Implicit Differentiation Equation Of The Normal To A Curve

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Introduction

In calculus, implicit differentiation is a powerful technique used to find the derivative of a function that is not explicitly defined in terms of one variable. This method is particularly useful when dealing with equations where it is difficult or impossible to isolate one variable. In this article, we will explore how to use implicit differentiation to determine the equation of the normal to the curve x² + y² + 3xy - 8x - 3y + 6 = 0 at the point (1, 1). This involves finding the derivative dy/dx, evaluating it at the given point to find the slope of the tangent, and then using the negative reciprocal of this slope to find the slope of the normal. Finally, we will use the point-slope form of a line to write the equation of the normal.

Understanding Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function where y is not explicitly defined as a function of x. In other words, we have an equation that relates x and y, but it is not in the form y = f(x). For example, the equation x² + y² = 25 represents a circle, and it is not straightforward to express y as a function of x over the entire domain. In such cases, implicit differentiation allows us to find dy/dx without having to solve for y explicitly. The key idea behind implicit differentiation is the chain rule. When we differentiate a term involving y with respect to x, we must remember that y is itself a function of x. Therefore, we differentiate the term with respect to y and then multiply by dy/dx. This process allows us to find the relationship between the rates of change of x and y.

For instance, consider the term y². When differentiating this with respect to x, we treat y as a function of x. Applying the chain rule, we get d/dx (y²) = 2y * (dy/dx). Similarly, for a term like 3xy, we need to use the product rule along with the chain rule. The derivative of 3xy with respect to x is 3(x * (dy/dx) + y * 1), which simplifies to 3x(dy/dx) + 3y. By applying these rules carefully, we can differentiate complex implicit equations and solve for dy/dx. This derivative represents the slope of the tangent line to the curve at any point (x, y) that satisfies the original equation. The power of implicit differentiation lies in its ability to handle equations that are not easily solved for one variable, making it an indispensable tool in calculus.

Step-by-Step Solution

1. Implicitly Differentiate the Equation

Our first step is to differentiate the given equation x² + y² + 3xy - 8x - 3y + 6 = 0 with respect to x. We apply the rules of differentiation, including the chain rule and the product rule where necessary. Differentiating each term, we get:

  • d/dx (x²) = 2x
  • d/dx (y²) = 2y (dy/dx)
  • d/dx (3xy) = 3(x * (dy/dx) + y * 1) = 3x (dy/dx) + 3y (using the product rule)
  • d/dx (-8x) = -8
  • d/dx (-3y) = -3 (dy/dx)
  • d/dx (6) = 0

Combining these, the derivative of the entire equation is:

2x + 2y (dy/dx) + 3x (dy/dx) + 3y - 8 - 3 (dy/dx) = 0

This equation relates the derivatives of x and y, allowing us to solve for dy/dx.

2. Solve for dy/dx

Now, we need to isolate dy/dx in the equation we obtained in the previous step. We start by grouping the terms that contain dy/dx on one side and the remaining terms on the other side:

(2y + 3x - 3) (dy/dx) = -2x - 3y + 8

Next, we divide both sides by (2y + 3x - 3) to solve for dy/dx:

dy/dx = (-2x - 3y + 8) / (2y + 3x - 3)

This expression gives us the slope of the tangent line to the curve at any point (x, y) that satisfies the original equation. It's a crucial step in finding the equation of the normal because the normal is perpendicular to the tangent.

3. Evaluate dy/dx at the Point (1, 1)

To find the slope of the tangent at the specific point (1, 1), we substitute x = 1 and y = 1 into the expression for dy/dx:

dy/dx |_(1,1) = (-2(1) - 3(1) + 8) / (2(1) + 3(1) - 3) = (-2 - 3 + 8) / (2 + 3 - 3) = 3 / 2

So, the slope of the tangent line at the point (1, 1) is 3/2. This value is essential for finding the slope of the normal, which is perpendicular to the tangent.

4. Find the Slope of the Normal

The normal to the curve at a point is perpendicular to the tangent at that point. The slopes of perpendicular lines are negative reciprocals of each other. Therefore, if the slope of the tangent is m_t, the slope of the normal m_n is given by:

m_n = -1 / m_t

In our case, the slope of the tangent at (1, 1) is 3/2, so the slope of the normal is:

m_n = -1 / (3/2) = -2/3

This is the slope of the line we are trying to find.

5. Determine the Equation of the Normal

Now that we have the slope of the normal (-2/3) and a point it passes through ((1, 1)), we can use the point-slope form of a line to find the equation of the normal. The point-slope form is:

y - y_1 = m(x - x_1)

where (x_1, y_1) is the point and m is the slope. Plugging in the values, we get:

y - 1 = (-2/3)(x - 1)

To write this in slope-intercept form (y = mx + b), we can simplify:

y - 1 = (-2/3)x + 2/3 y = (-2/3)x + 2/3 + 1 y = (-2/3)x + 5/3

Alternatively, we can write the equation in the general form Ax + By + C = 0 by multiplying through by 3 and rearranging:

3y = -2x + 5 2x + 3y - 5 = 0

Thus, the equation of the normal to the curve at the point (1, 1) is 2x + 3y - 5 = 0.

Conclusion

In summary, we have successfully determined the equation of the normal to the curve x² + y² + 3xy - 8x - 3y + 6 = 0 at the point (1, 1) using implicit differentiation. The key steps involved implicitly differentiating the equation, solving for dy/dx, evaluating dy/dx at the given point to find the slope of the tangent, finding the negative reciprocal of this slope to get the slope of the normal, and finally, using the point-slope form to write the equation of the normal. This process highlights the power and utility of implicit differentiation in solving calculus problems involving implicitly defined functions. The equation of the normal to the curve at the point (1, 1) is 2x + 3y - 5 = 0, which we found by carefully applying each step of the method. This example illustrates a common application of implicit differentiation and provides a clear methodology for tackling similar problems.