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

2024
Views:

70

Pages:

3
Author:

Herbosambx

Question 1 Solve the following system of equations and show all work: 𝑦 = 𝑥 2 + 3 y=x 2 +3 𝑦 = 𝑥 + 5 y=x+5 Answer We have to solve this: 𝑦 = 𝑥 2 + 3 y=x 2 +3 𝑦 = 𝑥 + 5 y=x+5 𝑥 2 + 3 = 𝑥 + 5 x 2 +3=x+5 𝑥 2 − 𝑥 − 2 = 0 x 2 −x−2=0 𝑥 2 − 2 𝑥 + 𝑥 − 2 = 0 x 2 −2x+x−2=0 ( 𝑥 − 2 ) ( 𝑥 + 1 ) = 0 ⇒ ( 𝑥 + 1 ) ( 𝑥 − 2 ) = 0 (x−2)(x+1)=0⇒(x+1)(x−2)=0 𝑥 + 1 = 0 or 𝑥 − 2 = 0 x+1=0orx−2=0 𝑥 = − 1 , 2 x=−1,2 Question 2 If 𝑓 ( 2 ) = 3 f(2)=3 and 𝑓 ′ ( 2 ) = 5 f ′ (2)=5, find an equation of: (a) the tangent line, and (b) the normal line to the graph of 𝑦 = 𝑓 ( 𝑥 ) y=f(x) at the point where 𝑥 = 2 x=2. Answer From part (a) and the fact that 𝑓 ′ ( 2 ) = 5 f ′ (2)=5, we know that the slope of the tangent line is 5. So the slope of the normal line is: 𝑚 = − 1 5 m=− 5 1 Since we also have 1 point ( 2 , 𝑓 ( 2 ) ) = ( 2 , 3 ) (2,f(2))=(2,3), we can use the equation of the line to determine the normal line at 𝑥 = 2 x=2: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) y−y 1 =m(x−x 1 ) 𝑦 − 3 = − 1 5 ( 𝑥 − 2 ) y−3=− 5 1 (x−2) 𝑦 = − 1 5 𝑥 + 2 5 + 3 y=− 5 1 x+ 5 2 +3 𝑦 = − 1 5 𝑥 + 17 5 y=− 5 1 x+ 5 17 Result: The normal line to the graph of 𝑦 = 𝑓 ( 𝑥 ) y=f(x) at 𝑥 = 2 x=2 is: 𝑦 = − 1 5 𝑥 + 17 5 y=− 5 1 x+ 5 17 Question 3 In a typical lightning strike, 2.5 C 2.5C flows from cloud to ground in 0.20 ms 0.20ms. What is the current during the strike? Answer You must use the following equation to solve this issue: 𝐼 = 𝑄 𝑡 I= t Q 𝐼 = 2.5 C 0.0002 s = 12500 A ± 5 I= 0.0002s 2.5C =12500A±5 Significant Figures: 𝐼 = 13000 Amps I=13000Amps. Question 4 How do you solve the system of equations 𝑥 − 𝑦 = 7 x−y=7 and − 2 𝑥 + 5 𝑦 = − 8 −2x+5y=−8? Answer 1 Your goal here is to remove one of the variables from the problem. You can see that the first equation has 𝑥 x and the second equation has − 2 𝑥 −2x. If we double the first equation, we get: 2 𝑥 − 2 𝑦 = 14 2x−2y=14 Then we simply add that to the second equation: − 2 𝑥 + 5 𝑦 = − 8 −2x+5y=−8 + 2 𝑥 − 2 𝑦 = 14 +2x−2y=14 3 𝑦 = 6 3y=6 Divide both sides by 3: 𝑦 = 2 y=2 Last, just plug 2 2 in for 𝑦 y in either equation (I). Question 5 Determine whether the ordered pair is a solution to the given system of linear equations: 3 𝑥 − 𝑦 = 1 2 𝑥 + 3 𝑦 = 8 3x−y=1 2x+3y=8 Answer Step 1: Determine whether the ordered pair is a solution to the given system of linear equations. Step 2: Consider the system of equations as: 3 𝑥 − 𝑦 = 1 (1) 3x−y=1(1) 2 𝑥 + 3 𝑦 = 8 (2) 2x+3y=8(2) Verify if ordered pair ( 1 , 2 ) (1,2) is a solution of (1) and (2): Substitute 𝑥 = 1 x=1 and 𝑦 = 2 y=2 in equation (1): 3 ( 1 ) − 2 = 1 3(1)−2=1 3 − 2 = 1 3−2=1 1 = 1 (True) 1=1(True) Hence, ( 1 , 2 ) (1,2) satisfies the equation (1). Substitute 𝑥 = 1 x=1 and 𝑦 = 2 y=2 in equation (2): 2 ( 1 ) + 3 ( 2 ) = 8 2(1)+3(2)=8 2 + 6 = 8 2+6=8 8 = 8 (True) 8=8(True) Hence, the ordered pair ( 1 , 2 ) (1,2) is a solution to the given system of linear equations.