Negative numbers and coordinate plane #4 - Questions and Answers

Question 1 A football team loses 3 yards in 3 consecutive plays. What is the total yardage gained? Answer Since there is actually a loss in yardage, then we represent the gain as a negative number. Hence, the total yardage gained is -3 yards. Question 2 Determine the size of the largest area rectangle whose base is on the x-axis and whose other two vertices are above the x-axis and on the parabola. (Round your answers to the nearest hundredth.) 𝑦 = 6 − 𝑥 2 y=6−x 2 Answer Area of the rectangle is 𝐴 = 𝑥 𝑦 A=xy. Substitute 𝑦 = 6 − 𝑥 2 y=6−x 2 in the formula of the area: 𝐴 = 𝑥 ( 6 − 𝑥 2 ) = 6 𝑥 − 𝑥 3 A=x(6−x 2 )=6x−x 3 Differentiate the area with respect to 𝑥 x and equate to 0: 𝐴 ′ = 𝑑 𝑑 𝑥 ( 6 𝑥 − 𝑥 3 ) = 6 − 3 𝑥 2 6 − 3 𝑥 2 = 03 𝑥 2 = 6 𝑥 2 = 2 𝑥 = ± 2 A ′ = dx d ​ (6x−x 3 )=6−3x 2 6−3x 2 =03x 2 =6x 2 =2x=± 2 ​ Substitute 𝑥 = ± 2 x=± 2 ​ in 𝑦 = 6 − 𝑥 2 y=6−x 2 to get the value of height: 𝑦 = 6 − 𝑥 2 = 6 − ( 2 ) 2 = 4 y=6−x 2 =6−( 2 ​ ) 2 =4 Hence, the dimensions of the rectangle are: 2 2 × 4 2 2 ​ ×4 Question 3 Find the slope and y-intercept of the line. Graph the line. 5 𝑥 + 𝑦 = 5 5x+y=5 Answer Let’s rephrase it as a slope-based equation. Slope equation type: 𝑦 = 𝑚 𝑥 + 𝑏 y=mx+b where 𝑚 m equals the slope and 𝑏 b equals the y-intercept of the line. Subtract 5 𝑥 5x from both sides of the equation: 𝑦 = − 5 𝑥 + 5 y=−5x+5 The slope is 𝑚 = − 5 m=−5, and the y-intercept is 𝑏 = 5 b=5. Question 4 Find an equation of the largest sphere with center (5, 4, 9) that is contained in the first octant. Answer We need to fill in the parts of the equation of the sphere: ( 𝑥 − ℎ ) 2 + ( 𝑦 − 𝑘 ) 2 + ( 𝑧 − 𝑙 ) 2 = 𝑟 2 (x−h) 2 +(y−k) 2 +(z−l) 2 =r 2 Here ( ℎ , 𝑘 , 𝑙 ) (h,k,l) is the center, and 𝑟 r is the radius. Given the center ( 5 , 4 , 9 ) (5,4,9), the sphere must have the largest radius that still stays within the first octant. This means we look at the distance from the center to each of the three bounding planes (the xy-plane, xz-plane, and yz-plane). ( 𝑥 − 5 ) 2 + ( 𝑦 − 4 ) 2 + ( 𝑧 − 9 ) 2 = 4 2 (x−5) 2 +(y−4) 2 +(z−9) 2 =4 2 Question 5 Find the point on the hyperbola 𝑥 𝑦 = 8 xy=8 that is closest to the point (3, 0). Answer Given 𝑥 𝑦 = 8 xy=8: 𝑦 = 8 𝑥 y= x 8 ​ . Any point on the hyperbola can be written as ( 𝑥 , 8 𝑥 ) (x, x 8 ​ ). The distance between ( 𝑥 , 8 𝑥 ) (x, x 8 ​ ) and (3, 0) is given by: 𝑑 = ( 𝑥 − 3 ) 2 + ( 8 𝑥 − 0 ) 2 d= (x−3) 2 +( x 8 ​ −0) 2 ​ Differentiate 𝑑 d with respect to 𝑥 x and minimize: 𝑑 2 = ( 𝑥 − 3 ) 2 + 64 𝑥 2 2 𝑑 ⋅ 𝑑 ′ = 2 ( 𝑥 − 3 ) − 128 𝑥 3 = 0 d 2 =(x−3) 2 + x 2 64 ​ 2d⋅d ′ =2(x−3)− x 3 128 ​ =0 Solve for 𝑥 x: 𝑥 = 4   or   𝑥 = − 4 x=4orx=−4 The corresponding point is ( 4 , 2 ) (4,2). Вы сказали: ChatGPT сказал: Extracted Text: Question 1 Find the point on the line 𝑦 = 3 𝑥 + 4 y=3x+4 that is closest to the origin. Answer Let’s consider an arbitrary point 𝑃 ( 𝑥 , 𝑦 ) P(x,y). The distance from 𝑃 P to the origin is: 𝑑 = 𝑥 2 + 𝑦 2 d= x 2 +y 2 ​ Note that this is the objective function. Since 𝑃 P is on the line 𝑦 = 3 𝑥 + 4 y=3x+4, the coordinates of 𝑃 P satisfy this equation. This is the constraint for the objective function in 𝑑 d. Therefore, the objective function can be written as a function of one variable: 𝑑 ( 𝑥 ) = 𝑥 2 + ( 3 𝑥 + 4 ) 2 d(x)= x 2 +(3x+4) 2 ​ From the first derivative test, we have that: 𝑑 ′ ( 𝑥 ) = 0 d ′ (x)=0 𝑑 𝑑 𝑥 ( 𝑥 2 + ( 3 𝑥 + 4 ) 2 ) = 0 dx d ​ (x 2 +(3x+4) 2 )=0 2 𝑥 + 2 ( 3 𝑥 + 4 ) ( 3 ) = 0 2x+2(3x+4)(3)=0 2 𝑥 + 18 𝑥 + 24 = 0 2x+18x+24=0 20 𝑥 = − 24 20x=−24 𝑥 = − 6 5 x=− 5 6 ​ Since 𝑑 ′ ( 𝑥 ) < 0 d ′ (x)<0 for 𝑥 < − 6 5 x<− 5 6 ​ and 𝑑 ′ ( 𝑥 ) > 0 d ′ (x)>0 for 𝑥 > − 6 5 x>− 5 6 ​ , the minimum distance from 𝑃 P to the origin is achieved when 𝑥 = − 6 5 x=− 5 6 ​ . Then, 𝑦 = 3 𝑥 + 4 = 3 ( − 6 5 ) + 4 y=3x+4=3(− 5 6 ​ )+4 𝑦 = − 18 + 20 5 = 2 5 y= 5 −18+20 ​ = 5 2 ​ We found that: 𝑃 ( − 6 5 , 2 5 ) P(− 5 6 ​ , 5 2 ​ ) is the point on the line 𝑦 = 3 𝑥 + 4 y=3x+4 closest to the origin. Question 2 Express the number in terms of 𝑖 i. − 144 −144 ​ Answer 1: − 144 −144 ​ 2: 144 ⋅ − 1 144 ​ ⋅ −1 ​ Result = 12 𝑖 12i. Question 3 Find the points of intersection of the line 𝑥 = 3 + 2 𝑡 , 𝑦 = 5 + 7 𝑡 , 𝑧 = − 3 + 𝑡 x=3+2t,y=5+7t,z=−3+t, that is, 𝐼 ( 𝑡 ) = ( 3 + 2 𝑡 , 5 + 7 𝑡 , − 3 + 𝑡 ) I(t)=(3+2t,5+7t,−3+t), with the coordinate planes. Answer The coordinate planes’ equations are provided below: 𝑦 𝑧 : 𝑥 = 0 , 𝑧 𝑥 : 𝑦 = 0 , 𝑥 𝑦 : 𝑧 = 0 yz:x=0,zx:y=0,xy:z=0. 