Equations #14 - Questions and Answers

Question 1 Find the second derivative of 𝑦 y with respect to 𝑥 x from the parametric equations given. 𝑥 = 𝑡 2 − 1 ,   𝑦 = 𝑡 2 + 𝑡 x=t 2 −1,y=t 2 +t Answer Step 1: Given 𝑥 = 𝑡 2 − 1 ,   𝑦 = 𝑡 2 + 𝑡 x=t 2 −1,y=t 2 +t Step 2: Solution 𝑑 𝑥 𝑑 𝑡 = 2 𝑡 ,   𝑑 𝑦 𝑑 𝑡 = 2 𝑡 + 1 dt dx ​ =2t, dt dy ​ =2t+1 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑦 𝑑 𝑡 𝑑 𝑥 𝑑 𝑡 = 2 𝑡 + 1 2 𝑡 dx dy ​ = dt dx ​ dt dy ​ ​ = 2t 2t+1 ​ 𝑑 2 𝑦 𝑑 𝑥 2 = 𝑑 𝑑 𝑥 ( 𝑑 𝑦 𝑑 𝑥 ) = 𝑑 𝑑 𝑡 ( 2 𝑡 + 1 2 𝑡 ) ⋅ 1 𝑑 𝑥 𝑑 𝑡 dx 2 d 2 y ​ = dx d ​ ( dx dy ​ )= dt d ​ ( 2t 2t+1 ​ )⋅ dt dx ​ 1 ​ First, differentiate 2 𝑡 + 1 2 𝑡 2t 2t+1 ​ : 𝑑 𝑑 𝑡 ( 2 𝑡 + 1 2 𝑡 ) = 2 ( 2 𝑡 ) − ( 2 𝑡 + 1 ) ( 2 ) ( 2 𝑡 ) 2 = 2 𝑡 ( 2 ) − ( 2 𝑡 + 1 ) ( 2 ) 4 𝑡 2 = − 2 4 𝑡 2 = − 1 2 𝑡 2 dt d ​ ( 2t 2t+1 ​ )= (2t) 2 2(2t)−(2t+1)(2) ​ = 4t 2 2t(2)−(2t+1)(2) ​ = 4t 2 −2 ​ =− 2t 2 1 ​ Now substitute: 𝑑 2 𝑦 𝑑 𝑥 2 = − 1 2 𝑡 2 ⋅ 1 2 𝑡 = − 1 4 𝑡 3 dx 2 d 2 y ​ = 2t 2 −1 ​ ⋅ 2t 1 ​ =− 4t 3 1 ​ Final Answer: 𝑑 2 𝑦 𝑑 𝑥 2 = − 1 4 𝑡 3 dx 2 d 2 y ​ =− 4t 3 1 ​ Question 2 What are the parametric equations for the intersection of the planes 𝑥 − 𝑦 − 𝑧 = 1 x−y−z=1 and 2 𝑥 + 3 𝑦 + 𝑧 = 2 2x+3y+z=2? Answer To find a point on this line, set 𝑧 = 0 z=0, and use the given equations to solve for 𝑥 x and 𝑦 y: Step 1: 𝑥 − 𝑦 − 𝑧 = 1    ⟹    𝑥 − 𝑦 = 1 x−y−z=1⟹x−y=1 2 𝑥 + 3 𝑦 + 𝑧 = 2    ⟹    2 𝑥 + 3 𝑦 = 2 2x+3y+z=2⟹2x+3y=2 With 𝑧 = 0 z=0: 𝑥 − 𝑦 = 1     (1) x−y=1(1) 2 𝑥 + 3 𝑦 = 2     (2) 2x+3y=2(2) Solve for 𝑥 x and 𝑦 y: From (1): 𝑥 = 𝑦 + 1 x=y+1. Substitute into (2): 2 ( 𝑦 + 1 ) + 3 𝑦 = 2 2(y+1)+3y=2 2 𝑦 + 2 + 3 𝑦 = 2    ⟹    5 𝑦 = 0    ⟹    𝑦 = 0 2y+2+3y=2⟹5y=0⟹y=0 Then 𝑥 = 1 x=1. So the point on the line is ( 1 , 0 , 0 ) (1,0,0). Step 2: The direction of the line is given by the cross product of the normal vectors of the planes. The normal vector of 𝑥 − 𝑦 − 𝑧 = 1 x−y−z=1 is ( 1 , − 1 , − 1 ) (1,−1,−1), and the normal vector of 2 𝑥 + 3 𝑦 + 𝑧 = 2 2x+3y+z=2 is ( 2 , 3 , 1 ) (2,3,1). Cross product: 𝑛 1 × 𝑛 2 = ∣ 𝑖 𝑗 𝑘 1 − 1 − 1 2 3 1 ∣ = 𝑖 ( − 1 − ( − 3 ) ) − 𝑗 ( 1 − ( − 2 ) ) + 𝑘 ( 3 − ( − 2 ) ) n 1 ​ ×n 2 ​ = ​ i 1 2 ​ j −1 3 ​ k −1 1 ​ ​ =i(−1−(−3))−j(1−(−2))+k(3−(−2)) = 𝑖 ( 2 ) − 𝑗 ( 3 ) + 𝑘 ( 5 ) = ( 2 , − 3 , 5 ) =i(2)−j(3)+k(5)=(2,−3,5) Step 3: The parametric equations of the line are: 𝑥 = 1 + 2 𝑡 ,   𝑦 = − 3 𝑡 ,   𝑧 = 5 𝑡 x=1+2t,y=−3t,z=5t Question 3 In converting 𝑟 = 5 r=5 from a polar equation to a rectangular equation, describe what should be done to both sides of the equation and why this should be done. Answer Step 1: Given: 𝑟 = 5 r=5 Step 2: The relationship between polar and rectangular coordinates is: 𝑥 = 𝑟 cos ⁡ 𝜃 ,   𝑦 = 𝑟 sin ⁡ 𝜃 ,   𝑟 2 = 𝑥 2 + 𝑦 2 x=rcosθ,y=rsinθ,r 2 =x 2 +y 2 Step 3: Solution: Square both sides: 𝑟 2 = 25    ⟹    𝑥 2 + 𝑦 2 = 25 r 2 =25⟹x 2 +y 2 =25 Final Answer: The rectangular equation is: 𝑥 2 + 𝑦 2 = 25 x 2 +y 2 =25 Question 4 Use the given conditions to write an equation for the line in point-slope form and in slope-intercept form. Passing through ( − 2 , − 8 ) (−2,−8) and parallel to the line whose equation is 𝑦 = − 3 𝑥 + 4 y=−3x+4. Answer Step 1: The slope of the necessary line is equal to − 3 −3 because parallel lines have the same slope. 𝑚 = − 3 m=−3 Step 2: Using the point-slope form: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) y−y 1 ​ =m(x−x 1 ​ ) 𝑦 − ( − 8 ) = − 3 ( 𝑥 − ( − 2 ) ) y−(−8)=−3(x−(−2)) 𝑦 + 8 = − 3 ( 𝑥 + 2 ) y+8=−3(x+2) Step 3: Simplify to slope-intercept form: 𝑦 = − 3 𝑥 − 6 − 8    ⟹    𝑦 = − 3 𝑥 − 14 y=−3x−6−8⟹y=−3x−14 Final Answer: Point-slope form: 𝑦 + 8 = − 3 ( 𝑥 + 2 ) y+8=−3(x+2) Slope-intercept form: 𝑦 = − 3 𝑥 − 14 y=−3x−14 Question 5 Convert polar equation to a rectangular equation. Then determine the graph's slope and y-intercept. 𝑟 sin ⁡ ( 𝜃 − 𝜋 4 ) = 2 rsin(θ− 4 π ​ )=2 Answer Step 1: Given polar equation: 𝑟 sin ⁡ ( 𝜃 − 𝜋 4 ) = 2 rsin(θ− 4 π ​ )=2 Step 2: To convert polar equation to a rectangular equation: 𝑟 = 𝑥 2 + 𝑦 2 ,   𝑥 = 𝑟 cos ⁡ 𝜃 ,   𝑦 = 𝑟 sin ⁡ 𝜃 r= x 2 +y 2 ​ ,x=rcosθ,y=rsinθ Use the trigonometric identity: sin ⁡ ( 𝜃 − 𝜋 4 ) = sin ⁡ 𝜃 cos ⁡ 𝜋 4 − cos ⁡ 𝜃 sin ⁡ 𝜋 4 sin(θ− 4 π ​ )=sinθcos 4 π ​ −cosθsin 4 π ​ = 2 2 ( sin ⁡ 𝜃 − cos ⁡ 𝜃 ) = 2 2 ​ ​ (sinθ−cosθ) Substitute: 𝑟 ( 2 2 ) ( 𝑦 − 𝑥 ) = 2    ⟹    2 ( 𝑦 − 𝑥 ) = 4 r( 2 2 ​ ​ )(y−x)=2⟹ 2 ​ (y−x)=4 Divide by 2 2 ​ : 𝑦 − 𝑥 = 2 2 y−x=2 2 ​ 𝑦 = 𝑥 + 2 2 y=x+2 2 ​ Final Answer: Rectangular equation: 𝑦 = 𝑥 + 2 2 y=x+2 2 ​ Slope: 1 1, Y-intercept: 2 2 2 2 ​ .