Math Problem Statement

Let P be a point on the graph of y = x^3. The tangent line at P will intersect the graph at the point Q. (a) If P has x-coordinate a, then what is the equation of the tangent line at P? (b) The line in (a) will intersect the curve y = x^3 in two places; one of these is P and the other point is Q. (c) What is the ratio of the slope of the curve at Q to the slope at P?

Solution

Let's analyze the problem in the image you've uploaded step by step:

Problem Overview:

  • We are given a point PP on the curve y=x3y = x^3.
  • The tangent line at point PP intersects the curve again at another point QQ.
  • We're tasked with finding certain information about this intersection.

Part (a): Equation of the Tangent Line at PP

Given that PP has an xx-coordinate of aa, we want to find the equation of the tangent line at PP.

  • The function is y=x3y = x^3.
  • The slope of the tangent line at any point is given by the derivative of yy with respect to xx, so: dydx=3x2\frac{dy}{dx} = 3x^2 At PP, where x=ax = a, the slope mm is: m=3a2m = 3a^2 The equation of a tangent line is generally given by: yy1=m(xx1)y - y_1 = m(x - x_1) At point P=(a,a3)P = (a, a^3), we plug in the values to get: ya3=3a2(xa)y - a^3 = 3a^2(x - a) Simplifying: y=3a2x2a3y = 3a^2x - 2a^3 So, the equation of the tangent line is y=3a2x2a3y = 3a^2x - 2a^3.

Part (b): Intersection of the Tangent Line with the Curve

We are told that the tangent line intersects the curve y=x3y = x^3 at two points: one is PP, and we need to find the coordinates of the other point QQ.

  • Set the equation of the tangent line equal to the curve's equation: 3a2x2a3=x33a^2x - 2a^3 = x^3 Rearrange this into a cubic equation: x33a2x+2a3=0x^3 - 3a^2x + 2a^3 = 0 We know x=ax = a is a solution (since PP is on both the tangent line and the curve). So, we can factor (xa)(x - a) out of the cubic equation: (xa)(x2+ax2a2)=0(x - a)(x^2 + ax - 2a^2) = 0 Now solve for the remaining factor x2+ax2a2=0x^2 + ax - 2a^2 = 0. Using the quadratic formula: x=a±a2+8a22x = \frac{-a \pm \sqrt{a^2 + 8a^2}}{2} Simplifying: x=a±3a2x = \frac{-a \pm 3a}{2} So, x=ax = a (which we already know) or x=2ax = -2a.

Thus, the coordinates of the second intersection point QQ are (2a,8a3)(-2a, -8a^3).

Part (c): Ratio of the Slopes at PP and QQ

We are asked to find the ratio of the slope of the curve at QQ to the slope at PP.

  • The slope at any point on the curve is given by dydx=3x2\frac{dy}{dx} = 3x^2.
    • At PP (where x=ax = a), the slope is 3a23a^2.
    • At QQ (where x=2ax = -2a), the slope is: 3(2a)2=3(4a2)=12a23(-2a)^2 = 3(4a^2) = 12a^2
    • The ratio of the slope at QQ to the slope at PP is: 12a23a2=4\frac{12a^2}{3a^2} = 4

Thus, the ratio is 44.


Let me know if you would like any further clarifications!

5 Related Questions:

  1. How would the result change if the curve were y=xny = x^n instead of y=x3y = x^3?
  2. What happens to the tangent line equation if PP is at x=0x = 0?
  3. How can you solve higher-degree polynomial intersections like this problem more efficiently?
  4. What is the significance of the ratio of the slopes at different points of a curve geometrically?
  5. How would this problem differ if we worked with an implicit curve equation like y2=x3y^2 = x^3?

Tip:

Always check if the equation you're dealing with is factorable by known solutions to simplify your work. In this case, knowing x=ax = a as a root helped us quickly factor the cubic equation.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Polynomial Functions
Algebra

Formulas

Slope of tangent line: dy/dx = 3x^2
Tangent line equation: y - y1 = m(x - x1)
Quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a

Theorems

Derivative of a Polynomial
Factor Theorem

Suitable Grade Level

Grades 11-12