/*! 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 73 Find \(f\). $$f^{\prime}(t)=\s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 73

Find \(f\). $$f^{\prime}(t)=\sqrt{t}+\frac{1}{\sqrt{t}}, \quad f(4)=0$$

Short Answer

Expert verified
The function is \(f(t) = \frac{2}{3} t^{3/2} + 2 t^{1/2} - \frac{28}{3}\).

Step by step solution

01

Integrate the derivative

To find the function \(f(t)\), integrate the given derivative \(f^{\prime}(t)\).\[\int f^{\prime}(t) \, dt = \int \left( \sqrt{t} + \frac{1}{\sqrt{t}} \right) \, dt\]Integrate each term separately.
02

Integrate the first term

Integrate \(\sqrt{t}\). Recall that \(\sqrt{t} = t^{1/2}\).\[\int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} + C = \frac{2}{3} t^{3/2} + C_1\]
03

Integrate the second term

Integrate \(\frac{1}{\sqrt{t}}\). Recall that \(\frac{1}{\sqrt{t}} = t^{-1/2}\).\[\int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} + C = 2 t^{1/2} + C_2\]
04

Combine the results

Combine the integrations. Note that the constants of integration can be combined into a single constant.\[f(t) = \frac{2}{3} t^{3/2} + 2 t^{1/2} + C\]
05

Apply the initial condition

Use the initial condition \(f(4) = 0\) to find the constant \(C\).\[f(4) = \frac{2}{3} (4)^{3/2} + 2 \cdot (4)^{1/2} + C = 0\]Evaluate \((4)^{3/2}\) and \((4)^{1/2}\) and solve for \(C\).
06

Simplify the equation with the initial condition

Simplify:\[f(4) = \frac{2}{3} \cdot 8 + 2 \cdot 2 + C = 0\]\[\frac{16}{3} + 4 + C = 0\]\[\frac{16}{3} + \frac{12}{3} + C = 0\]\[\frac{28}{3} + C = 0\]\[C = -\frac{28}{3}\]Thus, the full function is:\[f(t) = \frac{2}{3} t^{3/2} + 2 t^{1/2} - \frac{28}{3}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Integral Calculus
To find the original function from its derivative using integral calculus, you need to integrate the derivative. Integration is essentially the reverse process of differentiation.
When given a derivative like in this exercise, the aim is to find the primitive function, also known as the antiderivative.
For the given derivative \(f^{\prime}(t) = \sqrt{t} + \frac{1}{\sqrt{t}}\), you'll integrate each term separately.
Initial Conditions
When you're solving an indefinite integral, a constant of integration, typically denoted as \(C\), appears.
It's because the derivative of a constant is zero, meaning there are infinitely many possible antiderivatives.
Initial conditions help pinpoint the specific solution.
In our exercise, the initial condition is \(f(4) = 0\), which tells us the exact value of \(C\).
Applying the initial conditions involves substituting the given values into the integrated function to solve for \(C\). This ensures the function exactly matches the conditions provided.
Function Reconstruction
Function reconstruction means finding the original function from its derivative.
After integrating the derivative and obtaining a general form of the function, you then use initial conditions to determine any constants of integration.
In our exercise, we combined the integrations of individual terms and then solved for \(C\) using the condition \(f(4) = 0\).
This gave us the specific function \(f(t) = \frac{2}{3} t^{3/2} + 2 t^{1/2} - \frac{28}{3}\).
Each term in this function corresponds to parts of the original derivative, reconstructed through integration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the area under the graph of each function over the given interval. $$y=e^{x} ; \quad[-1,5]$$

A firm has the marginalprofit function \(\frac{d P}{d x}=\frac{9000-3000 x}{\left(x^{2}-6 x+10\right)^{2}}\) (GRAPH CAN'T COPY) Find the total-profit function given that \(P=1500\)dollars at \(x=3\).

Evaluate. $$\int_{4}^{16}(x-1) \sqrt{x} d x$$

The annual rate of change in the national credit market debt (in billions of dollars per year) can be modeled by the function $$D^{\prime}(t)=857.98+829.66 t-197.34 t^{2}+15.36 t^{3}$$ where \(t\) is the number of years since \(1995 .\) (Source: Federal Reserve System.) Use the preceding information. By how much did the credit market debt increase between 1996 and \(2000 ?\)

V. King Manufacturing buys a new machine for 250,000 dollars. The marginal revenue from the sale of products produced by the machine after \(t\) years is given by \(R^{\prime}(t)=4000 t\) The salvage value of the machine, in dollars, after \(t\) years is given by \(V(t)=200,000-25,000 e^{0.1 t}\) The total profit from the machine, in dollars, after \(t\) years is given by $$P(t)=\left(\begin{array}{c}\text { Revenue } \\\\\text { from } \\\\\text { sale of } \\\\\text { product } \end{array}\right)+\left(\begin{array}{c}\text { Revenue } \\\\\text { from } \\\\\text { sale of } \\\\\text { machine }\end{array}\right)-\left(\begin{array}{c}\text { cost } \\\\\text { of } \\\\\text { machine } \end{array}\right)$$ The company knows that \(R(0)=0\) a) Find \(P(t)\) b) Find \(P(10)\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.