Questions and Answers #9 Modeling

Questions and Answers Sheet 9 Modeling Question #1 a) The interval notation for the expression 𝑥 ≤ 5 . Answer: Interval notetion is a simplified form of writing the solution to an inequality or system of inequalities, using the bracket and perenthesis symbols in lieu of the inequality symbols. Intervals with parentheses are called open intervals, meaning the variable cannot have the value of the endpoints. Formula used: Inequality Interval notation xa ( a, ) xa (−, a) xa [ a,  ) xa (−, a] a xb (a, b) a xb [a, b) a xb (a, b] a xb [a, b] Description x is greater than a x x is less than a x is greater than or equal to a x is less than or equal to a x is strictly between a and b x is between a and b, including a x is between a and b, including b x is between a and b, including a and b Conclusion: The interval notation for the expression ≤ 5 is given by (−∞, 5]. Question #2 The expression √7 √28 Answer: 𝑎 √𝑎 Concept used: Given expression can be simplified using the property √ = , 𝑎, 𝑏 ≥ 0. 𝑏 √𝑏 Calculation: Given expression is √7 √28 =√ =√ = =√ 7 4×7 Factor 28 √1 √2×2 1 2 Factor 4 Square root property Therefore, √7 √28 Conclusion: √7 √28 = 1 2 = 1 2 𝑎 √𝑎 . Since 7, 28>0, therefore, by the property, √ = , 𝑏 √𝑏 1 4 √4 1 = 7 28 √7 √28 Question #3 Radical expression by factoring out largest perfect n-th power 4 Given: √64 Answer: Calculation: Step 1 Split into factors 4 4 √64 = √2 × 2 × 2 × 2 × 2 × 2 4 = √(2 × 2 × 2 × 2) × 2 × 2 4 = √(2)4 × 2 × 2 4 = √(2)4 × 4 2 is the perfect 4th power factor Step 2 𝑛 𝑛 𝑛 Using product property √𝑎 ⋅ 𝑏 = √𝑎 ⋅ √𝑏 4 4 4 √(2)4 × 4 = √(2)4 ⋅ √4 Step 3 1 𝑎 Using rule √𝑥 = 𝑥 𝑎 to take the perfect 4th power factor out of the radical. 4 4 √(2)4 = 24 = 21 = 2 Step 4 4 4 4 √(2)4 ⋅ √4 = 2 √4 Therefore, we have 4 4 √64 = 2 √4 4 4 Conclusion: √64 = 2 √4 Question #4 a) Find the mass density of a proton, modeling it as a solid sphere of radius 1.00 × 10 − 15𝑚. b) What If ? Consider a classical model of an electron as a uniform solid sphere with the same density as the proton. Find its radius. c) Imagine that this electron possesses spin angular momentum 𝑙𝑤 = ℎ 2 because of classical rotation about the z axis. Determine the speed of a point on the equator of the electron. d) State how this speed compares with the speed of light. Answer: Step 1 a) Let's us write the mass in terms of the density 𝜌 and volume V: 𝑚 = 𝜌𝑉 The volume of a spere is given by 4 𝑉 = 𝑟3 𝜋 3 so that we have 4 𝑚 = 𝜌 𝑟3𝜋 ⇒ 𝜌 = 3 3𝑚 4𝑟 3 𝜋 Using the mass of the proton and the given radius we obtain 𝜌= 3⋅(1.67⋅−27 𝑘𝑔) 4(1.00⋅10−15 𝑚)3 𝜋 ⇒ 𝜌 = 3.99 ⋅ 1017 𝑘𝑔 𝑚3 b) We can express the radius from eq. (1): 1 𝑟=( 3𝑚 3 ) 4𝜌𝜋 Inserting the electron mass and the density found in part (a) we get 1 3 3 ⋅ (9.11 ⋅ 10−31 𝑘𝑔) ) 𝑘𝑔 4 (3.99 ⋅ 1017 3 ) 𝜋 𝑚 ⇒ 𝑟 = 8.17 ⋅ 10−17 𝑚 𝑟=( Step 2 c) The angular speed can be written as 𝜔 = 𝐿𝑧 = 𝐼𝜔 = 𝐼 𝑣 𝑟 The problem states that this is equal to 𝑣 ℎ 𝑟 2 𝐼 = ⇒𝑣= ℎ 2 𝑣 𝑟 so that we have : 𝑟ℎ 2𝐼 We found the