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(a) Find the slope of the tangent line to the curve \(y=x-x^{3}\) at the point \((1,0)\) (i) using Definition 1 (b) Find an equation of the tangent line in part (a). (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at \((1,0)\) until the curve and the line appear to coincide.

Short Answer

Expert verified
The slope of the tangent line is -3, and the equation is \(y = -3x + 3\).

Step by step solution

01

Understanding the Derivative at a Point

The slope of the tangent line to a curve at a given point can be found using the derivative of the function. According to Definition 1, the derivative at a point \((a, f(a))\) is given by the limit \(f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h}\).
02

Applying Definition 1

To find the slope of the tangent line to the curve \(y = x - x^3\) at \((1, 0)\), let \(f(x) = x - x^3\). Calculate \(f'(1)\) by finding the limit \(f'(1) = \lim_{{h \to 0}} \frac{(1 + h) - (1 + h)^3 - (1 - 1^3)}{h}\).
03

Simplifying the Expression

Expand \((1 + h)^3\) as \(1 + 3h + 3h^2 + h^3\). Substitute into the limit equation: \(\frac{(1 + h) - (1 + 3h + 3h^2 + h^3) - 1}{h}\). Simplify to \(-3h - 3h^2 - h^3\) and then simplify the fraction to \(-3 - 3h - h^2\).
04

Taking the Limit

Now, compute the limit: \(\lim_{{h \to 0}} (-3 - 3h - h^2) = -3\). Therefore, the slope of the tangent line at \((1, 0)\) is \(-3\).
05

Writing the Equation of the Tangent Line

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m = -3\) and \((x_1, y_1) = (1, 0)\), we have \(y - 0 = -3(x - 1)\). Simplifying, the equation of the tangent line is \(y = -3x + 3\).
06

Graphing the Curve and Tangent Line

Graph the curve \(y = x - x^3\) and the equation \(y = -3x + 3\). Zoom into the point \((1, 0)\) by adjusting the scale and centering the view around \((1, 0)\) until the curve and tangent line appear to coincide. Use smaller intervals on both axes for precise viewing.

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Key Concepts

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

Derivative
The concept of a derivative is central in calculus and helps us understand how a function changes at any given point. The derivative at a specific point gives us the slope of the tangent line to the curve at that point. This slope is a measure of how steep the curve is.

Mathematically, the derivative is defined as a limit. For a function \( f(x) \), the derivative at a point \( a \) is given by: \[ f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h} \] This expression tells us that when the change \( h \) in \( x \) approaches zero, the ratio of the change in \( y \) (\( f(a+h) - f(a) \)) to the change in \( x \) gives the slope of the tangent line.

Using a derivative to find slopes is not only applicable to lines but also to any differentiable function, providing insight into the behavior of physical, economic, and various other types of systems.
  • Calculating Derivatives: Use the limit process.
  • Applications: Physics (velocity), Economics (marginal cost).
Tangent Line
A tangent line to a curve at a particular point provides a local linear approximation to the curve near that point. Imagine a curve as a road winding. The tangent line is like a straight path that follows the twisting curves at one precise point without veering off it.

The slope of this line is crucial. It matches the rate of change of the curve at that specific point. When we calculate a derivative, we effectively find this slope. Once we have the slope, we can use it to construct the equation of the line.

For example, in the problem, we calculated the tangent line to the curve \( y = x - x^3 \) at \( \(1,0\) \) with a slope of \(-3\). The line is then described by \(y = -3x + 3\).

The equation of a tangent line is useful for approximating function values and analyzing the local behavior of functions.
  • Key Concept: Local linear approximation.
  • Uses: Immediate insights into function behavior.
Limit
Limits are at the heart of calculus and deal with approaching values. When we talk about limits, we're focusing on what happens as we get arbitrarily close to a specific point.

Formally, the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\), written as \( \lim_{{x \rightarrow a}} f(x) = L \). This means that as \(x\) gets closer and closer to \(a\), \(f(x)\) gets closer to \(L\).

In derivative calculations, limits allow us to zoom in on the precise instantaneous rate of change. For instance, when finding a derivative, we take a limit as \(h\) approaches zero. The limit allows us to calculate the exact slope of the tangent line, offering insights into the curve's behavior at that point.

Understanding limits aids in learning continuity, solving indeterminate forms, and explaining intuitive yet precise mathematical arguments.
  • Application: Solving problems involving continuity, derivatives.
  • Understanding Change: Capture the essence of "approaching" a value.
Point-Slope Form
The point-slope form is a useful equation for lines when you know a point on the line and the slope. It's especially handy for writing the equation of the tangent line. The format is given by:

\[ y - y_1 = m(x - x_1) \] where: - \( m \) is the slope, - \((x_1, y_1)\) is a known point on the line. This formula is perfect for tangent lines since we often calculate the slope (the derivative) and have a point on the curve.

As in the exercise, with a slope \( m = -3 \) and the point \((1, 0)\), the equation became \( y = -3x + 3 \). It's straightforward and quick to use, ensuring clarity and simplicity when dealing with lines.

Besides calculus, point-slope form is common in coordinate geometry for line equations and helpful in analyzing how lines interact and connect in a given space.
  • Simplicity in Form: Keeps calculations direct.
  • Widely Used: Especially useful in geometry.

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Most popular questions from this chapter

Show that the sum of the \(x\) - and \(y\) -intercepts of any tangent line to the curve \(\sqrt{x}+\sqrt{y}=\sqrt{c}\) is equal to \(c .\)

On page 431 of Physics: Calculus, 2 \(\mathrm{d}\) ed. by Eugene Hecht (Pacific Grove, \(\mathrm{CA}, 2000 ),\) in the course of deriving the formula \(T=2 \pi \sqrt{L / g}\) for the period of a pendulum of length \(L,\) the author obtains the equation \(a_{T}=-g \sin \theta\) for the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta ;\) they differ by less than 2\(\%\) out to about \(20^{\circ} . "\) (a) Verify the linear approximation at 0 for the sine function: \(\sin x \approx x\) (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

$$ \begin{array}{l}{\text { (b) If } f(x)=|\sin x|, \text { find } f^{\prime}(x) \text { and sketch the graphs of } f} \\ {\text { and } f ! \text { Where is } f \text { not differentiable? }} \\ {\text { (c) If } g(x)=\sin |x|, \text { find } g^{\prime}(x) \text { and sketch the graphs of } g} \\ {\text { and } g^{\prime} \text { . Where is } g \text { not differentiable? }}\end{array} $$

The radius of a circular disk is given as 24 \(\mathrm{cm}\) with a maximum error in measurement of 0.2 \(\mathrm{cm} .\) (a) Use differentials to estimate the maximum error in the calculated area of the disk. (b) What is the relative error? What is the percentage error?

The equation \(y^{\prime \prime}+y^{\prime}-2 y=\sin x\) is called a differential equation because it involves an unknown function \(y\) and its derivatives \(y^{\prime}\) and \(y^{\prime \prime}\) . Find constants \(A\) and \(B\) such that the function \(y=A \sin x+B \cos x\) satisfies this equation. (Dif- ferential equations will be studied in detail in Section 7.7 )

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