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Chapter 1: Problem 15

Find the slope of the tangent line to the graph of \(y=x^{2}\) at the point indicated and then write the corresponding equation of the tangent line. $$\left(\frac{1}{3}, \frac{1}{9}\right)$$

Short Answer

Expert verified
The slope is \( \frac{2}{3} \), and the equation of the tangent line is \( y = \frac{2}{3}x - \frac{1}{9} \).

Step by step solution

01

Identify the Function and Point

The function given is \( y = x^2 \) and the point at which the tangent line is to be found is \( \left( \frac{1}{3}, \frac{1}{9} \right) \).
02

Find the Derivative of the Function

The slope of the tangent line is given by the derivative of the function. For \( y = x^2 \), the derivative is given by \( \frac{dy}{dx} = 2x \).
03

Calculate the Slope at the Given Point

Substitute \( x = \frac{1}{3} \) into the derivative to find the slope at the given point: \[ \text{Slope} = 2 \left( \frac{1}{3} \right) = \frac{2}{3} \].
04

Use the Point-Slope Form Equation

The equation of the tangent line can be written using the point-slope form: \[ y - y_1 = m(x - x_1) \]. Here, \( m \) is the slope and \( (x_1, y_1) \) is the given point. Thus, \[ y - \frac{1}{9} = \frac{2}{3}(x - \frac{1}{3}) \].
05

Simplify the Equation

Distribute and simplify the equation: \[ y - \frac{1}{9} = \frac{2}{3}x - \frac{2}{9} \]. Add \( \frac{1}{9} \) to both sides to get: \[ y = \frac{2}{3}x - \frac{1}{9} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

derivative
The derivative is a fundamental concept in calculus. It measures how a function changes as its input changes. Think of it as the slope of the function at any given point. For the function given in the exercise, which is \( y = x^2 \), the derivative is calculated. To do this, we use the power rule. The power rule states that if you have a function of the form \( f(x) = x^n \), its derivative is \( f'(x) = n \cdot x^{(n-1)} \). So, applying this rule to our function \( y = x^2 \), we get the derivative: \[\frac{dy}{dx} = 2x\]. This derivative function tells us the slope of the curve at any point \( x \).
slope
Slope is a measure of the steepness of a line. In the context of tangent lines, the slope is the rate at which the function changes at a specific point. When you find the derivative of a function, you are finding an expression that gives the slope of the tangent line at any point along the function. In our exercise, we have already found the derivative \( \frac{dy}{dx} = 2x \). To find the slope of the tangent line at the specific point \( \left( \frac{1}{3}, \frac{1}{9} \right) \), substitute \( x = \frac{1}{3} \) into the derivative: \[\text{Slope} = 2 \left( \frac{1}{3} \right) = \frac{2}{3}\]. This is the slope of the tangent line at our given point.
point-slope form
The point-slope form is a method used to write the equation of a line when you know the slope and a point on the line. The formula for the point-slope form is: \[ y - y_1 = m(x - x_1) \]. Here, \( m \) represents the slope, and \( (x_1, y_1) \) represents a specific point on the line. In our exercise, we have determined the slope \( m = \frac{2}{3} \) and the point \( \left( \frac{1}{3}, \frac{1}{9} \right) \). Plugging these values into the point-slope formula, we get: \[ y - \frac{1}{9} = \frac{2}{3}(x - \frac{1}{3}) \]. This equation represents the tangent line in its raw form before simplification.
tangent line equation
To find the equation of the tangent line, we simplify the point-slope form equation. Starting with: \[ y - \frac{1}{9} = \frac{2}{3}(x - \frac{1}{3}) \]. Next, distribute \( \frac{2}{3} \) to both terms in parentheses: \[ y - \frac{1}{9} = \frac{2}{3}x - \frac{2}{9} \]. Finally, add \( \frac{1}{9} \) to both sides of the equation to isolate \( y \): \[ y = \frac{2}{3}x - \frac{1}{9} \]. This is the simplified equation of the tangent line at the point \( \left( \frac{1}{3}, \frac{1}{9} \right) \). The tangent line touches the curve of \( y = x^2 \) at exactly one point, and it has the same slope as the curve at that point.

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Most popular questions from this chapter

Using the sum rule and the constant-multiple rule, show that for any functions \(f(x)\) and \(g(x).\) $$\frac{d}{d x}[f(x)-g(x)]=\frac{d}{d x} f(x)-\frac{d}{d x} g(x).$$

The tangent line to the curve \(y=\frac{1}{3} x^{3}-4 x^{2}+18 x+22\) is parallel to the line \(6 x-2 y=1\) at two points on the curve. Find the two points.

Use limits to compute \(f^{\prime}(x) .\) $$f(x)=3 x+1$$

(a) Draw two graphs of your choice that represent a function \(y=f(x)\) and its vertical shift \(y=f(x)+3.\) (b) Pick a value of \(x\) and consider the points \((x, f(x))\) and \((x, f(x)+3) .\) Draw the tangent lines to the curves at these points and describe what you observe about the tangent lines. (c) Based on your observation in part (b), explain why $$\frac{d}{d x} f(x)=\frac{d}{d x}(f(x)+3)$$

Use limits to compute the following derivatives. $$f^{\prime}(3), \text { where } f(x)=x^{2}+1$$
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