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

2024
Views:

479

Pages:

6
Author:

Javon Neal

Question 1 The base of 𝑆 S is an elliptical region with boundary curve 9 𝑥 2 + 4 𝑦 2 = 36 9x 2 +4y 2 =36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. Answer The given equation of the ellipse can be rewritten as: 𝑥 2 4 + 𝑦 2 9 = 1 4 x 2 + 9 y 2 =1 One cross-section (red line segment) is the hypotenuse ℎ h of an isosceles right triangle. The side length 𝑠 s of the triangle is: 𝑠 2 + 𝑠 2 = ℎ 2 ⟹ 2 𝑠 2 = ℎ 2 ⟹ 𝑠 = ℎ 2 s 2 +s 2 =h 2 ⟹2s 2 =h 2 ⟹s= 2 h The area of the isosceles right triangle is: 𝐴 = 1 2 𝑠 2 = 1 2 ( ℎ 2 ) 2 = ℎ 2 4 A= 2 1 s 2 = 2 1 ( 2 h ) 2 = 4 h 2 Substitute ℎ = 2 9 − 9 𝑥 2 h=2 9−9x 2 for the top half of the ellipse. Use symmetry and double the integral for the entire solid: 𝑉 = 2 ∫ 0 2 1 4 ( 36 − 9 𝑥 2 ) 𝑑 𝑥 V=2∫ 0 2 4 1 (36−9x 2 )dx 𝑉 = 1 2 ∫ 0 2 ( 36 − 9 𝑥 2 ) 𝑑 𝑥 = 1 2 [ 36 𝑥 − 3 𝑥 3 ] 0 2 = 1 2 [ 72 − 24 − 0 ] = 24 V= 2 1 ∫ 0 2 (36−9x 2 )dx= 2 1 [36x−3x 3 ] 0 2 = 2 1 [72−24−0]=24 Result: 𝑉 = 24 V=24. Question 2 Find the point on the line 𝑦 = 5 𝑥 + 2 y=5x+2 that is closest to the origin. Answer Let the point be ( 𝑥 , 5 𝑥 + 2 ) (x,5x+2). We minimize: 𝑑 = 𝑥 2 + ( 5 𝑥 + 2 ) 2 d=x 2 +(5x+2) 2 𝑑 ′ = 2 𝑥 + 10 ( 5 𝑥 + 2 ) = 0 d ′ =2x+10(5x+2)=0 2 𝑥 + 50 𝑥 + 20 = 0 ⟹ 52 𝑥 = − 20 ⟹ 𝑥 = − 10 26 = − 5 13 2x+50x+20=0⟹52x=−20⟹x=− 26 10 =− 13 5 𝑦 = 5 ( − 5 13 ) + 2 = 13 13 = 3 13 y=5(− 13 5 )+2= 13 13 = 13 3 Point: ( − 5 13 , 3 13 ) (− 13 5 , 13 3 ). Question 3 Find the area of the region enclosed by one loop of the curve 𝑟 = sin ⁡ 12 𝜃 r=sin12θ. Find the area of the region that lies inside the first curve outside the second curve 𝑟 = 14 cos ⁡ 𝜃 , 𝑟 = 7 r=14cosθ,r=7. Answer For 𝑟 = sin ⁡ 12 𝜃 r=sin12θ, one loop is 0 ≤ 𝜃 ≤ 𝜋 12 0≤θ≤ 12 π : 𝐴 = 1 2 ∫ 0 𝜋 / 12 ( sin ⁡ 12 𝜃 ) 2 𝑑 𝜃 = 1 2 ∫ 0 𝜋 / 12 1 − cos ⁡ 24 𝜃 2 𝑑 𝜃 A= 2 1 ∫ 0 π/12 (sin12θ) 2 dθ= 2 1 ∫ 0 π/12 2 1−cos24θ dθ 𝐴 = 1 4 [ 𝜃 − sin ⁡ 24 𝜃 24 ] 0 𝜋 / 12 = 𝜋 48 A= 4 1 [θ− 24 sin24θ ] 0 π/12 = 48 π For the second part: 𝐴 = 1 2 ∫ 0 𝜋 / 3 ( ( 14 cos ⁡ 𝜃 ) 2 − 7 2 ) 𝑑 𝜃 A= 2 1 ∫ 0 π/3 ((14cosθ) 2 −7 2 )dθ 𝐴 = 1 2 ∫ 0 𝜋 / 3 ( 196 cos ⁡ 2 𝜃 − 49 ) 𝑑 𝜃 = … ⟹ 𝐴 = 49 3 + 49 𝜋 / 3 A= 2 1 ∫ 0 π/3 (196cos 2 θ−49)dθ=…⟹A=49 3 +49π/3 Question 4 Is 𝑦 = − 7 𝑥 − 10 y=−7x−10 defined as a function of 𝑥 x? Answer Let 𝑥 = − 2 x=−2, then 𝑦 = − 7 ( − 2 ) − 10 = 24 y=−7(−2)−10=24. The definition of a function states each input gives a unique output. Thus, 𝑦 = − 7 𝑥 − 10 y=−7x−10 is a function of 𝑥 x. Question 5 Given linear equation 𝑦 = 3 + 4 𝑥 y=3+4x: Calculate slope and y-intercept. Determine whether the line slopes upward, slopes downward, or is horizontal. Use two points to graph the equation. Answer Slope ( 𝑚 = 4 m=4), y-intercept ( 𝑏 = 3 b=3). Line slopes upward ( 𝑚 > 0 m>0). Points: ( 0 , 3 ) (0,3), ( − 0.75 , 0 ) (−0.75,0). Graph: (Refer to image).