/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 (a) Let \(y=1 / x\). Find \(d y\... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Let \(y=1 / x\). Find \(d y\) and \(\Delta y\) at \(x=1\) with \(d x=\Delta x=-0.5\) (b) Sketch the graph of \(y=1 / x\), showing \(d y\) and \(\Delta y\) in the picture.

Short Answer

Expert verified
\( dy = 0.5 \), \( \Delta y = 1 \), sketch shows curvature and tangent.

Step by step solution

01

Define the function and variables

We are working with the function \( y = \frac{1}{x} \), and we know that at \( x = 1 \), both \( dx = \Delta x = -0.5 \). We need to find the values of \( dy \) and \( \Delta y \).
02

Calculate the differential \( dy \)

The differential \( dy \) is given by \( dy = f'(x) \cdot dx \), where \( f'(x) \) is the derivative of \( y \) with respect to \( x \). The function is \( y = \frac{1}{x} \), so \( f'(x) = -\frac{1}{x^2} \). At \( x = 1 \), we substitute to find:\[dy = \left(-\frac{1}{1^2}\right) \cdot (-0.5) = 0.5\]
03

Calculate \( \Delta y \) using the original function

The change in \( y \), \( \Delta y \), is calculated using the difference \( \Delta y = y(x+\Delta x) - y(x) \). Here, we will find:\[y(1-0.5) = \frac{1}{0.5} = 2, \quad y(1) = 1\]\[\Delta y = 2 - 1 = 1\]
04

Sketch the function graph

Draw the hyperbola representing \( y = \frac{1}{x} \). At the point \( x = 1 \), draw a tangent line to the curve to illustrate the differential \( dy \). This tangent shows the estimated change in \( y \) when moving from \( x = 1 \) by \( dx = -0.5 \). Draw a vertical line from \( x = 0.5 \) up to the curve to show the actual change \( \Delta y \). The length of the tangent line segment represents \( dy \), and the vertical segment on the curve represents \( \Delta y \).

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Key Concepts

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

Derivatives
Derivatives in differential calculus help us determine the rate at which a function is changing at any given point. Essentially, the derivative provides a formula that tells you the slope of the tangent line to the curve at a specific point. In mathematical terms, for the function \( y = \frac{1}{x} \), the derivative \( f'(x) \) is calculated as\[ f'(x) = -\frac{1}{x^2} \].
This shows how the function changes as \( x \) changes. At \( x = 1 \), the derivative evaluates to \(-1\), meaning the slope of the tangent line is \(-1\) at this point. This slope helps us approximate changes in \( y \) for small changes in \( x \) through the differential \( dy \). When \( dx = -0.5 \), which signifies a small decrease in \( x \), the resulting \( dy \) derived from \( f'(x) \cdot dx \) is \( 0.5 \).
Thus, the derivative not only provides an instantaneous description of the curve's behavior at a point but also guides us in estimating changes over small intervals.
Tangent Line
A tangent line is a straight line that touches a curve at a single point and matches the curve's slope at that point. This line provides a good approximation of the curve near the point of tangency. For the function \( y = \frac{1}{x} \), at \( x = 1 \), the slope of the tangent line is \(-1\) as determined by the derivative.
The equation of the tangent line to the function at \( x = 1 \) can be realized using the point-slope form \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency, which is \( (1,1) \). For this case, the equation is:\[ y - 1 = -1(x - 1) \].
Following through, this simplifies to \( y = -x + 2 \).
This tangent provides a linear approximation for \( y = \frac{1}{x} \) near \( x = 1 \). When \( x \) decreases by \( 0.5 \), the tangent line predicts a \( 0.5 \) increase in \( y \), closely aligning with the differential \( dy \).
The tangent line is quite useful for visualizing the immediate rate of change and making quick approximations.
Hyperbola
The function \( y = \frac{1}{x} \) describes a hyperbola, which is a type of curve defined by the inverse relationship between \( x \) and \( y \). This means as \( x \) increases, \( y \) decreases considerably, and vice versa. The graph of \( y = \frac{1}{x} \) features two asymptotic arms for positive and negative values of \( x \), avoiding crossing the x or y axes.
For drawing the hyperbola, begin with the knowledge that the curve approaches the x-axis as \( x \) goes to infinity and the y-axis as \( x \) approaches zero, but never actually touches either axis.
At the specific point \( x = 1 \), the curve makes an important shift, showcasing significant changes visually through both \( dy \) and \( \Delta y \). Defining the tangent at \( x = 1 \) and the vertical change \( \Delta y \) showcases the nature of the hyperbola graphically, emphasizing the curve's behavior under positive and negative shifts in \( x \).
Sketching these elements on the hyperbola can visually articulate how minor changes in \( x \) can result in differences between approximate and actual changes in \( y \), easily observed when \( x = 0.5 \).

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