1. Consider in xy plane 𝑧 = 0 z=0: Substitute 0 for 𝑧 z in the equation of line: 𝑧 = − 3 + 𝑡 = 0 ⇒ 𝑡 = 3 z=−3+t=0⇒t=3 At 𝑡 = 3 t=3: 𝑥 = 3 + 2 ( 3 ) = 9 x=3+2(3)=9 𝑦 = 5 + 7 ( 3 ) = 26 y=5+7(3)=26 Point is ( 9 , 26 , 0 ) (9,26,0). 2. Consider in yz plane 𝑥 = 0 x=0: Substitute 0 for 𝑥 x: 𝑥 = 3 + 2 𝑡 = 0 ⇒ 𝑡 = − 3 2 x=3+2t=0⇒t=− 2 3 ​ At 𝑡 = − 3 2 t=− 2 3 ​ : 𝑦 = 5 + 7 ( − 3 2 ) = − 11 2 y=5+7(− 2 3 ​ )=− 2 11 ​ 𝑧 = − 3 + ( − 3 2 ) = − 9 2 z=−3+(− 2 3 ​ )=− 2 9 ​ Point is ( 0 , − 11 2 , − 9 2 ) (0,− 2 11 ​ ,− 2 9 ​ ). 3. Consider in zx plane 𝑦 = 0 y=0: Substitute 0 for 𝑦 y: 𝑦 = 5 + 7 𝑡 = 0 ⇒ 𝑡 = − 5 7 y=5+7t=0⇒t=− 7 5 ​ At 𝑡 = − 5 7 t=− 7 5 ​ : 𝑥 = 3 + 2 ( − 5 7 ) = 11 7 x=3+2(− 7 5 ​ )= 7 11 ​ 𝑧 = − 3 + ( − 5 7 ) = − 26 7 z=−3+(− 7 5 ​ )=− 7 26 ​ Point is ( 11 7 , 0 , − 26 7 ) ( 7 11 ​ ,0,− 7 26 ​ ). Thus, the points of intersections on the coordinate planes are as follows: 𝑥 𝑦 : ( 9 , 26 , 0 ) xy:(9,26,0) 𝑦 𝑧 : ( 0 , − 11 2 , − 9 2 ) yz:(0,− 2 11 ​ ,− 2 9 ​ ) 𝑧 𝑥 : ( 11 7 , 0 , − 26 7 ) zx:( 7 11 ​ ,0,− 7 26 ​ ) Question 4 Find the average value of 𝐹 ( 𝑥 , 𝑦 , 𝑧 ) F(x,y,z) over the given region. 𝐹 ( 𝑥 , 𝑦 , 𝑧 ) = 𝑥 2 + 9 F(x,y,z)=x 2 +9 over the cube in the first octant bounded by the coordinate planes and the planes 𝑥 = 2 , 𝑦 = 2 , 𝑧 = 2 x=2,y=2,z=2. Answer Let 𝐷 D be the volume of the cube bounded by the coordinate planes and the planes 𝑥 = 2 , 𝑦 = 2 , 𝑧 = 2 x=2,y=2,z=2. Then 𝐷 = 2 ⋅ 2 ⋅ 2 = 8 D=2⋅2⋅2=8. Calculate average value of 𝑥 2 + 9 x 2 +9 over cube 𝐷 D using formula: Average value = volume of  𝐷 ∫ ∫ ∫ 𝐷 𝐹 𝑑 𝑉 Average value= ∫∫∫ D ​ FdV volume of D ​ Volume of 𝐷 D is equal to 2 3 = 8 2 3 =8. Average value = 1 8 ∫ 0 2 ∫ 0 2 ∫ 0 2 ( 𝑥 2 + 9 ) 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 Average value= 8 1 ​ ∫ 0 2 ​ ∫ 0 2 ​ ∫ 0 2 ​ (x 2 +9)dzdydx Integration: = 1 8 ∫ 0 2 ∫ 0 2 [ ( 𝑥 2 + 9 ) 𝑧 ] 0 2 𝑑 𝑦 𝑑 𝑥 = 1 8 ∫ 0 2 ∫ 0 2 2 ( 𝑥 2 + 9 ) 𝑑 𝑦 𝑑 𝑥 = 8 1 ​ ∫ 0 2 ​ ∫ 0 2 ​ [(x 2 +9)z] 0 2 ​ dydx= 8 1 ​ ∫ 0 2 ​ ∫ 0 2 ​ 2(x 2 +9)dydx = 1 8 ∫ 0 2 [ 2 ( 𝑥 2 + 9 ) 𝑦 ] 0 2 𝑑 𝑥 = 1 8 ∫ 0 2 4 ( 𝑥 2 + 9 ) 𝑑 𝑥 = 8 1 ​ ∫ 0 2 ​ [2(x 2 +9)y] 0 2 ​ dx= 8 1 ​ ∫ 0 2 ​ 4(x 2 +9)dx = 1 8 [ 4 𝑥 3 3 + 36 𝑥 ] 0 2 = 1 8 ( 32 3 + 72 ) = 1 8 ⋅ 288 + 32 3 = 1 8 ⋅ 320 3 = 40 3 = 8 1 ​ [ 3 4x 3 ​ +36x] 0 2 ​ = 8 1 ​ ( 3 32 ​ +72)= 8 1 ​ ⋅ 3 288+32 ​ = 8 1 ​ ⋅ 3 320 ​ = 3 40 ​ Final answer: 31 3 3 31