Equations #7 - Questions and Answers

Question 1 Find the area of the region bounded by the graphs of the equation: 𝑦 = 5 𝑥 2 + 2 , 𝑥 = 0 , 𝑥 = 2 , 𝑦 = 0 y=5x 2 +2,x=0,x=2,y=0 Answer The value of the integral is actually represented by the area of the region enclosed by the graphs: ∫ 0 2 ( 5 𝑥 2 + 2 )   𝑑 𝑥 ∫ 0 2 ​ (5x 2 +2)dx Step-by-step solution: ∫ 0 2 ( 5 𝑥 2 + 2 )   𝑑 𝑥 = ∫ 0 2 5 𝑥 2   𝑑 𝑥 + ∫ 0 2 2   𝑑 𝑥 ∫ 0 2 ​ (5x 2 +2)dx=∫ 0 2 ​ 5x 2 dx+∫ 0 2 ​ 2dx = [ 5 𝑥 3 3 ] 0 2 + [ 2 𝑥 ] 0 2 =[ 3 5x 3 ​ ] 0 2 ​ +[2x] 0 2 ​ = ( 5 ( 2 3 ) 3 − 5 ( 0 3 ) 3 ) + ( 2 ( 2 ) − 2 ( 0 ) ) =( 3 5(2 3 ) ​ − 3 5(0 3 ) ​ )+(2(2)−2(0)) = 40 3 + 4 = 3 40 ​ +4 = 52 3 = 3 52 ​ Answer: 52 3 3 52 ​ Question 2 A rectangular package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross-section) of 108 inches. Find the dimensions of the package of maximum volume that can be sent. (Assume the cross-section is square.) Answer 1 Let: 108 = 4 𝑥 + 𝑦 108=4x+y 108 − 4 𝑥 = 𝑦 108−4x=y The area of the square cross-section is: 𝐴 ( 𝑥 ) = 𝑥 2 A(x)=x 2 And the volume of the rectangular package is: 𝑉 ( 𝑥 ) = 𝑥 2 𝑦 V(x)=x 2 y Substituting 𝑦 = 108 − 4 𝑥 y=108−4x: 𝑉 ( 𝑥 ) = 𝑥 2 ( 108 − 4 𝑥 ) V(x)=x 2 (108−4x) 𝑉 ( 𝑥 ) = 108 𝑥 2 − 4 𝑥 3 V(x)=108x 2 −4x 3 To find the maximum volume, take the derivative and set it to zero: 𝑑 𝑉 𝑑 𝑥 = 216 𝑥 − 12 𝑥 2 dx dV ​ =216x−12x 2 216 𝑥 − 12 𝑥 2 = 0 216x−12x 2 =0 𝑥 ( 216 − 12 𝑥 ) = 0 x(216−12x)=0 𝑥 = 0   or   𝑥 = 18 x=0orx=18 Since 𝑥 = 0 x=0 does not make sense, 𝑥 = 18 x=18. Substitute 𝑥 = 18 x=18 into 𝑦 = 108 − 4 𝑥 y=108−4x: 𝑦 = 108 − 4 ( 18 ) = 36 y=108−4(18)=36 Dimensions: Length = 36 ,   Width = 18 ,   Height = 18 Length=36,Width=18,Height=18 Question 3 Find the point on the line 𝑦 = 3 𝑥 + 4 y=3x+4 that is closest to the origin. Answer 1 Step 1 The given line equation is: 𝑦 = 3 𝑥 + 4 y=3x+4 Step 2 Compare 𝑦 = 3 𝑥 + 4 y=3x+4 with 𝑦 = 𝑚 𝑥 + 𝑐 y=mx+c, where 𝑚 = 3 m=3. The slope of a line perpendicular to this line is: − 1 3 − 3 1 ​ Step 3 The equation of the perpendicular line passing through the origin is: 𝑦 = − 1 3 𝑥 y=− 3 1 ​ x Step 4 Set the two line equations equal to find the intersection point: 3 𝑥 + 4 = − 1 3 𝑥 3x+4=− 3 1 ​ x 3 𝑥 + 1 3 𝑥 = − 4 3x+ 3 1 ​ x=−4 10 𝑥 3 = − 4 3 10x ​ =−4 𝑥 = − 12 10 = − 1.2 x=− 10 12 ​ =−1.2 Step 5 Substitute 𝑥 = − 1.2 x=−1.2 into 𝑦 = 3 𝑥 + 4 y=3x+4: 𝑦 = 3 ( − 1.2 ) + 4 = − 3.6 + 4 = 0.4 y=3(−1.2)+4=−3.6+4=0.4 Closest Point: ( − 1.2 , 0.4 ) (−1.2,0.4) Question 4 How do you solve for 𝑥 x in 𝑎 𝑥 − 𝑐 = 𝑏 ax−c=b? Answer 1 Step 1 Add 𝑐 c to both sides to isolate the term with 𝑥 x: 𝑎 𝑥 − 𝑐 + 𝑐 = 𝑏 + 𝑐 ax−c+c=b+c 𝑎 𝑥 = 𝑏 + 𝑐 ax=b+c Step 2 Divide by 𝑎 a: 𝑥 = 𝑏 + 𝑐 𝑎 x= a b+c ​ Answer: 𝑥 = 𝑏 + 𝑐 𝑎 x= a b+c ​ Question 5 A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villain (mass 70.0 kg), who is standing directly under the chandelier. Assume the stuntman’s center of mass moves downward 5.0 m. (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is 0.250, how far do they slide? Answer 1 (a) Using conservation of energy: 𝑚 𝑠 𝑔 ℎ = 1 2 ( 𝑚 𝑠 + 𝑚 𝑣 ) 𝑣 2 m s ​ gh= 2 1 ​ (m s ​ +m v ​ )v 2 Solving for 𝑣 v: 𝑣 = 2 𝑚 𝑠 𝑔 ℎ 𝑚 𝑠 + 𝑚 𝑣 v= m s ​ +m v ​ 2m s ​ gh ​ ​ Substitute values: 𝑣 = 2 ( 80.0 ) ( 9.8 ) ( 5.0 ) 80.0 + 70.0 v= 80.0+70.0 2(80.0)(9.8)(5.0) ​ ​ 𝑣 ≈ 6.62   m/s v≈6.62m/s (b) Using work-energy principle to find the distance: 𝜇 𝑚 𝑔 𝑑 = 1 2 𝑚 𝑣 2 μmgd= 2 1 ​ mv 2 Solve for 𝑑 d: 𝑑 = 𝑣 2 2 𝜇 𝑔 d= 2μg v 2 ​ Substitute values: 𝑑 = ( 6.62 ) 2 2 ( 0.250 ) ( 9.8 ) d= 2(0.250)(9.8) (6.62) 2 ​ 𝑑 ≈ 19.5   m d≈19.5m