Explanation Given: According to special relativity, the stone’s mass depends on its velocity divided by c, and the maximum height is given by h ( c ) = c c 2 g 2 + 1 4 − c 2 g A stone tossed up vertically from ground level returns to Earth 1s later Using Newton’s laws, the stone’s maximum height is h = g 8 Einstein’s theory of special relativity reduces to Newtonian mechanics in the limit as c → ∞ . Thus, by applying the limit lim c → ∞ h ( c ) on the equation h ( c ) = c c 2 g 2 + 1 4 − c 2 g , we can find the stone’s maximum height using Einstein’s theory of special relativity. The given equation has a radical; this can be simplified by multiplying and dividing by its conjugate. After simplification, to apply limit at infinity, the infinity properties can be used. Calculations: Description lim c → ∞ h ( c ) ; Step 1: The given equation has a radical; this can be simplified by multiplying and dividing by its conjugate. h ( c ) = c c 2 g 2 + 1 4 − c 2 g = ( c c 2 / g 2 + 1 / 4 − c 2 / g ) ( c c 2 / g 2 + 1 / 4 + c 2 / g ) c c 2 / g 2 + 1 / 4 + c 2 / g Step 2: Using the algebraic identity ( a + b ) ( a − b ) = a 2 − b 2 , we get h ( c ) = ( c c 2 / g 2 + 1 / 4 ) 2 − ( c 2 / g ) 2 c c 2 / g 2 + 1 / 4 + c 2 / g = c 2 ( c 2 / g 2 + 1 / 4 ) − c 4 / g 2 c c 2 / g 2 + 1 / 4 + c 2 / g Step 3: Remove the brackets by distributing c 2 times to each term within the brackets. h ( c ) = c 4 / g 2 + c 2 / 4 − c 4 / g 2 c c 2 / g 2 + 1 / 4 + c 2 / g = c 4 / g 2 + c 2 / 4 − c 4 / g 2 c c 2 / g 2 + 1 / 4 + c 2 / g Step 4: Simplify the denominator by taking out the term c 2 g 2 that is distributed within the square-root. Since the terms is absent in the second term, it is multiplied with the inverse term g 2 c 2

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Chapter 2.7, Problem 42E

To determine

To prove:

limc→∞h(c)=g/8

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Chapter 2.7, Problem 41E

Chapter 2.7, Problem 43E

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Physicists have observed that Einstein's theory of special relativity reduces to Newtonian mechanics in the limit as c → 00, where c is the speed of light. This is illustrated by a stone tossed up vertically from ground level so that it returns to Earth 1 s later. Using Newton's Laws, we find that the stone's maximum height is h = 8/8 m (g = 9.8 m/s²). According to special relativity, the stone's mass depends on its velocity divided by c, and the maximum height is h(c) = c/c2/g² +1/4 – c² /g Prove that lim h(c) = g/8. c00

7. The number of visible sun spots varies rhythmically from 10 to 110 per year over a period of eleven years. The 110 max last occurred in the year 2014. a.) b.) Write a sinusoidal function N (t) to model that number of sun spots per year, where t is in years past the year 2014. What number of sun spots does your function predict for the year 2018?

QU 5 a). A drug in the blood stream decays by according to the equation y'=ky, where y is the concentration of the drug in the bloodstream. If the half- life of the drug is 4 hours, what fraction of the drug remains after 6 hours? b). Using gamma function, evaluate so x e-2x dx c). Shrinks have discovered that there is a limit to the number of meaningless words that a person can learn, no matter how much study time is available, and that the rate of learning is proportional to the number of words not yet learnt. If this limit is 200 words, then the rate of learning satisfies the differential equation dW = k(200 – W) dt where k is a constant and W(t) is the number of words mastered in the first t minutes i). Explain why k is positive ii). Determine the solution to differential equation

Chapter 2 Solutions

Calculus - Standalone book

Ch. 2.1 - Prob. 1PQ Ch. 2.1 - Prob. 2PQ Ch. 2.1 - Prob. 3PQ Ch. 2.1 - Prob. 4PQ Ch. 2.1 - Prob. 5PQ Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E

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