Problem 242 - Calculus Volume 1

Question

Find the equation of the tangent line to \(y=\left(3 x+\frac{1}{x}\right)^{2}\) at the point (1,16) . Use a calculator to graph the function and the tangent line together.

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 16x \).

Step 1: Understand the Problem

We need to find the equation of the tangent line to the given function \( y = \left(3x + \frac{1}{x}\right)^2 \) at the point (1, 16). This involves calculating the derivative to find the slope at \( x = 1 \) and using this slope to write the equation of the tangent line.

Step 2: Differentiate the Function

First, rewrite the function as \( y = u^2 \), where \( u = 3x + \frac{1}{x} \). Use the chain rule to differentiate: \( \frac{dy}{dx} = 2u \cdot \frac{du}{dx} \). Calculate \( \frac{du}{dx} \) as follows: \( \frac{d}{dx}(3x) = 3 \) and \( \frac{d}{dx}\left( \frac{1}{x} \right) = -\frac{1}{x^2} \). Thus, \( \frac{du}{dx} = 3 - \frac{1}{x^2} \). Substitute to find \( \frac{dy}{dx} = 2\left(3x + \frac{1}{x}\right) \left( 3 - \frac{1}{x^2} \right) \).

Step 3: Calculate the Slope at x = 1

Substitute \( x = 1 \) into the derivative to find the slope of the tangent line. First, calculate \( u \) at \( x = 1 \): \( u = 3(1) + \frac{1}{1} = 4 \). The derivative is thus \( \frac{dy}{dx} = 2(4)\left(3 - \frac{1}{1^2}\right) = 8 \cdot 2 = 16 \). Therefore, the slope of the tangent line at \( x = 1 \) is 16.

Step 4: Write the Equation of the Tangent Line

Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m = 16 \) and the point is \((1, 16)\). Substitute these values: \( y - 16 = 16(x - 1) \). Simplify the equation to the slope-intercept form: \( y = 16x \). Thus, the equation of the tangent line is \( y = 16x \).

Step 5: Graph the Function and Tangent Line

Using a calculator or graphing software, plot the function \( y = \left(3x + \frac{1}{x}\right)^2 \) and the tangent line equation \( y = 16x \). Both curves should intersect at the point (1, 16), with the tangent line just touching the function at this point, indicating it is tangential to the curve.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it at that point. This special line represents the immediate "path" a curve is following. Imagine being on a roller coast right when it starts to "turn." At that exact point, the tangent line hints at where the coaster is headed next. When we say we want to find the equation of a tangent line, we want to find the math equation that captures this precise moment and its direction.

• It usually relies on a point the line touches - in our case, (1, 16).
• You'll also need the slope of the curve at that exact point.

Knowing how to determine a tangent line helps in many math and real-world problems, especially when predicting short-term behavior or changes.

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It's like having a speedometer that tells you how fast something is moving at any moment. In simpler words, the derivative represents the slope of a function at any given point.

• It allows us to find the slope of a tangent line.
• It's at the heart of many calculations in calculus, unlocking how functions grow or shrink.

In this exercise, to find out how steep our curve is at a certain point (1, 16), we differentiated our function, which gives us the ability to calculate the slope of the tangent line right then and there.

Chain Rule

The chain rule is a tool we can use when we need to differentiate a composite function, which is a function inside another function. Think of it like layers of an onion - you work through each layer to get to the derivative.

• It's especially useful when differentiating complex, layered functions.
• The rule tells us that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

When dealing with our function, employing the chain rule helped unravel its complexity, allowing us to correctly calculate the derivative needed for our tangent line's slope.

Point-Slope Form

To nail down the equation of our tangent line, the point-slope form of a line is a handy formula. It uses one specific point on the line and the slope to build the equation. The formula is given by: \[ y - y_1 = m(x - x_1) \] where \(m\) is the slope, and \((x_1, y_1)\) is the known point on the line.

• This form is straightforward when you know the slope and a point.
• It's often a starting point before transitioning to easier-to-read forms like slope-intercept.

In our example, with a known point (1, 16) and the slope of 16, the point-slope form quickly led us to the exact equation of the tangent line - a vital step in sketching and predicting the curve at that point.