radius in part (b) but what about the moment of inertia? Observe that we have modeled an electron as a solid shere which means that its moment of inertia can be expressed as: 2 𝐼 = 𝑚𝑟 2 5 Therefore, we can write the speed as 𝑣= 𝑟ℎ 2 5 2 𝑚𝑟 2 = 5ℎ 4𝑚𝑟 Inserting the numbers we get 5 ⋅ (1.055 ⋅ 10−34 𝐽 ⋅ 𝑠) 𝑣= 4 ⋅ (9.11 ⋅ 10−31 𝑘𝑔)(8.17 ⋅ 10−17 𝑚) 𝑚 ⇒ 𝑣 = 1.77 ⋅ 1017 𝑠 Step 3 The speed found above is greater than the speed of light. This alone indicates that the model is incorrect. Question #5 The differince quotient for function 𝑓(𝑥) = 𝑚𝑥 + 𝑏 Answer: Formula used: Difference quotient of a function f(x) can be found as: Difference quotient = 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ Calculation: First of all, we find an expression for f(x+h). In order to find the expression for f(x+h), we replace x by 𝑥 + ℎ𝑖𝑛𝑓(𝑥). 𝑓(𝑥 + ℎ) = 𝑚(𝑥 + ℎ) + 𝑏 𝑓(𝑥 + ℎ) = 𝑚𝑥 + 𝑚ℎ + 𝑏 On plugging the expression for f(x+h) and f(x) in the difference quotient formula, we get: Difference quotient = Difference quotient = 𝑚𝑥+𝑚ℎ+𝑏−(𝑚𝑥+𝑏) ℎ 𝑚𝑥+𝑚ℎ+𝑏−𝑚𝑥−𝑏 ℎ Difference quotient = 𝑚ℎ ℎ Difference quotient =m. Conclusion: On plugging the given linear function in the difference quotient formula, we get the difference quotient of the function f(x)=mx+b as m. Question #6 Average rate of change 𝑓(𝑥) = 𝑥 2 − 3𝑥 from x=-3 to x=-1 Answer: Given: The function f is given as 𝑓(𝑥) = 𝑥 2 − 3𝑥 , x values are given as x=-3, x=-1 We have to find the average rate of change of 'f' between x=-3 and x=-1 . For that initially we have to find f(x) for x=-3 and x=-1. Then we have to find the average rate of change of f(x) between x=-3 and x=-1 . Formula used: 𝐴(𝑥) = 𝑓(𝑥2 )−𝑓(𝑥1 ) 𝑥2 −𝑥1 Calculation: Description f ( x) = x 2 − 3x Step 1: Substitutex = −3 in f ( x) f (−3) = (3− −3(−3) = 9 + 9 = 18 Step 2:Substitutex = −1in f ( x) f (−1) = (1− −3(−1) = 1 + 3 = 4 Step 3: Substitute the values of f ( −3) and f ( −1) in the average rate of change formula to find the final answer. A( x) = f ( x2 ) − f ( x1 ) x2 − x1 A( x) = f (−1) − f (−3) −1 − (−3) = 4 − 18 A( x) −1 + 3 = 4 − 18 −14 = = −7 A( x) = −7 −1 + 3 2 Conclusion: Average rate of change of 𝑓(𝑥) = 𝑥 2 − 3𝑥 from 𝑥 = −3 to 𝑥 = −1 is equal to -7. Question #7 The interval notation of the given inequality GIven: 2𝑥−5 2 < 5𝑥+1 5 Answer: Formula used: The interval notation for 𝑥 > 𝑎 is (𝑎, ∞) The interval notation for 𝑥 ≥ 𝑎 is [𝑎, ∞) The interval notation for The interval notation for The interval notation for The interval notation for The interval notation for The interval notation for Calculations: Here 2𝑥−5 2 < 𝑥 < 𝑎 is (−∞, 𝑎) 𝑥 ≤ 𝑎 is (−∞, 𝑎] 𝑎 < 𝑥 is (𝑎, 𝑏) 𝑎 ≤ 𝑥 ≤ 𝑏 is [𝑎, 𝑏] 𝑎 ≤ 𝑥 < 𝑏 is [𝑎, 𝑏) 𝑎 < 𝑥 ≤ 𝑏 is (𝑎, 𝑏] 5𝑥+1 5 First we shall solve it using algebraic method. Cross multiplying we get 5(2𝑥 − 5) < 2(5𝑥 + 1) Distributing on both sides we get 5(2𝑥) − 5(5) ≤ 2(5𝑥) + 2(1) 10𝑥 − 25 ≤ 10𝑥 + 2 Subtracting 10x from both sides 10𝑥 − 25 − 10𝑥 ≤ 10𝑥 + 2 − 10𝑥 −25 ≤ 2 Hence −25 < 2 is always true. Hence the value of x can be any real number. The