Problem 29 - Calculus Early Transcendentals

Question

At what points on the curve \(x=2 t^{3}, y=1+4 t-t^{2}\) does the tangent line have slope 1\(?\)

Step-by-Step Solution

Answer

The points are \( \left( \frac{16}{27}, \frac{25}{9} \right) \) and \( (-2, -4) \).

Step 1: Find the derivatives

To find the slope of the tangent line, first, find the derivatives of the parametric equations. We have: \ \( \frac{dx}{dt} = \frac{d}{dt}(2t^3) = 6t^2 \). \ \( \frac{dy}{dt} = \frac{d}{dt}(1 + 4t - t^2) = 4 - 2t \).

Step 2: Express slope of tangent line

The slope of the tangent line for parametric equations \(x(t)\) and \(y(t)\) is given by \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Substitute the derivatives from Step 1: \ \(\frac{dy}{dx} = \frac{4 - 2t}{6t^2}\).

Step 3: Set the slope equal to 1

We are given that the slope of the tangent line is 1, so set \(\frac{dy}{dx} = 1\): \ \(\frac{4 - 2t}{6t^2} = 1\).

Step 4: Solve for t

Solve the equation \(\frac{4 - 2t}{6t^2} = 1\) for \(t\): \ Multiply both sides by \(6t^2\) to get: \ \(4 - 2t = 6t^2\). \ Rearrange to get a quadratic equation: \ \(6t^2 + 2t - 4 = 0\). \ Simplify: \ \(3t^2 + t - 2 = 0\).

Step 5: Solve the quadratic equation

Solve \(3t^2 + t - 2 = 0\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), and \(c = -2\): \ \(t = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm 5}{6}\). \ This gives \(t = \frac{4}{6} = \frac{2}{3}\) and \(t = \frac{-6}{6} = -1\).

Step 6: Find the points on the curve

Substitute \(t = \frac{2}{3}\) and \(t = -1\) back into the parametric equations to find the points. \ For \(t = \frac{2}{3}\): \ \(x = 2\left(\frac{2}{3}\right)^3 = \frac{16}{27}\), \(y = 1 + 4\left(\frac{2}{3}\right) - \left(\frac{2}{3}\right)^2 = \frac{25}{9}\). \ For \(t = -1\): \ \(x = 2(-1)^3 = -2\), \(y = 1 + 4(-1) - (-1)^2 = -4\).

Step 7: Conclude the points

Thus, the points on the curve where the tangent line has a slope of 1 are: \( \left( \frac{16}{27}, \frac{25}{9} \right) \) and \( (-2, -4) \).

Key Concepts

Derivatives

In parametric equations, derivatives help us understand the rate at which one variable changes with respect to another. In our context, we have two functions: one for the x-coordinate, \(x = 2t^3\), and one for the y-coordinate, \(y = 1 + 4t - t^2\). To find the equations of the tangent lines, we first need to compute the derivatives of these functions with respect to the parameter \(t\).
This involves using the rules of differentiation, a process that tells us how a function changes at any given point.

• The derivative of \(x\) with respect to \(t\), \(\frac{dx}{dt}\), is \(6t^2\).
• The derivative of \(y\) with respect to \(t\), \(\frac{dy}{dt}\), is \(4 - 2t\).

These derivatives are the building blocks for finding the slope of the tangent line in parametric equations.

Slope of Tangent Line

The slope of the tangent line to a parametric curve is a fundamental concept in calculus. It gives us the steepness of the line at any point on the curve. For a parametric equation, the slope of the tangent line is expressed as the ratio of the change in \(y\) to the change in \(x\):\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

This formula allows us to calculate how much \(y\) changes for a small change in \(x\), providing us the gradient of the curve at a specific point.
By substituting our derivatives, \(\frac{dy}{dt} = 4 - 2t\) and \(\frac{dx}{dt} = 6t^2\), we find the slope of the tangent line:\[\frac{dy}{dx} = \frac{4-2t}{6t^2}\].
This result must be considered at the points where the slope is equal to the value given in the exercise, in this case, 1.

Quadratic Equation

When we set the slope of the tangent line equal to 1, we need to solve a quadratic equation: \(\frac{4 - 2t}{6t^2} = 1\). Simplifying this equation involves a few algebraic steps.
First, multiply through by \(6t^2\) to eliminate the denominator:

• \(4 - 2t = 6t^2\)

After rearranging the terms, we have \(6t^2 + 2t - 4 = 0\), which simplifies to \(3t^2 + t - 2 = 0\).
This is a standard quadratic equation, and we solve it using the quadratic formula:
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\],
where \(a = 3\), \(b = 1\), and \(c = -2\).
This formula gives us the roots \(t = \frac{2}{3}\) and \(t = -1\), critical to finding the exact points on the curve where the slope of the tangent line is 1.

Tangent Line

The tangent line beautifully touches the curve at just one unique point, having the same slope as the curve at that point. To find where this line intersects the curve with a slope of 1, we use the parameter values obtained: \(t = \frac{2}{3}\) and \(t = -1\).

• For \(t = \frac{2}{3}\), substituting into the parametric equations gives us the point \(\left( \frac{16}{27}, \frac{25}{9} \right)\).
• For \(t = -1\), substituting these values results in the point \((-2, -4)\).

These points are significant because they denote where the tangent line, with a slope of 1, meets the curve. Understanding the concept of the tangent line helps us see how a curve behaves at any individual point.