/*! 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 29 Use linear approximations to est... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 29

Use linear approximations to estimate the following quantities. Choose a value of a to produce a small error. $$1 / \sqrt[3]{510}$$

Short Answer

Expert verified
Question: Estimate the value of $$\frac{1}{\sqrt[3]{510}}$$ using linear approximations. Answer: Approximately $$\frac{2057}{16464}$$

Step by step solution

01

Choose 'a' to minimize the error

Choose a value of 'a' that is easy to compute the cube root and is close to 510. In this case, the closest value is $$a=512$$, as $$\sqrt[3]{512} = 8$$.
02

Define the function

Define the function $$f(x) = \frac{1}{\sqrt[3]{x}}$$ that we're trying to approximate.
03

Compute the derivative

Compute the derivative of the function $$f'(x)$$ to find the slope of the tangent line. Using the chain rule and power rule, we find that: $$f'(x)= -\frac{1}{3x^{\frac{4}{3}}}$$
04

Calculate the slope at 'a'

Calculate the slope of the tangent line at the point a = 512 using the derivative function found in Step 3: $$f'(512)= -\frac{1}{3(512)^{\frac{4}{3}}} = -\frac{1}{3(4096)} = -\frac{1}{12288}$$
05

Find equation of the tangent line

Find the equation of the tangent line to the function at the point (a, f(a)): $$L(x) = f(a) + f'(a)(x-a) = \frac{1}{\sqrt[3]{512}} - \frac{1}{12288}(x-512) = \frac{1}{8} - \frac{1}{12288}(x-512)$$
06

Estimate the value using the tangent line

Substitute the value x=510 in the tangent line equation found in Step 5 to find the approximation for the given expression: $$L(510) = \frac{1}{8} - \frac{1}{12288}(510-512) = \frac{1}{8} + \frac{2}{12288} \approx \frac{2057}{16464}$$ So, the approximate value of $$\frac{1}{\sqrt[3]{510}}$$ using linear approximations is $$\frac{2057}{16464}$$.

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!

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

Concavity of parabolas Consider the general parabola described by the function \(f(x)=a x^{2}+b x+c .\) For what values of \(a, b,\) and \(c\) is \(f\) concave up? For what values of \(a, b,\) and \(c\) is \(f\) concave down?

A large tank is filled with water when an outflow valve is opened at \(t=0 .\) Water flows out at a rate, in gal/min, given by \(Q^{\prime}(t)=0.1\left(100-t^{2}\right),\) for \(0 \leq t \leq 10\). a. Find the amount of water \(Q(t)\) that has flowed out of the tank after \(t\) minutes, given the initial condition \(Q(0)=0\) b. Graph the flow function \(Q,\) for \(0 \leq t \leq 10\) c. How much water flows out of the tank in 10 min?

A mass oscillates up and down on the end of a spring. Find its position \(s\) relative to the equilibrium position if its acceleration is \(a(t)=\sin (\pi t),\) and its initial velocity and position are \(v(0)=3\) and \(s(0)=0,\) respectively.

Give an argument to support the claim that if a function is concave up at a point, then the tangent line at that point lies below the curve near that point.

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\csc ^{2} \theta+1\right) d \theta$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

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.