + \beta x^3 \) and the equation of a line \( 3x + y = 8 \) which is tangent to the curve at the specific point (2, 2). We need to find the value of \( \alpha - \beta \).

Key Concepts for Tangency

When a line is tangent to a curve at a point, two conditions must be met:

• The point of tangency lies on both the curve and the tangent line.
• The slope of the tangent line at the point of tangency is equal to the slope of the curve (the value of the derivative \( \frac{dy}{dx} \)) at that point.

Step 1: Use the Point of Tangency (2, 2) on the Curve

Since the point (2, 2) lies on the curve \( y^2 = \alpha + \beta x^3 \), we can substitute \( x=2 \) and \( y=2 \) into the curve's equation:

\( (2)^2 = \alpha + \beta (2)^3 \)

\( 4 = \alpha + 8\beta \)

Let's call this Equation (1): \( \alpha + 8\beta = 4 \)

Step 2: Find the Slope of the Tangent Line

The equation of the tangent line is given as \( 3x + y = 8 \). To find its slope, we can rearrange the equation into the standard slope-intercept form \( y = mx + c \), where \( m \) is the slope.

\( y = -3x + 8 \)

The slope of the tangent line is \( m = -3 \).

Step 3: Find the Derivative of the Curve

To find the slope of the curve at any point \( (x, y) \), we need to find the derivative \( \frac{dy}{dx} \). We will differentiate the curve's equation \( y^2 = \alpha + \beta x^3 \) implicitly with respect to \( x \).

\( \frac{d}{dx}(y^2) = \frac{d}{dx}(\alpha + \beta x^3) \)

Using the chain rule on the left side and the power rule on the right side:

\( 2y \frac{dy}{dx} = 0 + 3\beta x^2 \)

Now, solve for \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{3\beta x^2}{2y} \)

Step 4: Equate the Derivative at (2, 2) to the Tangent Slope

The value of the derivative \( \frac{dy}{dx} \) at the point of tangency (2, 2) must equal the slope of the tangent line, which is -3. Substitute \( x=2 \) and \( y=2 \) into the derivative expression:

\( \left. \frac{dy}{dx} \right|_{(2,2)} = \frac{3\beta (2)^2}{2(2)} = \frac{3\beta \cdot 4}{4} = 3\beta \)

Now, set this equal to the tangent slope:

\( 3\beta = -3 \)

Solving for \( \beta \):

\( \beta = \frac{-3}{3} \)

\( \beta = -1 \)

Step 5: Solve for \( \alpha \) and \( \beta \)

We have two pieces of information:

Equation (1): \( \alpha + 8\beta = 4 \)

From Step 4: \( \beta = -1 \)

Substitute the value of \( \beta \) into Equation (1):

\( \alpha + 8(-1) = 4 \)

\( \alpha - 8 = 4 \)

Solving for \( \alpha \):

\( \alpha = 4 + 8 \)

\( \alpha = 12 \)

So, we have found the values \( \alpha = 12 \) and \( \beta = -1 \).

Step 6: Calculate \( \alpha - \beta \)

Finally, we need to find the value of \( \alpha - \beta \):

\( \alpha - \beta = 12 - (-1) \)

\( \alpha - \beta = 12 + 1 \)

\( \alpha - \beta = 13 \)

The value of \( \alpha - \beta \) is 13.

Revision Table: Tangent and Derivatives

Additional Information: Implicit Differentiation Steps

Implicit differentiation is crucial when dealing with equations where \( y \) is not simply \( f(x) \), like \( y^2 = \alpha + \beta x^3 \). Here are the general steps:

1. Differentiate both sides of the equation with respect to \( x \).
2. Remember to apply the chain rule when differentiating terms involving \( y \). For example, \( \frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx} \).
3. Treat constants (like \( \alpha \) and \( \beta \)) as usual; their derivative with respect to \( x \) is 0 if they are just constants.
4. After differentiating, collect all terms containing \( \frac{dy}{dx} \) on one side of the equation.
5. Factor out \( \frac{dy}{dx} \).
6. Solve for \( \frac{dy}{dx} \) by dividing by the factor collected in the previous step.

In this problem, we differentiated \( y^2 = \alpha + \beta x^3 \). Differentiating \( y^2 \) with respect to \( x \) gives \( 2y \frac{dy}{dx} \). Differentiating \( \alpha \) (a constant) gives 0. Differentiating \( \beta x^3 \) with respect to \( x \) gives \( 3\beta x^2 \). This led to \( 2y \frac{dy}{dx} = 3\beta x^2 \), from which we found \( \frac{dy}{dx} = \frac{3\beta x^2}{2y} \).