University:
College Entrance Preparation
Course:
ACT Preparation (Mathematics)
Academic year:

2021
Views:

469

Pages:

2
Author:

FrextWhestembki

Parametric Equations Quiz Review Name: 1. A curve in the xy-plane is defined parametrically by the equations x = t² + t and y = t² - t. For what value of t is the tangent line to the curve horizontal? dy/dx = (2t - 1) / (2t + 1) 2t - 1 = 0 t = 1/2 2. A particle moves along the curve defined by the parametric equations x(t) = 2t and y(t) = 36 - t² for time t, 0 ≤ t ≤ 6. Write the equation of the line tangent to the graph at (2t, 36 - t²) in terms of t and x. dy/dx = (-2t) / 2 = -t y - (36 - t²) = -t(x - 2t) y - 36 + t² = -tx + 2t² y = t² - tx + 36 3. For t > 0, there is a point on the curve where the line tangent to the curve has slope 2. At what time t is the object at this point? 1 1. cvhs.johnston.k12.nc.us cvhs.johnston.k12.nc.us dy/dx = (sin(t² + 1)) / (tan(t/ln(t))) = 2 4. Given the parametric equations: x = t + 1 y = t² + 2 Sketch a graph of this curve and derive a corresponding xy equation. t = x - 1 y = (x - 1)² + 2 y = x² - 2x + 1 + 2 y = x² - 2x + 3 [Image of a graph with x and y axes, and a parabola plotted] 5. Generate parametric equations for the line segment from (-4, 2) to (-2, 1). x = -4 + 2t y = 2 - t 6. Generate parametric equations for a circle of radius 5 with center at (2, 6). x = 2 + 5cos(t) y = 6 + 5sin(t) 7. Find all points of intersection between the following parametric equations. x = t + 3 y = t² x = 1 - s y = 2 - s² t + 3 = 1 - s t² = 2 - s² t + s = -2 t² + s² = 2 Solving these equations, we get: t = -1, s = -1 Therefore, the point of intersection is: x = 2, y = 1 8. Given the parametric equations below, find dy/dx at a) t = 1/2, b) t = π, and c) at (-2, 0). x = 2cos(t) y = 3sin(t) a) dy/dx = (dy/dt) / (dx/dt) = (3cos(t)) / (-2sin(t)) dy/dx(π/2) = (3cos(π/2)) / (-2sin(π/2)) = 0 / -2 = 0 b) dy/dx(π) = (3cos(π)) / (-2sin(π)) = -3 / 0 = undefined c) To find t at (-2, 0): 2cos(t) = -2 cos(t) = -1 t = π Therefore, dy/dx at (-2, 0) is undefined.