Problem 30 - Calculus Early Transcendentals

Question

Find equations of the tangents to the curve \(x=3 t^{2}+1\) \(y=2 t^{3}+1\) that pass through the point \((4,3)\)

Step-by-Step Solution

Answer

The tangent lines are \( y=x-1 \) and \( y=-2x+11 \).

Step 1: Parametric Equations

The given parametric equations are \( x = 3t^2 + 1 \) and \( y = 2t^3 + 1 \). We'll use these to find the derivatives needed for the tangent line equations.

Step 2: Find Derivatives

To find the tangent, first find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). Using these, we calculate \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).\[\frac{dy}{dt} = 6t^2, \quad \frac{dx}{dt} = 6t\] Thus, \[\frac{dy}{dx} = \frac{6t^2}{6t} = t\].

Step 3: Equation of Tangent Line

The slope of the tangent at any point is \( t \), so the equation of the tangent line is \( y - (2t^3 + 1) = t(x - (3t^2 + 1)) \). This simplifies to \( y = t(x - 3t^2 - 1) + 2t^3 + 1 \).

Step 4: Solve System of Equations

To find the \( t \) values where the tangents pass through the point \((4, 3)\), substitute \( x = 4 \) and \( y = 3 \) into the tangent equation: \[3 = t(4 - 3t^2 - 1) + 2t^3 + 1\]. This reduces to \[3 = t(3 - 3t^2) + 2t^3 + 1\].

Step 5: Simplify and Solve

The equation becomes: \[3 = 3t - 3t^3 + 2t^3 + 1\], which simplifies to \[2 = 3t - t^3\]. Reorganize as \[t^3 - 3t + 2 = 0\]. Solve this cubic equation for \( t \).

Step 6: Find Tangent Points

Factor \( t^3 - 3t + 2 = 0 \): Notice \( t = 1 \) is a root by substitution. After division of \( (t-1) \) out, the remaining quadratic is \( t^2 + t - 2 \) which factors to \( (t+2)(t-1) \). Thus the roots are \( t = 1, -2\).

Step 7: Tangent Equations for \( t=1 \) and \( t=-2 \)

For \( t = 1 \), the point is \( (4, 3) \) and \( t = -2 \) gives another point on the curve. Calculate the tangent line equations: For \( t=1 \), \( y=x-1 \). For \( t=-2 \), \( y=-2x+11 \).

Key Concepts

Tangent Lines

Tangent lines play a crucial role in calculus and geometry. They represent the straight line that touches a curve at a single point without crossing over. In practical terms, the tangent line provides the best linear approximation to the curve at that point.
These lines are especially useful when you want to understand the behavior of a curve in the immediate vicinity of a point. In this exercise, you find the tangents to a parametric curve defined by two equations:

• The equation for x: \( x = 3t^2 + 1 \)
• The equation for y: \( y = 2t^3 + 1 \)

To determine the tangent line, you must find the slope at any given point. This requires calculating the derivative of the parametric equations to obtain \( \frac{dy}{dx} \). This gives the rate of change of y with respect to x, essentially the slope of the tangent line.
Once you know the slope, the equation of the tangent line can be constructed using point-slope form:\[ y - y_0 = m(x - x_0) \]where \( m \) is the slope \( \frac{dy}{dx} \) evaluated at the desired point, and \( (x_0, y_0) \) is the point on the curve.

Derivatives

Derivatives are a foundational concept in calculus, representing the rate at which a quantity changes. For curves defined by parametric equations, you usually require the derivative of y with respect to x, \( \frac{dy}{dx} \), to find tangent lines.
To achieve this, you calculate the derivatives of both parametric equations with respect to their parameter t:

• Derivative of x: \( \frac{dx}{dt} = 6t \)
• Derivative of y: \( \frac{dy}{dt} = 6t^2 \)

Then, use these derivatives to find the slope of the curve at any point:\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2}{6t} = t \]
This expression \( \frac{dy}{dx} = t \) simplifies the process of finding the slope of the tangent line at any point by directly substituting the value of t. This step-by-step derivative calculation provides insights into how the curve behaves over time or at specific points.

Cubic Equation Solutions

Solving a cubic equation involves finding the values of x that make the equation true. Cubic equations typically have the form \( ax^3 + bx^2 + cx + d = 0 \). In this exercise, the key cubic equation to solve is \( t^3 - 3t + 2 = 0 \).
Cubic equations can be tricky, but you can often find at least one real root through guessing or trial and error. In this instance, \( t = 1 \) is identified as a root. Once you find one root, you can factor the cubic equation to simplify it into a quadratic equation, which is easier to solve.
For this problem, factoring out the root \( t - 1 \) results in:

• Remaining quadratic: \( t^2 + t - 2 \)
• Solution: Factors to \( (t + 2)(t - 1) \)

This reveals all roots of the cubic equation: \( t = 1, -2 \). Understanding how to handle cubic equations and their factors helps clarify how to find specific tangent-lines points on complex curves.