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Chapter 4: Problem 31

Find all possible functions with the given derivative. \begin{equation}\quad \text { a. }y^{\prime}=x \quad \text { b. } y^{\prime}=x^{2} \quad \text { c. } y^{\prime}=x^{3}\end{equation}

Short Answer

Expert verified
(a) \(y = \frac{x^2}{2} + C \), (b) \(y = \frac{x^3}{3} + C \), (c) \(y = \frac{x^4}{4} + C \).

Step by step solution

01

Analyze the Problem

We need to find functions whose derivatives are given. This involves finding the antiderivative or the indefinite integral of each given function.
02

Solve Part (a)

For part (a), the derivative of the function is given by \( y' = x \). To find the function \( y \), we need to integrate \( x \). The integral of \( x \) with respect to \( x \) is \( \int x \, dx = \frac{x^2}{2} + C \), where \( C \) is a constant of integration.
03

Solve Part (b)

For part (b), we start with the expression \( y' = x^2 \). Integrating \( x^2 \) with respect to \( x \) gives \( \int x^2 \, dx = \frac{x^3}{3} + C \), where \( C \) is a constant of integration.
04

Solve Part (c)

For part (c), the derivative of the function is \( y' = x^3 \). We need to integrate \( x^3 \) to find \( y \). The integral is \( \int x^3 \, dx = \frac{x^4}{4} + C \), where \( C \) is the constant of integration.

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Key Concepts

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

Antiderivative
An antiderivative of a function is essentially the reverse of taking a derivative. If you have a function that you know is a derivative, you can find the original function by calculating its antiderivative. For example, given a derivative like \( y' = x \), we find the original function \( y \) by integrating \( x \).
The process of finding antiderivatives involves indefinite integrals. These integrals provide us with a family of functions that can have the same derivative. The essential idea is that there are many functions that could have led to a given derivative, differing only by a constant. This is why the antiderivative is tied closely to the concept of indefinite integration.
  • The notation for an antiderivative is usually the integral symbol \( \int \).
  • It requires the function we are working with and the variable of integration.
  • The result includes a family of functions, as it is not just a single one.
Antiderivatives play a crucial role in integral calculus and are used widely to solve a large array of problems, especially those involving motion, area, and accumulated quantities.
Constant of Integration
When dealing with indefinite integrals, you'll often come across the term "constant of integration". Imagine you've derived a function, and now you want to find what the original function was. There could be various functions that differentiate to the same result.
This inherent ambiguity necessitates the constant of integration, usually represented as \( C \) in mathematics. Adding \( C \) to the solution recognizes that there might have been a constant in the original function which disappeared upon differentiation.
  • If we integrate \( x \) to find \( y \), we compute \( \int x \ dx = \frac{x^2}{2} + C \).
  • The constant \( C \) summarizes any constant value that could be part of the original function \( y \).
  • It is crucial as it represents an infinite number of possible solutions leading to the same derivative.
Understanding the role of \( C \) is important for solving problems that incorporate real-world data, where initial conditions might specify a particular \( C \) value.
Integral Calculus
Integral calculus involves finding the integral of a function, which can help us determine the area under curves, accumulated changes, and many more practical applications. Two main types of integrals exist: definite and indefinite integrals.
Indefinite integrals do not have limits of integration and represent a family of antiderivatives. This is why they include the constant of integration \( C \). For instance, if \( y' = x^2 \), its indefinite integral is \( \int x^2 \ dx = \frac{x^3}{3} + C \).
  • Indefinite integrals help in reconstructing the original functions from their derivatives.
  • In real-world scenarios like physics, they can describe quantities like displacement, based on velocity.
Integral calculus is an indispensable tool in many fields of study, allowing us to model dynamic systems and solve complex problems by reversing the process of differentiation.

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Most popular questions from this chapter

For \(x>0,\) sketch a curve \(y=f(x)\) that has \(f(1)=0\) and \(f^{\prime}(x)=1 / x .\) Can anything be said about the concavity of such a curve? Give reasons for your answer.

Motion on a line The positions of two particles on the \(s\) -axis are \(s_{1}=\sin t\) and \(s_{2}=\sin (t+\pi / 3),\) with \(s_{1}\) and \(s_{2}\) in meters and \(t\) in seconds. \begin{equation} \begin{array}{l}{\text { a. At what time(s) in the interval } 0 \leq t \leq 2 \pi \text { do the particles }} \\ \quad {\text { meet? }} \\ {\text { b. What is the farthest apart that the particles ever get? }} \\ {\text { c. When in the interval } 0 \leq t \leq 2 \pi \text { is the distance between the }} \\ \quad {\text { particles changing the fastest? }}\end{array} \end{equation}

Distance between two ships At noon, ship \(A\) was 12 nautical miles due north of ship \(B .\) Ship \(A\) was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship \(B\) was sailing east at 8 knots and continued to do so all day. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the distance }} \\ \quad{\text { } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad{\text { noon? One hour later? }} \\\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad{\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal asymptote }} \\ \quad {\text { in the first quadrant. This in turn suggests that } d s / d t} \\ \quad {\text { approaches a limiting value as } t \rightarrow \infty . \text { What is this value? }} \\ \quad {\text { What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? \begin{equation}\begin{array}{l}{\text { Graph } f(x)=x \cos x \text { and its second derivative together for }} \\ {0 \leq x \leq 2 \pi . \text { Comment on the behavior of the graph of } f \text { in }} \\ {\text { relation to the signs and values of } f^{\prime \prime} \text {.}}\end{array}\end{equation}

A cone is formed from a circular piece of material of radius 1 meter by removing a section of angle \(\theta\) and then joining the two straight edges. Determine the largest possible volume for the cone.
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