Calculating the Slope of a Curve: A Comprehensive Guide
Understanding the slope of a curve is a fundamental concept in calculus. This guide will explore how to find the slope of a specific curve and discuss the importance of ensuring that a given point lies on the curve before attempting to calculate the slope.
Introduction to the Curve
Consider the equation of a curve: y3 - xy - 3x 1. This equation represents a relationship between the variables x and y. Our goal is to find the slope of the curve at a specific point. Before calculating the slope, it is essential to confirm whether the point of interest lies on the curve.
Validation of the Point
Suppose we want to find the slope at the point (0, -1). To verify this, substitute the point into the equation:
```plaintext -1^3 - 0 - 3(0) 1 -1 1 ```The result is -1 1, which is mathematically incorrect. Therefore, the point (0, -1) does not lie on the curve. Calculating the slope at a point that doesn't satisfy the curve equation is meaningless. Thus, there is no tangent at (0, -1).
Derivation of the Slope Formula
If the goal is to find the general derivative (slope) of the curve, we can proceed as follows:
Take the derivative of both sides with respect to x:
```plaintext 3y^2 frac{dy}{dx} - yx - 3 0 ```After manipulation, this equation simplifies to:
```plaintext 3y^2 - x frac{dy}{dx} y 3 ```Solving for (frac{dy}{dx}):
```plaintext frac{dy}{dx} frac{y 3}{3y^2 - x} ```This formula represents the slope of the curve at any point (x, y) where the point satisfies the original equation of the curve.
Identifying Points on the Curve
Let us check if the point (0, -1) satisfies the curve equation:
```plaintext (-1)^3 - 0 - 3(0) 1 -1 ≠ 1 ```As calculated earlier, the point (0, -1) does not satisfy the equation, confirming that we cannot use this point to calculate the slope.
Next, we will find the points that lie on the curve:
Substitute (x 0) into the equation:
```plaintext y^3 - 0 - 3(0) 1 y^3 1 y 1 ```Therefore, the point (0, 1) lies on the curve.
Substitute (y -1) into the equation:
```plaintext (-1)^3 - x - 3x 1 -1 - 4x 1 4x -2 x -frac{1}{2} ```Therefore, the point ((- frac{1}{2}, -1)) also lies on the curve.
For the y-intercept, substitute (x 0) into the equation:
```plaintext y^3 - 0 - 3(0) 1 y^3 1 y 1 ```However, we need the point where the curve intersects the y-axis, which is when (x 0):
```plaintext y^3 - 0 - 3(0) 1 y^3 1 y 1 ```Thus, the point ((0, 1)) is also the y-intercept.
If we calculate the slope at ((0, 1)):
```plaintext frac{dy}{dx} frac{-1 times 3}{3(-1)^2 - 0} frac{dy}{dx} frac{3}{3} frac{dy}{dx} 1 ```The slope at ((0, 1)) is 1, and the equation of the tangent line is:
```plaintext y 1x 1 y x 1 ```At the point ((- frac{1}{2}, -1)), the slope is:
```plaintext frac{dy}{dx} frac{-frac{1}{2} times 3}{3(-frac{1}{2})^2 - (-1)} frac{dy}{dx} frac{-frac{3}{2}}{3 times frac{1}{4} 1} frac{dy}{dx} frac{-frac{3}{2}}{frac{3}{4} 1} frac{dy}{dx} frac{-frac{3}{2}}{frac{7}{4}} frac{dy}{dx} frac{-frac{3}{2}}{frac{7}{4}} times frac{4}{4} frac{dy}{dx} frac{-6}{7} ```The slope at ((- frac{1}{2}, -1)) is (-frac{6}{7}), and the equation of the tangent line is:
```plaintext y -frac{6}{7}x - 1 ```Finally, at the y-intercept ((0, 1)), the slope is 1, and the tangent line is:
```plaintext y 1x 1 y x 1 ```These calculations show the importance of verifying that a point lies on the curve before attempting to find the slope or tangent line.
Conclusion
Understanding the procedure for calculating the slope of a curve is crucial in calculus, but it is equally important to ensure that the points of interest satisfy the curve's equation. This guide has demonstrated the steps and importance of such validation in arriving at accurate results.