Lapse rates in the atmosphere Refer to Example 2. Concurrent measurements indicate that at an elevation of 6.1 km, the temperature is -10.3° C and at an elevation of 3.2km , the temperature is 8.0°C . Based on the Mean Value Theorem, can you conclude that the lapse rate exceeds the threshold value of 7°C/ km at some intermediate elevation? Explain.
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Differentials
Problem 4.8.19
Textbook Question
{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.
f(x) = cos⁻¹ x - x; x₀ = 0.75

1
Step 1: Understand Newton's Method. Newton's method is an iterative process used to find successively better approximations to the roots (or zeroes) of a real-valued function. The formula for Newton's method is: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mi>x</mi><mo>-</mo><mfrac><mrow><mi>f</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>f</mi><mo>'</mo><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mfrac></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mo>(</mo><mi>x</mi><mo>)</mo></math> is the derivative of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>(</mo><mi>x</mi><mo>)</mo></math>.
Step 2: Calculate the derivative of the function. Given <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>(</mo><mi>x</mi><mo>)</mo><mo>=</mo><mo>cos</mo><mo>-</mo><mo>1</mo><mo>(</mo><mi>x</mi><mo>)</mo><mo>-</mo><mi>x</mi></math>, find <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mo>(</mo><mi>x</mi><mo>)</mo></math>. The derivative of <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>cos</mo><mo>-</mo><mo>1</mo><mo>(</mo><mi>x</mi><mo>)</mo></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mfrac><mn>1</mn><msqrt><mrow><mn>1</mn><mo>-</mo><msup><mi>x</mi><mn>2</mn></msup></mrow></msqrt></mfrac></math>, and the derivative of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn></math>. Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mo>(</mo><mi>x</mi><mo>)</mo><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><msqrt><mrow><mn>1</mn><mo>-</mo><msup><mi>x</mi><mn>2</mn></msup></mrow></msqrt></mfrac><mo>-</mo><mn>1</mn></math>.
Step 3: Apply Newton's method using the initial approximation <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mn>0.75</mn></math>. Substitute <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mn>0.75</mn></math> into the Newton's formula: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mn>0.75</mn><mo>-</mo><mfrac><mrow><mi>f</mi><mo>(</mo><mn>0.75</mn><mo>)</mo></mrow><mrow><mi>f</mi><mo>'</mo><mo>(</mo><mn>0.75</mn><mo>)</mo></mrow></mfrac></math>. Calculate <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>(</mo><mn>0.75</mn><mo>)</mo></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mo>(</mo><mn>0.75</mn><mo>)</mo></math>.
Step 4: Compute the next approximation. Use the values from Step 3 to find the next approximation <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mi>x</mi><mo>-</mo><mfrac><mrow><mi>f</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>f</mi><mo>'</mo><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mfrac></math>. Repeat this process iteratively, updating <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi></math> each time.
Step 5: Continue iterations until two successive approximations agree to five decimal places. Create a table to track each iteration, showing the values of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi></math>, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>(</mo><mi>x</mi><mo>)</mo></math>, and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>'</mo><mo>(</mo><mi>x</mi><mo>)</mo></math> for each step. Stop when the difference between successive approximations is less than <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0.00001</mn></math>.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Newton's Method
Newton's Method is an iterative numerical technique used to find approximate roots of a real-valued function. It starts with an initial guess and refines it using the formula x₁ = x₀ - f(x₀)/f'(x₀), where f' is the derivative of f. This process continues until the difference between successive approximations is sufficiently small, indicating convergence to a root.
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Evaluating Composed Functions
Convergence Criteria
In numerical methods, convergence criteria determine when to stop the iterative process. For Newton's Method, this often involves checking if the absolute difference between two successive approximations is less than a specified tolerance level, such as 0.00001 for five decimal places. This ensures that the approximations are accurate enough for practical purposes.
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Derivatives
Function and Derivative Evaluation
To apply Newton's Method, one must evaluate both the function f and its derivative f' at each iteration. For the given function f(x) = cos⁻¹ x - x, the derivative f'(x) = -1/√(1-x²) - 1 must be computed. Accurate evaluation of these functions is crucial for the method's success, as errors in these calculations can lead to incorrect approximations of the root.
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Evaluating Composed Functions
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