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Faster than a calculator Use the approximation \((1+x)^{k} \approx\) \(1+k x\) to estimate the following a. \((1.0002)^{50} \quad\) b. \(\sqrt[3]{1.009}\)

Short Answer

Expert verified
a. 1.01; b. 1.003.

Step by step solution

01

Understand the Approximation Formula

The approximation formula \[(1+x)^k \approx 1 + kx\]holds for small values of \(x\). This means when \(x\) is close to zero, we can estimate \((1+x)^k\) using this simpler expression.
02

Apply the Formula to (1.0002)鈦碘伆

In this problem, let \(x = 0.0002\) and \(k = 50\). Apply the formula:\[(1+0.0002)^{50} \approx 1 + 50 \times 0.0002.\]Calculating the right-hand side:\[1 + 50 \times 0.0002 = 1 + 0.01 = 1.01.\]Therefore, \((1.0002)^{50} \approx 1.01.\)
03

Express Cube Root as Exponent

The expression \(\sqrt[3]{1.009}\) can be written as \((1.009)^{1/3}\). We will use our approximation formula on this expression.
04

Apply the Formula to Cube Root of 1.009

First, express \((1.009)^{1/3}\) in terms of \((1+x)^k\):- Here, let \(x = 0.009\) and \(k = \frac{1}{3}\).Now, apply the approximation:\[(1+0.009)^{1/3} \approx 1 + \frac{1}{3} \times 0.009.\]Calculating the right-hand side:\[1 + \frac{1}{3} \times 0.009 = 1 + 0.003 = 1.003.\]Therefore, \(\sqrt[3]{1.009} \approx 1.003.\)

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Key Concepts

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

Understanding Approximation
Approximation is a powerful tool in calculus that helps us estimate complex calculations easily. When we use approximation, we aim to simplify a problem by finding a nearby value that is easy to compute. One common approximation formula is \[(1+x)^k \approx 1 + kx.\]
This is particularly useful when dealing with small values of \(x\), as shown in our problem. The smaller \(x\) is, the closer the approximation is to the true value.
The formula works well by expanding \((1+x)^k\) into its series form and then simplifying it by truncating the series after the linear term. This gives a linear relation for estimates, keeping calculations simple and intuitive. By choosing an appropriate \(x\) and \(k\), we can apply it to various scenarios, including mathematical functions like exponents and roots.
Basics of Exponents
Exponents are an essential part of mathematics that describe how many times a number, known as the base, is multiplied by itself. For instance, \((a)^n\) means multiplying \(a\) by itself \(n\) times. It's a shortcut for repeated multiplication.
Exponents follow certain rules:
  • Power of a Power: \((a^m)^n = a^{mn}\)
  • Product of Powers: \(a^m \times a^n = a^{m+n}\)
  • Division of Powers: \(\frac{a^m}{a^n} = a^{m-n}\)
  • Negative Exponent: \(a^{-n} = \frac{1}{a^n}\)
In our context, exponents simplify expressions involving cube roots and allow us to use approximation techniques effectively. For instance, writing a cube root as an exponent, \(\sqrt[3]{x} = x^{1/3}\), enables us to apply formulas and approximations. It's crucial to correctly interpret expressions using exponents to avoid errors and ensure precise calculations.
Working with Cube Roots
Cube roots are a type of radical expression where we find a number \(x\) such that when it is multiplied by itself three times, it results in the given number. In mathematical terms, the cube root of \(x\) is \(x^{1/3}\).
The cube root can often be cumbersome to calculate, but expressing it in terms of fractions, as an exponent, makes it more manageable. This expression, \(x^{1/3}\), allows mathematicians to use exponent rules and approximation formulas to estimate values. For example, estimating \(\sqrt[3]{1.009}\) becomes a straightforward task using the approximation formula. By recognizing the cube root as an exponent \(\frac{1}{3}\), it's easy to calculate an estimate with less effort.
Understanding cube roots through exponents simplifies mathematical operations, making calculations of roots faster and more accurate when applying approximation techniques. This conversion is especially helpful in solving complex equations and determining solutions quickly.

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Most popular questions from this chapter

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2} $$

In Exercises \(81-86,\) find a parametrization for the curve. the ray (half line) with initial point \((2,3)\) that passes through the point \((-1,-1)\)

Which of the following could be true if \(f^{\prime \prime}(x)=x^{-1 / 3} ?\) $$ \begin{array}{ll}{\text { a. } f(x)=\frac{3}{2} x^{2 / 3}-3} & {\text { b. } f(x)=\frac{9}{10} x^{5 / 3}-7} \\ {\text { c. } f^{\prime \prime \prime}(x)=-\frac{1}{3} x^{-4 / 3}} & {\text { d. } f^{\prime}(x)=\frac{3}{2} x^{2 / 3}+6}\end{array} $$

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ x+\tan \left(\frac{y}{x}\right)=2, \quad P\left(1, \frac{\pi}{4}\right) $$

The derivative of \(\cos \left(x^{2}\right)\) Graph \(y=-2 x \sin \left(x^{2}\right)\) for \(-2 \leq x \leq 3 .\) Then, on the same screen, graph $$ y=\frac{\cos \left((x+h)^{2}\right)-\cos \left(x^{2}\right)}{h} $$ for \(h=1.0,0.7,\) and \(0.3 .\) Experiment with other values of \(h .\) What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

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