University:
Rice University
Course:
MATH 102 | Single Variable Calculus II
Academic year:

2023
Views:

15

Pages:

2
Author:

MathMonk

Questions and Answers Sheet 6 Parametric Equations Question #1 Convert each polar equation to rectangular form. = −𝜋/6 Answer: 1) 𝜃 = −𝜋/6 tan 𝜃 = tan(−𝜋/6) 2) tan 𝜃 = (sin 𝜃)/(cos 𝜃) (sin 𝜃)/(cos 𝜃) = −(√3)/3 3) Replace sin θ with y and cos θ with x 𝑦/𝑥 = −(√3)/3 4) Multiply both sides by x 𝑦 = −(√3)/3 × 𝑥 𝑦 = −(√3)/3 × 𝑥 + 0 This is the equation of a line with slope of – (√3)/3 and y-intercept of 0. Question #2 Which of the following polar coordinate pairs does not represent the point with rectangular coordinates (2,-2)? (A) (2√2; −135°) (B) (2√2; 225°) (C) (−2√2; −315°) (D) (−2√2; 45°) (E) (−2√2; 135°) Answer: A polar coordinate (r; θ) is given, then (𝑥; 𝑦) = (𝑟 × cos 𝜃 ; 𝑟 × sin 𝜃) A) (2√2 cos(−135°) , 2√2 sin(−135°)) = (2√2 × −√2/2; 2√2 × −√2/2) = (−2; −2) B) (2√2 𝑐𝑜𝑠 225°, 2√2 𝑠𝑖𝑛 225°) = (2√2 × −√2/2; 2√2 ×– √2/2) = (−2; −2) C) (−2√2𝑐𝑜𝑠(−315°), −2√2𝑠𝑖𝑛(−315°)) = (−2√2 × √2/2; −2√2 × √2/2) = (−2; −2) D) (−2√2 𝑐𝑜𝑠 45°, −2√2 𝑠𝑖𝑛 45°) = (−2√2 × √2/2; −2√2 × √2/2) = (−2; −2) E) (−2√2 𝑐𝑜𝑠 135°, −2√2 𝑠𝑖𝑛 135°) = (−2√2 × −√2/2; −2√2 × √2/2) = (2; −2) Choice (E) does not represent (-2; -2). Question #3 Find zw and z/w. Leave your answers in polar form. 𝑧 = 𝑐𝑜𝑠80° + 𝑖 𝑠𝑖𝑛 80° 𝑤 = 𝑐𝑜𝑠50° + 𝑖 𝑠𝑖𝑛 50° Answer: 𝑧 = 𝑐𝑜𝑠80° + 𝑖 𝑠𝑖𝑛80° and 𝑤 = 𝑐𝑜𝑠50° + 𝑖 𝑠𝑖𝑛50° 𝑧𝑥 = 𝑐𝑜𝑠(80° + 50°) + 𝑖 𝑠𝑖𝑛 (80° + 50°) = 𝑐𝑜𝑠130° + 𝑖 𝑠𝑖𝑛 130° 𝑧/𝑤 = 𝑐𝑜𝑠(80° − 50°) + 𝑖 𝑠𝑖𝑛 (80° − 50°) = 𝑐𝑜𝑠30° + 𝑖 𝑠𝑖𝑛 30° Question #4 Write the equation of the line tangent to 𝑓(𝑥) = 𝑡𝑎𝑛 𝑥 + 3 at (𝜋/4; 4) Answer: f(x) = tan x + 3 at (𝜋/4; 4) y = f(x) = tan + 3 𝑑/𝑑𝑥 × 𝑓(𝑥) = 𝑑/𝑑𝑥(𝑡𝑎𝑛 𝑥 + 3) 𝑓 ′(𝑥) = 𝑠𝑒𝑐 2 𝑥 + 0 (𝑑𝑦)/(𝑑𝑥) = 𝑠𝑒𝑐 2 𝑥 2 Slope 𝑚 = (𝑑𝑦)/(𝑑𝑥) = 𝑠𝑒𝑐 2 × 𝜋/4 = 𝑠𝑒𝑐 2 (𝜋/4) = (√2) = 2 (𝑦– 4) = 2(𝑥– 𝜋/𝑥) 𝑦 = 2𝑥 – 𝜋/2 + 4 𝑦 = 2𝑥– (𝜋 + 8)/2 Question #5 Find all values of x such that the tangent line to 𝑓(𝑥) = (𝑥 2 − 9)2 is horizontal. Answer: 𝑓(𝑥) = (𝑥 2 – 9)2 y = f(x) (dy)/(dx) = 0; (f(x) = 0) (𝑑𝑦)/(𝑑𝑥) = 𝑑/𝑑𝑥(𝑥 2 – 9)2 = 2(𝑥 2 – 9)𝑑/𝑑𝑥(𝑥 2 – 9) = 2(𝑥 2 – 9)(2𝑥) = 4𝑥(𝑥 2 – 9) (dy)/(dx) = 0, then 4𝑥(𝑥 2 – 9) = 0 𝑥 2– 9 = 0 𝑥2 = 9 x = +-3 and 4x = 0 x=0 x values: -3; 3; 0.