interval notation for the value of x shall be (−∞, ∞) Conclusion: The interval notation for the value of 2𝑥−5 2 < 5𝑥+1 5 shall be (−∞, ∞) Question #8 Evaluare the expression A combination is an arrangement of items but not in particular order. Any arrangement of r objects selected from a single group of n possible objects. The order of arrangement is not important in combination. Here in our problem, (𝑛 = 4), (𝑟 = 2), we need to substitute it in the formula. Answer: Formula used: Combination formula 𝐶(𝑛, 𝑟) = 𝑛! (𝑛 − 𝑟)! 𝑟! , where n is the total number of objects and r is the number of objects seslected. Calculation: Description 𝐶(4,2) Step1:Substitute the value of n and r in the equation formula. 𝐶(4,2) = 4! (4 − 2)! 2! = 4! (2)! 2! = 4 × 3 × 2! 2! × 2 × 1 = 2 × 31𝐶(4,2) = 6 Conclusion: The combination of the expression is 6. Question #9 The fourth term of (2𝑥 + 𝑦)5 A binomial is a polynomial which contains two terms. Algebraically, sum or difference of two monomials is a binomial. Binomial theorem is used to expand the binomials to any given power without direct multiplication. According to the theorem, (𝑥 + 𝑦)𝑛 = ∑ 𝑘 = 0𝑛𝑛𝐶𝑘𝑥𝑛 − 𝑘𝑦𝑘 where 𝑛𝐶𝑘 = 𝐶(𝑛, 𝑘) = 𝑛! (𝑛 − 𝑘)! 𝑘! is the binomial coefficient in the theorem. It represents the number of ways in which k unordered outcomes can be selected from a total of n possibilities. The formula to find rth term is given as 𝑡𝑟 = 𝑛𝐶𝑟 − 1𝑥𝑛 − 𝑟 + 1𝑦𝑟 − 1 , where n is the exponent of the binomial, r is the number of the term to be found, x is the first term of the binomial and y is the second term of the binomial. Answer: Formula used: The formula to find rth term is given as 𝑡𝑟 = 𝑛𝐶𝑟 − 1𝑥𝑛 − 𝑟 + 1𝑦𝑟 − 1 , where n is the exponent of the exponent of the binomial, r is the number of the term to be found x is the first term of the binomial and y is the second term of the binomial. Calculation: Description 𝑡𝑟 = 𝑛𝐶𝑟 − 1𝑥𝑛 − 𝑟 + 1𝑦𝑟 − 1 Step 1: Here we have to assign values as. 𝑥 = 2𝑥 𝑎𝑛𝑑 𝑦 = 𝑦 Step 2: We have to find 4th term, that means r=4. For the given problem, 𝑛 = 5. We substitute n and r values to find the 4th term 𝑡4 = 5𝐶4 − 1(2𝑥)5 − 4 + 1𝑦4 − 1 = 5𝐶3(2𝑥)2𝑦3𝑡4 = 5 × 4 × 31 × 2 × 3(4𝑥2)𝑦3 = 10(4𝑥2)𝑦3𝑡4 = 40𝑥2𝑦3 Conclusion: The 4th term of (2𝑥 + 𝑦)5 𝑖𝑠 40𝑥2𝑦3 . Question #10 All positive integers n for which the given statement 3𝑛 > 6𝑛 is not true. Answer: Given: 3n>6n First we have to consider the given statement as P(n). P(n), Proof of induction is a mathematical statement involving a value n. We have to find all positive integers n for which the given statement 3𝑛 > 6𝑛 is not true. For that we have to Substitute values of n as 1,2,3 ... to check for which value P(n) is true. Calculation: Description 3n>6n Step 1: Substitute values of n as 1, 2, 3... to check for which value P(n) is true. 𝑃(𝑛): 3𝑛 > 6𝑛𝑃(1): 31 > 6(1) ⇒ 3 > 6 Is not true. 𝑃(2): 32 > 6(2) ⇒ 9 > 12 is not true. 𝑃(3): 33 > 6(3) ⇒ 27 > 18 is true Conclusion: All positive integers n for which the given statement 3𝑛 > 6𝑛 is not true are 1,2.