Math Problem Statement

Consider the curve given by   3 y f x x x    3

a. Find the equation of the line tangent to the curve at the point (1,3) b. Find the line normal to the curve at the point (1,3)

Solution

We are given the curve y=f(x)=x3x+3y = f(x) = x^3 - x + 3 and need to:

(a) Find the equation of the tangent line at the point (1,3)(1, 3)

To find the equation of the tangent line at a given point, we need the slope of the tangent, which is the derivative of f(x)f(x) evaluated at the point x=1x = 1.

  1. Find the derivative f(x)f'(x): f(x)=ddx(x3x+3)f'(x) = \frac{d}{dx}(x^3 - x + 3) Using basic differentiation rules: f(x)=3x21f'(x) = 3x^2 - 1

  2. Evaluate the derivative at x=1x = 1 to find the slope of the tangent line: f(1)=3(1)21=31=2f'(1) = 3(1)^2 - 1 = 3 - 1 = 2 So, the slope of the tangent line at x=1x = 1 is 22.

  3. Use the point-slope form of the line equation: The point-slope form of the equation of a line is: yy1=m(xx1)y - y_1 = m(x - x_1) where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line. Here, m=2m = 2 and the point is (1,3)(1, 3), so: y3=2(x1)y - 3 = 2(x - 1)

    Simplifying this: y3=2x2y - 3 = 2x - 2 y=2x+1y = 2x + 1 Therefore, the equation of the tangent line is y=2x+1y = 2x + 1.


(b) Find the equation of the normal line at the point (1,3)(1, 3)

The normal line is perpendicular to the tangent line. If the slope of the tangent line is mtm_t, then the slope of the normal line, mnm_n, is the negative reciprocal: mn=1mtm_n = -\frac{1}{m_t}

Since the slope of the tangent line is 2, the slope of the normal line is: mn=12m_n = -\frac{1}{2}

Now, we use the point-slope form of the line equation again, with slope mn=12m_n = -\frac{1}{2} and the point (1,3)(1, 3): y3=12(x1)y - 3 = -\frac{1}{2}(x - 1)

Simplifying this: y3=12x+12y - 3 = -\frac{1}{2}x + \frac{1}{2} y=12x+72y = -\frac{1}{2}x + \frac{7}{2} Therefore, the equation of the normal line is y=12x+72y = -\frac{1}{2}x + \frac{7}{2}.


Summary of Results:

  • The tangent line at (1,3)(1, 3) is y=2x+1y = 2x + 1.
  • The normal line at (1,3)(1, 3) is y=12x+72y = -\frac{1}{2}x + \frac{7}{2}.

Would you like further details or clarification on any of the steps? Here's a list of related questions for you to explore:

  1. How do you find the slope of the tangent to a curve?
  2. What is the geometric meaning of the normal line?
  3. How do you determine the derivative of more complex polynomial functions?
  4. How can you apply these concepts in real-world problems, like velocity or optimization?
  5. What are other forms of the equation of a line besides point-slope form?

Tip: Always double-check your differentiation and arithmetic when working with derivatives, as small errors in these steps can lead to incorrect results in both tangent and normal line equations!

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Tangent and Normal Lines

Formulas

Derivative of a polynomial function: f'(x) = 3x^2 - 1
Point-slope form of a line: y - y_1 = m(x - x_1)
Slope of normal line: m_n = -1/m_t

Theorems

Fundamental Theorem of Calculus (derivatives)
Perpendicular lines have negative reciprocal slopes

Suitable Grade Level

Grades 11-12