Understanding and Calculating the Curve's Parameters: A Step-by-Step Guide

In this article, we will delve into the process of finding the parameters of a cubic curve. Given the equation of a curve with the form yax^3 bx^2 c, we are provided with specific points and conditions to determine the values of a, b, and c. We will explore the concept of stationary points and gradients, providing a detailed step-by-step solution to this problem.

Understanding the Curve and its Parameters

The curve in question has the form yax^3 bx^2 c. This is a cubic function, which has a unique shape characterized by its ability to have multiple turning points. The key feature that we are interested in here is the existence of stationary points at specific x-values and the gradient at another x-value. A stationary point is a point where the gradient (derivative) of the curve is zero. The gradient at a particular point on the curve is given by the derivative of the function.

Derivative of the Curve

To solve for the parameters a, b, and c, we first need to calculate the derivative of the curve. The derivative of yax^3 bx^2 c is:

y' 3ax^2 2bx

Stationary Points and Gradients

We know that the curve has stationary points when x1 and at the origin (where x0). Stationary points occur where the gradient (derivative) of the curve is zero. Therefore, we can set up the following equations based on the conditions given:

Given that the origin is a stationary point, we have:

y'(0) 0

This condition is already satisfied by the form of the derivative, as long as b and c are appropriately chosen. Now, let's use the condition at the point x1 where the gradient is zero:

3a(1)^2 2b(1) 0

This simplifies to:

3a 2b 0 (1)

Next, we use the condition that the gradient at the point where x2 is 2:

3a(2)^2 2b(2) 2

This simplifies to:

12a 4b 2 (2)

System of Equations and Solving for Parameters

Now we have a system of two equations with two unknowns:

3a 2b 0 (1)

12a 4b 2 (2)

Let's solve these equations step-by-step:

From equation (1):

2b -3a

12b -9a

Substitute 12b -9a into equation (2):

12a (-9a) 2

3a 2

a 2/3

Substitute a 2/3 back into equation (1):

3(2/3) 2b 0

2 2b 0

b -1

Now, we need to find the value of c. Since the curve has a stationary point at the origin, we know that y(0) 0. However, in this specific problem, we have the condition that the curve passes through the origin, which means c 0 will satisfy this condition.

Final Solution

Thus, we have determined the values of the parameters:

a 2/3, b -1, c 0

So the equation of the curve is:

y (2/3)x^3 - x^2

Conclusion

In summary, we have walked through the process of finding the parameters of a cubic curve given specific conditions. This problem involved understanding the definition of stationary points and gradients, setting up a system of equations based on given conditions, and solving the system to find the values of the curve's parameters.