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Chapter 3: Problem 22

\(19-22\) Compute \(\Delta y\) and dy for the given values of \(x\) and \(d x=\Delta x\) . Then sketch a diagram like Figure 5 showing the line segments with lengths dx, dy, and \(\Delta y\) . \(y=e^{x}, x=0, \quad \Delta x=0.5\)

Short Answer

Expert verified
\( dy = 0.5 \), \( \Delta y \approx 0.6487 \).

Step by step solution

01

Determine the function and its derivative

Given the function is \( y = e^{x} \). To find the derivative, we use the formula for differentiation of exponential functions: \( \frac{dy}{dx} = e^{x} \).
02

Calculate dy

To find \( dy \), which represents the differential change in \( y \) for an infinitesimal change in \( x \), we use the formula: \[ dy = \frac{dy}{dx} \cdot dx = e^{x} \cdot dx \] Substituting the given values, where \( x=0 \) and \( dx=\Delta x=0.5 \): \[ dy = e^{0} \cdot 0.5 = 1 \cdot 0.5 = 0.5 \]
03

Calculate \(\Delta y\)

\(\Delta y\) represents the actual change in \( y \) as \( x \) changes from the initial to the final value (\( 0 \) to \( 0.5 \)). We compute it as: \[ \Delta y = e^{x + \Delta x} - e^{x} \] Substituting the values, \[ \Delta y = e^{0.5} - e^{0} \approx 1.6487 - 1 = 0.6487 \]
04

Sketch a diagram

Draw a tangent line at \( x = 0 \). The line goes through the point \((0, 1)\) on the function curve \( y = e^{x} \). Mark the following: - \( dx = 0.5 \) along the x-axis increment from \( x = 0 \) to \( x = 0.5 \)- \( dy = 0.5 \) as the vertical change corresponding to \( dx \) on the tangent- \( \Delta y = 0.6487 \) as the actual vertical change on the curve from \( x = 0 \) to \( x = 0.5 \).This illustrates how \( dy \) approximates \( \Delta y \).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, focusing on finding the rate at which a function changes at any given point. This process involves calculating the derivative of a function, which provides the steepness or slope of the function's graph.
In the exercise provided, we work with the function \( y = e^{x} \). Differentiating \( e^{x} \) is straightforward, as the derivative of an exponential function \( e^{x} \) is itself, \( \frac{dy}{dx} = e^{x} \).
This derivative tells us how \( y \) changes concerning a small change in \( x \). It's crucial to grasp this concept because it forms the basis for understanding how functions behave and change.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. The base of natural exponential functions is \( e \), an irrational number approximately equal to 2.718.
An exponential function like \( y = e^{x} \), as seen in the exercise, grows rapidly as \( x \) increases. Exponential functions are unique because their rate of growth is proportional to their current value.
  • This property leads to applications in various fields, such as in compound interest calculations and population growth models.
  • The constant growth speed of exponential functions makes them predictable and easy to differentiate, with the derivative matching the original function.
Tangent Line
A tangent line is a straight line that touches a curve at only one point without crossing it. This line represents the instantaneous rate of change, or the slope, of the function at that point.
In the provided problem, the tangent line at \( x = 0 \) for the function \( y = e^{x} \) touches the curve at the point (0, 1).
  • The equation of the tangent line can be written using the point-slope form: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope.
  • The slope \( m \) at \( x = 0 \) is \( e^{0} = 1 \), so the tangent line shows how the function \( y = e^{x} \) is behaving at this initial point.
  • Approximating curve behavior with tangent lines helps in estimating changes, like \( dy \), close to the given point.
Differentials
Differentials are used in calculus to approximate small changes in functions. They provide a linear approximation of how a function behaves near a point.
In our example, \( dy \) is the differential calculated as \( dy = \frac{dy}{dx} \times dx \). This equation shows how much \( y \) is expected to change for a given small change in \( x \), assuming a constant rate dictated by the derivative.
  • For \( y = e^{x} \) at \( x = 0 \) with \( dx = 0.5 \), the calculation gives \( dy = 0.5 \), indicating a predicted change.
  • Though \( dy \) is a good approximation, it rarely equals the actual change \( \Delta y \).
  • Comparing \( dy \) with \( \Delta y \) highlights how the tangent approximates the curve, emphasizing the precision of differential calculations.

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Most popular questions from this chapter

A water trough is 10 \(\mathrm{m}\) long and a cross-section has the shape of an isosceles trapezoid that is 30 \(\mathrm{cm}\) wide at the bottom, 80 \(\mathrm{cm}\) wide at the top, and has height 50 \(\mathrm{cm} .\) If the trough is being filled with water at the rate of 0.2 \(\mathrm{m}^{3} / \mathrm{min}\) , how fast is the water level rising when the water is 30 \(\mathrm{cm}\) deep?

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 \(\mathrm{m}\) higher than the bow of the boat. If the rope is pulled in at a rate of \(1 \mathrm{m} / \mathrm{s},\) how fast is the boat approaching the dock when it is 8 \(\mathrm{m}\) from tho dock?

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure \(P\) and volume V satisfy the equation \(\mathrm{PV}=\mathrm{C}\) , where \(\mathrm{C}\) is a constant. Suppose that at a certain instant the volume is \(600 \mathrm{cm}^{3},\) the pressure is \(150 \mathrm{kPa},\) and the pressure is increasing at a rate of 20 \(\mathrm{kPa} / \mathrm{min.}\) At what rate is the volume decreasing at this instant?

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If \(p(x)\) is the total value of the production when there are \(x\) workers in a plant, then the average productivity of the workforce at the plant is $$A(x)=\frac{p(x)}{x}$$ (a) Find \(A^{\prime}(x) .\) Why does the company want to hire more workers if \(A^{\prime}(x)>0 ?\) (b) Show that \(A^{\prime}(x)>0\) if \(p^{\prime}(x)\) is greater than the average productivity.
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