How to Find the Equations of the Tangent and Normal Lines to the Curve (y^2 x^3 1) at the Point ((1, 1))

The process of determining the equations of the tangent and normal lines to a curve at a specific point involves several key steps, especially when dealing with implicitly defined curves. This article will guide you through the process using the curve (y^2 x^3 1) at the point ((1, 1)).

Step-by-Step Guide

Step 1: Find the Derivative Using Implicit Differentiation

Given the curve (y^2 x^3 1), we start by implicitly differentiating both sides of the equation with respect to (x).

Let's begin with the implicit differentiation:

[frac{d}{dx}(y^2) frac{d}{dx}(x^3 1)]

Applying the chain rule to the left-hand side:

[2y frac{dy}{dx} 3x^2]

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

[frac{dy}{dx} frac{3x^2}{2y}]

Substituting the point ((1, 1)) into the derivative, we get:

[frac{dy}{dx} frac{3(1)^2}{2(1)} frac{3}{2}]

Step 2: Write the Equation of the Tangent Line

The slope of the tangent line at the point ((1, 1)) is (frac{3}{2}). We use the point-slope form of the line equation: [y - y_1 m(x - x_1)].

Substituting the slope and point ((1, 1)):

[y - 1 frac{3}{2}(x - 1)]

Simplifying:

[y - 1 frac{3}{2}x - frac{3}{2} 1]

[y frac{3}{2}x - frac{1}{2}]

Step 3: Find the Equation of the Normal Line

The normal line is perpendicular to the tangent line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:

[m_{text{normal}} -frac{1}{frac{3}{2}} -frac{2}{3}]

Using the point-slope form again for the normal line:

[y - 1 -frac{2}{3}(x - 1)]

Simplifying:

[y - 1 -frac{2}{3}x frac{2}{3}]

[y -frac{2}{3}x frac{5}{3}]

Summary of Results

Tangent Line Equation

The equation of the tangent line at the point ((1, 1)) is:

[y frac{3}{2}x - frac{1}{2}]

Normal Line Equation

The equation of the normal line at the point ((1, 1)) is:

[y -frac{2}{3}x frac{5}{3}]

Alternative Approach Using Shift to Origin

If we shift the coordinate system to the origin by letting (x u 1) and (y v 1), the given curve (y^2 x^3 1) becomes:

[(v 1)^2 (u 1)^3 1]

Expanding to the first degree terms:

[v^2 u^3 3u^2 3u 1] [v^2 u^3 3u^2 3u]

Taking the first-degree terms, we get:

[2v 3u]

Rewriting the equation in terms of (x) and (y):

[2y - 2 3x - 3]

Simplifying:

[2y - 3x 1 0]

This is the equation of the tangent line.

For the normal line:

[3v -2u]

Rewriting in terms of (x) and (y):

[3y - 2x 5 0]

This is the equation of the normal line.

Conclusion

This article has provided a clear and detailed process for finding the equations of the tangent and normal lines to the curve (y^2 x^3 1) at the point ((1, 1)). The steps involve implicit differentiation, using the point-slope form of the line equation, and understanding the concept of perpendicularity to find the normal line. By following these steps, you can solve similar problems for other curves and points.