# Lecture 10

22 de May de 2020
1 de 11

### Lecture 10

• 2. DEFINITION: PARAMETRIC EQUATIONS If there are functions f and g with a common domain T, the equations x = f(t) and y = g(t), for t in T, are parametric equations of the curve consisting of all points ( f(t), g(t) ), for t in T. The variable t is the parameter. The equations x = t + 2 and y = 3t – 1 for example are parametric equations and t is the parameter. The equations define a graph. If t is assigned a value, corresponding values are determined for x and y. The pair of values for x and y constitute the coordinates of a point of the graph. The complete graph consists of the set of all points determined in this way
• 3. as t varies through all its chosen values. We can eliminate t between the equations and obtain an equation involving x and y. Thus, solving either equation for t and substituting in the other, we get 3x – y = 7 The graph of this equation, which also the graph of the parametric equations, is a straight line. Example 1: Sketch the graph of the parametric equations x = 2 + t and y = 3 – t2 . t -3 -2 -1 0 1 2 3 x -1 0 1 2 3 4 5 y -6 -1 2 3 2 -1 -6
• 5. Example 2: Eliminate the parameter between x = t + 1 and y = t2 + 3t + 2 and sketch the graph. Solution: Solving x = t + 1 for t, we have t = x – 1. Substitute into y = t2 + 3t + 2, then y = (x – 1)2 + 3(x – 1) + 2 y = x2 – 2x + 1 + 3x – 3 + 2 y = x2 + x Reducing to the standard form, y + ¼ = x2 + x + ¼ y + ¼ = (x + ½)2 , a parabola with V(-½,-¼) opening upward
• 7. Example 3: Eliminate the parameter between x = sin t and y = cos t and sketch the graph. Solution: Squaring both sides of the parametric equations, we have x2 = sin2 t and y2 = cos2 t And adding the two equations will give us x2 + y2 = sin2 t + cos2 t But sin2 t + cos2 t = 1 Therefore x2 + y2 = 1 , a circle with C(0, 0) and r = 1
• 9. Example 4: Find the parametric representation for the line through (1, 5) and (-2, 3). Solution: Letting (1, 5) and (-2, 3) be the first and second points, respectively, of x = x1 + r(x2 – x1) and y = y1 + r(y2 – y1) We then have x = 1 + r(-2 – 1) and y = 5 + r(3 – 5) x = 1 – 3r y = 5 – 2r
• 10. Example 5: Eliminate the parameter between x = sin t + cos t and y = sin t. Solution: Solving sin2 t + cos2 t = 1 for cos t, we have Substitute into x = sin t + cos t , then x = sin t + But y = sin t and y2 = sin2 t Therefore x = y + x – y = Squaring both sides (x – y)2 = 1 – y2 tsin1tcos 2 −= tsin1 2 − 2 y1− 2 y1−
• 11. Exercises: Eliminate the parameter and sketch the curve. • x = t2 + 1, y = t + 1 • x = t2 + t – 2 , y = t + 2 • x = cos θ , y = cos2 θ + 8 cos θ • x = 4 cos θ , y = 7 sin θ • x = cos θ , y = sin 2θ • x = 1 + cos 2θ , y = 1 – sin θ