University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

492

Pages:

4
Author:

Herbosambx

Question 1 Find the intersection of two separate planes that forms the line 𝐿 = 𝑥 + 𝑡 , 𝑦 = 2 − 𝑡 , 𝑧 = 3 + 2 𝑡 L=x+t,y=2−t,z=3+2t. Write equations for each plane in the form 𝐴 𝑥 + 𝐵 𝑦 + 𝐶 𝑧 = 𝐷 Ax+By+Cz=D. Answer Given: 𝐿 1 = 𝑥 + 𝑡 , 𝑦 = 2 − 𝑡 , 𝑧 = 3 + 2 𝑡 L 1 =x+t,y=2−t,z=3+2t Let: 𝑃 1 : 𝐴 1 𝑥 + 𝐵 1 𝑦 + 𝐶 1 𝑧 = 𝐷 1 𝑃 2 : 𝐴 2 𝑥 + 𝐵 2 𝑦 + 𝐶 2 𝑧 = 𝐷 2 P 1 :A 1 x+B 1 y+C 1 z=D 1 P 2 :A 2 x+B 2 y+C 2 z=D 2 The constants in the two planes equation determine if the two planes intersect with the line or not. Hence, we will choose two points on the line, and then make them on the two planes as follows: By putting 𝑡 = 0 t=0 in the line equation, we get the point 𝑃 1 = ( 1 , 2 , 3 ) P 1 =(1,2,3). By putting 𝑡 = 1 t=1 in the line equation, we get the point 𝑃 2 = ( 2 , 1 , 5 ) P 2 =(2,1,5). Substituting 𝑃 1 P 1 and 𝑃 2 P 2 into the equations 𝑃 1 : 𝐴 1 𝑥 + 𝐵 1 𝑦 + 𝐶 1 𝑧 = 𝐷 1 P 1 :A 1 x+B 1 y+C 1 z=D 1 and 𝑃 2 : 𝐴 2 𝑥 + 𝐵 2 𝑦 + 𝐶 2 𝑧 = 𝐷 2 P 2 :A 2 x+B 2 y+C 2 z=D 2 , we find the following system of equations: 𝐴 1 + 2 𝐵 1 + 3 𝐶 1 = 𝐷 1 2 𝐴 1 + 𝐵 1 + 5 𝐶 1 = 𝐷 1 A 1 +2B 1 +3C 1 =D 1 2A 1 +B 1 +5C 1 =D 1 Solving this system for 𝐴 1 = 1 , 𝐵 1 = 1 A 1 =1,B 1 =1, we get: 5 + 3 𝐶 1 = 𝐷 1 4 + 5 𝐶 1 = 𝐷 2 5+3C 1 =D 1 4+5C 1 =D 2 Finally, by using the values of constants, the two planes' equations are: 𝑃 1 : 2 𝑥 + 4 𝑦 + 𝑧 = 13 𝑃 2 : 𝑥 + 3 𝑦 + 𝑧 = 10 P 1 :2x+4y+z=13P 2 :x+3y+z=10 Question 2 A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/sec. a. How fast is the top of the ladder sliding down the wall then? b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle between the ladder and the ground changing then? Answer We are aware that the ladder is a certain length 𝑙 = 13 ft l=13ft. We also know that for 𝑥 = 12 x=12, the rate of change of the base is: 𝑑 𝑥 𝑑 𝑡 = 5 ft/sec dt dx =5ft/sec Part (a): From the given information, we infer that for a fixed 𝑙 , 𝑡 = 13 = 𝑥 2 + 𝑦 2 l,t=13= x 2 +y 2 : We want to find the speed of the ladder sliding down the wall, represented by 𝑑 𝑦 𝑑 𝑡 dt dy : 𝑥 2 + 𝑦 2 = 1 3 2 x 2 +y 2 =13 2 Differentiating both sides: 2 𝑥 𝑑 𝑥 𝑑 𝑡 + 2 𝑦 𝑑 𝑦 𝑑 𝑡 = 0 2x dt dx +2y dt dy =0 Substituting 𝑥 = 12 , 𝑑 𝑥 𝑑 𝑡 = 5 , 𝑦 = 1 3 2 − 1 2 2 = 5 x=12, dt dx =5,y= 13 2 −12 2 =5: 2 ( 12 ) ( 5 ) + 2 ( 5 ) 𝑑 𝑦 𝑑 𝑡 = 0 2(12)(5)+2(5) dt dy =0 120 + 10 𝑑 𝑦 𝑑 𝑡 = 0 120+10 dt dy =0 𝑑 𝑦 𝑑 𝑡 = − 12 ft/sec dt dy =−12ft/sec Part (b): The area of the triangle is: 𝐴 = 1 2 𝑥 𝑦 A= 2 1 xy Differentiating: 𝑑 𝐴 𝑑 𝑡 = 1 2 ( 𝑥 𝑑 𝑦 𝑑 𝑡 + 𝑦 𝑑 𝑥 𝑑 𝑡 ) dt dA = 2 1 (x dt dy +y dt dx ) Substituting 𝑥 = 12 , 𝑦 = 5 , 𝑑 𝑥 𝑑 𝑡 = 5 , 𝑑 𝑦 𝑑 𝑡 = − 12 x=12,y=5, dt dx =5, dt dy =−12: 𝑑 𝐴 𝑑 𝑡 = 1 2 [ ( 12 ) ( − 12 ) + ( 5 ) ( 5 ) ] dt dA = 2 1 [(12)(−12)+(5)(5)] 𝑑 𝐴 𝑑 𝑡 = 1 2 ( − 144 + 25 ) = − 59.5 ft 2 / sec dt dA = 2 1 (−144+25)=−59.5ft 2 /sec Part (c): The angle 𝜃 θ satisfies: cos ⁡ 𝜃 = 𝑥 𝑙 , 𝑙 = 13 cosθ= l x ,l=13 Differentiating: sin ⁡ 𝜃 𝑑 𝜃 𝑑 𝑡 = 𝑑 𝑥 𝑑 𝑡 𝑙 sinθ dt dθ = l dt dx Substitute values: sin ⁡ 𝜃 = 5 13 , 𝑑 𝑥 𝑑 𝑡 = 5 sinθ= 13 5 , dt dx =5 𝑑 𝜃 𝑑 𝑡 = 5 13 ⋅ 1 5 = 1 13 rad/sec dt dθ = 13 5 ⋅ 5 1 = 13 1 rad/sec Question 3 Solve the equation: ∣ 𝑥 − 3 ∣ = 2 ∣x−3∣=2 Answer 1 Step 1: Solve the equation ∣ 𝑥 − 3 ∣ = 2 ∣x−3∣=2. ∣ 𝑥 − 3 ∣ = 𝑎 ⇒ 𝑥 − 3 = 𝑎 or 𝑥 − 3 = − 𝑎 ∣x−3∣=a⇒x−3=aorx−3=−a 𝑥 − 3 = 2 or 𝑥 − 3 = − 2 x−3=2orx−3=−2 𝑥 = 5 or 𝑥 = 1 x=5orx=1 Question 4 Find a parametric representation of the solution set of the linear equation. 3 𝑥 − 1 2 𝑦 = 9 3x− 2 1 y=9 Answer 1 Given equation: 3 𝑥 − 1 2 𝑦 = 9 3x− 2 1 y=9 Let 𝑥 = 𝑡 x=t. Substitute into the equation: 3 𝑡 − 1 2 𝑦 = 9 3t− 2 1 y=9 Solve for 𝑦 y: 𝑦 = 6 𝑡 − 18 y=6t−18 Parametric solution set: { 𝑥 , 𝑦 } = { 𝑡 , 6 𝑡 − 18 } {x,y}={t,6t−18} Question 5 Find the general solution to the differential equation. 𝑦 ′ − 8 𝑦 = 0 y ′ −8y=0 Answer Rewrite as: 𝑑 𝑦 𝑑 𝑥 = 8 𝑦 dx dy =8y Separate variables and solve: ∫ 1 𝑦 𝑑 𝑦 = ∫ 8 𝑑 𝑥 ∫ y 1 dy=∫8dx ln ⁡ 𝑦 = 8 𝑥 + 𝐶 lny=8x+C 𝑦 = 𝐶 𝑒 8 𝑥 y=Ce 8x