Problem 54
Question
In Problems , find the normal line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=1-\pi x^{2}, \text { at } x=-1 $$
Step-by-Step Solution
Verified Answer
The normal line in standard form is \( x + 2\pi y = 2\pi \).
1Step 1: Determine the Derivative
First, find the derivative of the function to get the slope of the tangent line at the given point. Here, the function is \( y = 1 - \pi x^2 \). The derivative \( y' \) is computed as follows: \( y' = -2\pi x \).
2Step 2: Evaluate Derivative at the Point
Substitute \( x = -1 \) into the derivative to find the slope of the tangent line at the point. \( y'(-1) = -2\pi(-1) = 2\pi \). This slope is for the tangent line.
3Step 3: Find the Slope of the Normal Line
The slope of the normal line is the negative reciprocal of the tangent slope. Therefore, the slope of the normal line is \( -\frac{1}{2\pi} \).
4Step 4: Calculate the Point on the Curve
Substitute \( x = -1 \) into the original function to find the corresponding \( y \)-value for the point on the curve. \( y = 1 - \pi (-1)^2 = 1 - \pi \). So the point is \( (-1, 1 - \pi) \).
5Step 5: Determine the Equation of the Normal Line
Use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point. Here, the equation becomes \( y - (1 - \pi) = -\frac{1}{2\pi}(x + 1) \).
6Step 6: Convert to Standard Form
Transform the equation into standard form \( Ax + By = C \). Start with: \( y - 1 + \pi = -\frac{1}{2\pi}x - \frac{1}{2\pi} \). Rearrange and multiply through by \( 2\pi \) to eliminate the fraction: \( 2\pi y - 2\pi(1) + 2\pi^2 = -x - 1\). Finally, rearrange to get \( x + 2\pi y = 2\pi \).
Key Concepts
DerivativeTangent LineNormal LineEquation of a Line
Derivative
In calculus, the derivative is a fundamental concept that measures how a function changes as its input changes. It's like finding the rate of change or how steep the curve is at any given point. To visualize this, think of a mountain slope. Just like a steeper mountain means a more challenging climb, a larger derivative value indicates a steeper curve.
For a given function, the derivative is often denoted as \( y' \) or \( \frac{dy}{dx} \). In our example, we have the function \( y = 1 - \pi x^2 \). To find how this function behaves near any point, we calculate the derivative. Here, the derivative is \( y' = -2\pi x \).
Calculating this gives us the slope of the curve at any specific \( x \) value. For instance, at \( x = -1 \), substituting into the derivative \( y' = -2\pi(-1) \) results in \( 2\pi \), which is the slope of the tangent at that point.
For a given function, the derivative is often denoted as \( y' \) or \( \frac{dy}{dx} \). In our example, we have the function \( y = 1 - \pi x^2 \). To find how this function behaves near any point, we calculate the derivative. Here, the derivative is \( y' = -2\pi x \).
Calculating this gives us the slope of the curve at any specific \( x \) value. For instance, at \( x = -1 \), substituting into the derivative \( y' = -2\pi(-1) \) results in \( 2\pi \), which is the slope of the tangent at that point.
Tangent Line
A tangent line is a straight line that touches a curve at just one point. It's like a brief "kiss" between the line and the curve, representing the curve's slope at that exact point.
When we find the derivative of a function at a point, we can determine the tangent line's slope. In our example, at \( x = -1 \), we found the derivative (or slope of the tangent line) to be \( 2\pi \).
To fully describe this tangent line, we need a point and a slope. We already have the slope, and using \( x = -1 \) in the function \( y = 1 - \pi x^2 \), we find the point \( (-1, 1 - \pi) \). With these, we can write the equation of the tangent line using the point-slope form.
When we find the derivative of a function at a point, we can determine the tangent line's slope. In our example, at \( x = -1 \), we found the derivative (or slope of the tangent line) to be \( 2\pi \).
To fully describe this tangent line, we need a point and a slope. We already have the slope, and using \( x = -1 \) in the function \( y = 1 - \pi x^2 \), we find the point \( (-1, 1 - \pi) \). With these, we can write the equation of the tangent line using the point-slope form.
Normal Line
The normal line is the line that is perpendicular to the tangent line at the point where they touch the curve. It's as if the normal line is making a "right angle" with the tangent, giving us a whole new perspective on the curve.
To find this, we take the slope of the tangent line and find its negative reciprocal. In our example, with a tangent slope of \( 2\pi \), the normal slope becomes \(-\frac{1}{2\pi}\).
Using this slope and the same point \((-1, 1 - \pi)\), the normal line's equation can also be written in point-slope form, making it clear how sharply the curve bends at that spot.
To find this, we take the slope of the tangent line and find its negative reciprocal. In our example, with a tangent slope of \( 2\pi \), the normal slope becomes \(-\frac{1}{2\pi}\).
Using this slope and the same point \((-1, 1 - \pi)\), the normal line's equation can also be written in point-slope form, making it clear how sharply the curve bends at that spot.
Equation of a Line
The equation of a line can show us how a line runs through a given point with a specific slope. This is typically expressed in the format \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
When we derive a line's equation from the point-slope form \( y - y_1 = m(x - x_1) \), we can find how exactly the line interacts with the graph. For the normal line in our example, this leads us to \( y - (1 - \pi) = -\frac{1}{2\pi}(x + 1) \).
To convert this to a more standardized format, known as the standard form \( Ax + By = C \), we conduct algebraic manipulations to clear fractions and rearrange terms. For the normal line, this transformation results in the clear, neat equation \( x + 2\pi y = 2\pi \), perfectly aligning with how we conventionally visualize lines on a coordinate plane.
When we derive a line's equation from the point-slope form \( y - y_1 = m(x - x_1) \), we can find how exactly the line interacts with the graph. For the normal line in our example, this leads us to \( y - (1 - \pi) = -\frac{1}{2\pi}(x + 1) \).
To convert this to a more standardized format, known as the standard form \( Ax + By = C \), we conduct algebraic manipulations to clear fractions and rearrange terms. For the normal line, this transformation results in the clear, neat equation \( x + 2\pi y = 2\pi \), perfectly aligning with how we conventionally visualize lines on a coordinate plane.
Other exercises in this chapter
Problem 54
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