University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

144

Pages:

5
Author:

Herbosambx

Question 1 Line AB contains points A(4,5) and B(9,7). What is the slope of 𝐴 𝐵 ‾ AB ? Answer The slope of the line AB must be determined. 𝐴 = ( 4 , 5 ) , 𝐵 = ( 9 , 7 ) where ( 𝑥 1 , 𝑦 1 ) , ( 𝑥 2 , 𝑦 2 ) A=(4,5),B=(9,7)where(x 1 ,y 1 ),(x 2 ,y 2 ) slope = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 slope= x 2 −x 1 y 2 −y 1 = 7 − 5 9 − 4 = 2 5 = 9−4 7−5 = 5 2 Question 2 What values of 𝑏 b satisfy 3 ( 2 𝑏 + 3 ) 2 = 36 3(2b+3) 2 =36? Answer Solution: 3 ( 2 𝑏 + 3 ) 2 = 36 3(2b+3) 2 =36 ( 2 𝑏 + 3 ) 2 = 12 (2b+3) 2 =12 2 𝑏 + 3 = ± 12 2b+3=± 12 2 𝑏 + 3 = ± 2 3 2b+3=±2 3 𝑏 = − 3 + 2 3 2 and 𝑏 = − 3 − 2 3 2 b= 2 −3+2 3 andb= 2 −3−2 3 Question 3 How do you expand ( 𝑥 − 1 ) 3 (x−1) 3 ? Answer 1 Explanation: Note that ( 𝑧 + 𝑎 ) 3 = 𝑧 3 + 3 𝑧 2 𝑎 + 3 𝑧 𝑎 2 + 𝑎 3 . (z+a) 3 =z 3 +3z 2 a+3za 2 +a 3 . Let 𝑧 = 𝑥 , 𝑎 = − 1 z=x,a=−1: ( 𝑥 − 1 ) 3 = 𝑥 3 + 3 𝑥 2 ( − 1 ) + 3 𝑥 ( − 1 ) 2 + ( − 1 ) 3 (x−1) 3 =x 3 +3x 2 (−1)+3x(−1) 2 +(−1) 3 = 𝑥 3 − 3 𝑥 2 + 3 𝑥 − 1. =x 3 −3x 2 +3x−1. Answer 2 Using the Binomial Theorem: 𝑧 3 + 3 𝑧 2 ( − 1 ) + 3 𝑧 ( − 1 ) 2 + ( − 1 ) 3 z 3 +3z 2 (−1)+3z(−1) 2 +(−1) 3 = 𝑥 3 − 3 𝑥 2 + 3 𝑥 − 1 =x 3 −3x 2 +3x−1 Question 4 Prove that if 𝑚 m and 𝑛 n are integers and 𝑚 𝑛 mn is even, then 𝑚 m is even or 𝑛 n is even. Answer 1 Assume 𝑚 = 2 𝑘 + 1 , 𝑛 = 2 𝑙 , 𝑘 , 𝑙 ∈ 𝑍 m=2k+1,n=2l,k,l∈Z. Then 𝑚 𝑛 = ( 2 𝑘 + 1 ) ( 2 𝑙 ) = 2 ( 2 𝑘 𝑙 + 𝑙 ) , where 𝑟 = 2 𝑘 𝑙 + 𝑙 ∈ 𝑍 . mn=(2k+1)(2l)=2(2kl+l),wherer=2kl+l∈Z. Then 𝑚 𝑛 mn is even. Similarly, when 𝑚 = 2 𝑘 , 𝑛 = 2 𝑙 + 1 , 𝑘 , 𝑙 ∈ 𝑍 m=2k,n=2l+1,k,l∈Z: 𝑚 𝑛 = 2 𝑟 , 𝑟 ∈ 𝑍 . mn=2r,r∈Z. So 𝑚 𝑛 mn is even. Answer 2 Step 1: Proof by contradiction. Assume 𝑚 m and 𝑛 n are odd integers. 𝑚 = 2 𝑎 + 1 , 𝑛 = 2 𝑏 + 1. m=2a+1,n=2b+1. Calculate 𝑚 𝑛 = ( 2 𝑎 + 1 ) ( 2 𝑏 + 1 ) = 4 𝑎 𝑏 + 2 𝑎 + 2 𝑏 + 1. mn=(2a+1)(2b+1)=4ab+2a+2b+1. This is odd, contradicting the assumption. Hence, 𝑚 m or 𝑛 n must be even. Answer 3 By assuming the negation: 𝑚 = 2 𝑎 + 1 , 𝑛 = 2 𝑏 + 1. m=2a+1,n=2b+1. Conclude by contradiction that 𝑚 𝑛 mn cannot be even. Question 5 Find an equation of a parabola that has curvature 4 at the origin. Answer 1 Assume an equation 𝑓 ( 𝑥 ) = 𝑎 𝑥 2 , 𝑎 ≠ 0. f(x)=ax 2 ,a  =0. Calculate the curvature: 𝑘 = ∣ 𝑓 ′ ′ ( 𝑥 ) ∣ [ 1 + ( 𝑓 ′ ( 𝑥 ) ) 2 ] 3 / 2 k= [1+(f ′ (x)) 2 ] 3/2 ∣f ′′ (x)∣ 𝑓 ′ ( 𝑥 ) = 2 𝑎 𝑥 , 𝑓 ′ ′ ( 𝑥 ) = 2 𝑎 . f ′ (x)=2ax,f ′′ (x)=2a. 𝑘 = ∣ 2 𝑎 ∣ [ 1 + ( 2 𝑎 𝑥 ) 2 ] 3 / 2 k= [1+(2ax) 2 ] 3/2 ∣2a∣ At 𝑥 = 0 x=0, curvature is 𝑘 ( 0 ) = 4. k(0)=4. Solving, we get 𝑎 = 2. a=2. Thus, the parabola is 𝑓 ( 𝑥 ) = 2 𝑥 2 f(x)=2x 2 . Answer 2 Using 𝜅 = 1 ∣ 𝑎 ∣ , κ= ∣a∣ 1 , solve for 𝑎 = 1 4 a= 4 1 . Substitute to find 𝑓 ( 𝑥 ) = 1 4 𝑥 2 . f(x)= 4 1 x 2 .