Problem 41 - Thomas Calculus

Question

a. Find an equation for the line perpendicular to the tangent to the curve \(y=x^{3}-4 x+1\) at the point (2,1). b. What is the smallest slope on the curve? At what point on the curve does the curve have this slope? c. Find equations for the tangents to the curve at the points where the slope of the curve is 8 .

Step-by-Step Solution

Answer

a) y = -1/8x + 5/4. b) Slope -4 at (0,1). c) y = 8x - 15 and y = 8x + 25.

Step 1: Find the Derivative

To find the slope of the tangent to the curve, we need to find the derivative of the curve. The given curve is \(y = x^3 - 4x + 1\). The derivative is \(y' = 3x^2 - 4\). This will give us the slope of the tangent at any point \(x\).

Step 2: Calculate Slope at Specific Point

We need to find the slope of the tangent at the point \((2,1)\). Use the derivative found in Step 1: \(y' = 3(2)^2 - 4 = 12 - 4 = 8\). Hence, the slope of the tangent at \((2,1)\) is 8.

Step 3: Equation for Perpendicular Line

The slope of a line perpendicular to another line is the negative reciprocal of the original slope. Thus, the slope of the line perpendicular to the tangent at \((2,1)\) is \(-\frac{1}{8}\). Use point-slope form \(y - y_1 = m(x - x_1)\) with \((x_1, y_1) = (2, 1)\) and \(m = -\frac{1}{8}\): \(y - 1 = -\frac{1}{8}(x - 2)\). Simplify: \(y = -\frac{1}{8}x + \frac{1}{4} + 1\) resulting in \(y = -\frac{1}{8}x + \frac{5}{4}\).

Step 4: Find Critical Points of the Derivative

To find the minimum slope on the curve, solve \(y'' = 6x = 0\) for critical points, giving \(x = 0\). Evaluate \(y'\) at \(x = 0\): \(y' = 3(0)^2 - 4 = -4\). So, \(-4\) is the smallest slope, occurring at \((0, y)\), where \(y = 0^3 - 4(0) + 1 = 1\), i.e., point \((0, 1)\).

Step 5: Tangents with Specific Slope

Find where the slope of the tangent \(y' = 8\). Set \(3x^2 - 4 = 8\) which gives \(3x^2 = 12\), or \(x^2 = 4\). Thus, \(x = 2\) or \(x = -2\). For each \(x\), find \(y\): \(y = 2^3 - 4(2) + 1 = 1\) and \(y = (-2)^3 - 4(-2) + 1 = 9\). The points are \((2, 1)\) and \((-2, 9)\).

Step 6: Equations of Tangents

Using point-slope form \(y - y_1 = m(x - x_1)\), plug \((x_1, y_1, m) = (2, 1, 8)\) and \((-2, 9, 8)\). For \((2, 1)\): \(y - 1 = 8(x - 2)\), simplifying gives \(y = 8x - 15\). For \((-2, 9)\): \(y - 9 = 8(x + 2)\), simplifying gives \(y = 8x + 25\).

Key Concepts

Derivatives

In calculus, derivatives play a crucial role in understanding the behavior of functions. A derivative essentially measures how a function changes as its input changes. For a function represented as \( y = f(x) \), the derivative, denoted as \( y' \) or \( \frac{dy}{dx} \), captures the instantaneous rate of change of \( y \) with respect to \( x \). This rate of change is also the slope of the tangent line at any given point on the curve.

For our specific problem, the function is \( y = x^3 - 4x + 1 \). To find the derivative, apply the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). Hence, the derivative here is \( y' = 3x^2 - 4 \). This expression allows us to determine the slope of the tangent line at any point \( x \) on the curve.

Tangent Lines

Tangent lines are straight lines that touch a curve at exactly one point without crossing it. They serve as an excellent approximation of the curve near the point of tangency. The slope of the tangent line at a particular point is given by the derivative of the function at that point.

Let's consider the tangent line to our curve \( y = x^3 - 4x + 1 \) at the point \( (2, 1) \). By substituting \( x = 2 \) into the derivative \( y' = 3x^2 - 4 \), we find the slope at that point as 8. To form the equation of the tangent line, we use the point-slope form of a linear equation: \( y - y_1 = m(x - x_1) \). Here, \( (x_1, y_1) = (2, 1) \) and \( m = 8 \), resulting in the equation \( y = 8x - 15 \) for the tangent line.

Perpendicular Lines

A line is said to be perpendicular to another if it intersects it at a right angle. The relationship between the slopes of two perpendicular lines is that they are negative reciprocals of each other.

In our exercise, we've found the slope of the tangent line at the point \( (2, 1) \) to be 8. Therefore, the slope of a line perpendicular to this tangent would be \( -\frac{1}{8} \). Using the point-slope form again with this slope and the point \( (2, 1) \), we derive the equation for the perpendicular line: \( y = -\frac{1}{8}x + \frac{5}{4} \). This line intersects the tangent at \( (2, 1) \) and forms a right angle with it.

Critical Points

Critical points in calculus occur where the derivative of a function equals zero or is undefined. They are important because they can indicate potential maximums, minimums, or points of inflection on the curve.

For the function \( y = x^3 - 4x + 1 \), critical points are determined by setting \( y' = 3x^2 - 4 \) equal to zero. This results in \( 3x^2 = 4 \), or \( x^2 = \frac{4}{3} \). Solving gives \( x = \pm \sqrt{\frac{4}{3}} \). These \( x \)-values are the locations where the slope of the tangent is zero, suggesting potential local minimums or maximums. However, in Step 5 of the solution, \( y' = 8 \) instead, helps identify more significant points like natural changes in slope direction on the curve between negative and positive slopes, reinforcing actual slope trends.