University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

357

Pages:

4
Author:

Javon Neal

Question 1 Find the area enclosed by the curve 𝑥 = 𝑡 2 − 2 𝑡 x=t 2 −2t, 𝑦 = 𝑡 y= t , and the y-axis. Answer Consider the following parametric equations: 𝑥 = 𝑡 2 − 2 𝑡 and 𝑦 = 𝑡 x=t 2 −2tandy= t The objective is to find the area enclosed by the curve using the formula. The area under the curve is given by the parametric equations 𝑥 = 𝑓 ( 𝑡 ) x=f(t), 𝑦 = 𝑔 ( 𝑡 ) y=g(t), and is traversed once as 𝑡 t increases from 𝑎 a to 𝑏 b. The formula for the area is: 𝐴 = ∫ 𝑎 𝑏 𝑔 ( 𝑡 ) 𝑓 ′ ( 𝑡 ) 𝑑 𝑡 A=∫ a b g(t)f ′ (t)dt The curve intersects with the y-axis, so 𝑥 = 0 x=0: 𝑡 2 − 2 𝑡 = 0 𝑡 ( 𝑡 − 2 ) = 0 𝑡 = 0 or 𝑡 = 2 t 2 −2t=0t(t−2)=0t=0ort=2 Evaluate: 𝑓 ( 𝑡 ) = 𝑡 2 − 2 𝑡 , 𝑔 ( 𝑡 ) = 𝑡 , 𝑓 ′ ( 𝑡 ) = 2 𝑡 − 2 f(t)=t 2 −2t,g(t)= t ,f ′ (t)=2t−2 Now, find the area under the curve: 𝐴 = ∫ 0 2 𝑡 ( 2 𝑡 − 2 ) 𝑑 𝑡 A=∫ 0 2 t (2t−2)dt Simplify: 𝐴 = 2 ∫ 0 2 𝑡 3 / 2 − 𝑡 1 / 2 𝑑 𝑡 A=2∫ 0 2 t 3/2 −t 1/2 dt Integrate: 𝐴 = 2 [ 𝑡 5 / 2 5 / 2 − 𝑡 3 / 2 3 / 2 ] 0 2 A=2[ 5/2 t 5/2 − 3/2 t 3/2 ] 0 2 𝐴 = 2 [ 2 𝑡 5 / 2 5 − 2 𝑡 3 / 2 3 ] 0 2 A=2[ 5 2t 5/2 − 3 2t 3/2 ] 0 2 Substitute 𝑡 = 2 t=2: 𝐴 = 2 ( 2 5 / 2 ) 5 − 2 ( 2 3 / 2 ) 3 A= 5 2(2 5/2 ) − 3 2(2 3/2 ) Final result: 𝐴 = 8 2 15 A= 15 8 2 Question 2 True or False: The graph of the linear equation 2 𝑥 − 3 𝑦 = 6 2x−3y=6 is used to graph the linear inequality 2 𝑥 − 3 𝑦 > 6 2x−3y>6. Answer The given statement is False. The given equation of the line does not satisfy the linear inequality 2 𝑥 − 3 𝑦 > 6 2x−3y>6. It is not used to graph the given inequality. Question 3 Graph the linear equation 𝑦 = 3 𝑥 + 1 y=3x+1. Answer Since the equation is in the form of 𝑦 = 𝑚 𝑥 + 𝑏 y=mx+b, it is considered to be a linear equation. The graph will be a straight line. Assume 𝑥 = 0 x=0: 𝑦 = 3 ( 0 ) + 1 = 1 y=3(0)+1=1 The first point is ( 0 , 1 ) (0,1). Assume 𝑦 = 0 y=0: 0 = 3 𝑥 + 1 𝑥 = − 1 3 0=3x+1x=− 3 1 The second point is ( − 1 3 , 0 ) (− 3 1 ,0). Plotting the two points ( 0 , 1 ) (0,1) and ( − 1 3 , 0 ) (− 3 1 ,0), we get the graph. Question 4 Graph for 𝑦 = 1 2 𝑥 − 5 y= 2 1 x−5. Answer The equation 𝑦 = 1 2 𝑥 − 5 y= 2 1 x−5 has: Slope = 0.5 0.5 Passes through the point ( 0 , − 5 ) (0,−5). Graph the line passing through ( 0 , − 5 ) (0,−5) with a slope of 0.5 0.5. Question 5 How do you graph 𝑥 = 0 x=0? Answer 1 Any vertical line has the equation 𝑥 = 0 x=0. This is because it only has an x-intercept. Answer 2 This is the equation of the y-axis. It is a vertical line through the origin.