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SECTION 9.2

SECTION 9.2. CALCULUS WITH PARAMETRIC CURVES. TANGENTS. If dx / dt ≠ 0 , we can solve for dy / dx :. Example 1. A curve C is defined by the parametric equations x = t 2 , y = t 3 – 3 t. Show that C has two tangents at the point (3, 0) and find their equations.

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SECTION 9.2

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  1. SECTION 9.2 CALCULUS WITH PARAMETRIC CURVES

  2. TANGENTS • If dx/dt≠ 0, we can solve for dy/dx: 9.2

  3. Example 1 • A curve C is defined by the parametric equations x = t2, y = t3– 3t. • Show that C has two tangents at the point (3, 0) and find their equations. • Find the points on C where the tangent is horizontal or vertical. • Determine where the curve is concave upward or downward. • Sketch the curve. 9.2

  4. Example 1(a) SOLUTION • Notice that y = t3 – 3t = t(t2– 3) = 0 when t = 0 or t = ± . • Thus, the point (3, 0) on C arises from two values of the parameter: t = and t = – • This indicates that C crosses itself at (3, 0). 9.2

  5. Example 1(a) SOLUTION • Since • the slope of the tangent when t = ± is: dy/dx = ± 6/(2 ) = ± • So, the equations of the tangents at (3, 0) are: 9.2

  6. Example 1(b) SOLUTION • C has a horizontal tangent when dy/dx = 0,that is, when dy/dt = 0 and dx/dt≠ 0. • Since dy/dt = 3t2 – 3, this happens when t2 = 1, that is, t = ±1. • The corresponding points on C are (1, – 2) and (1, 2). • C has a vertical tangent when dx/dt = 2t = 0, that is, t = 0. • Note that dy/dt ≠ 0 there. • The corresponding point on C is (0, 0). 9.2

  7. Example 1(c) SOLUTION • To determine concavity, we calculate the second derivative: • The curve is concave upward when t > 0. • It is concave downward when t < 0. 9.2

  8. Example 1(d) SOLUTION • Using the information from (b) and (c), we sketch C. 9.2

  9. Example 2 • Find the tangent to the cycloid x = r(– sin ), y = r(1 – cos ) at the point where  = p/3. • See Example 7 in Section 9.1. • At what points is the tangent horizontal? When is it vertical? 9.2

  10. Example 2(a) SOLUTION • The slope of the tangent line is: 9.2

  11. Example 2(a) SOLUTION • When  = p/3, we haveand 9.2

  12. Example 2(a) SOLUTION • Hence, the slope of the tangent is . • Its equation is: 9.2

  13. Example 2(a) SOLUTION • The tangent is sketched in Figure 2. 9.2

  14. Example 2(b) SOLUTION • The tangent is horizontal when dy/dx = 0. • This occurs when sin  = 0 and 1 – cos ≠0, that is,  = (2n – 1)p, n an integer. • The corresponding point on the cycloid is ((2n – 1)pr, 2r). 9.2

  15. Example 2(b) SOLUTION • When  = 2np, both dx/d and dy/d are 0. • It appears from the graph that there are vertical tangents at those points. 9.2

  16. Example 2(b) SOLUTION • We can verify this by using l’Hospital’s Rule as follows: • A similar computation shows that dy/dx →–∞ as → 2np–. • So, indeed, there are vertical tangents when  = 2np, that is, when x = 2npr. 9.2

  17. Example 3 • Find the area under one arch of the cycloid x = r(– sin) y = r(1 – cos) • See Figure 3. 9.2

  18. Example 3 SOLUTION • One arch of the cycloid is given by 0 ≤  ≤ 2p. • Using the Substitution Rule withy = r(1 – cos ) and dx = r(1 – cos) d, we have the following result. 9.2

  19. Example 3 SOLUTION 9.2

  20. Theorem 5 Let a curve C is described by the parametric equations x = f(t), y = g(t), a≤t≤b, where f ’ and g’ are continuous on [a, b] and C is traversed exactly once as t increases from a to b. Then, the length of C is: 9.2

  21. Example 4 • Suppose we use the representation of the unit circle given in Example 2 in Section 9.1:x = cos ty = sin t 0 ≤ t ≤ 2p • Then, dx/dt = –sin t and dy/dt = cos t • So, as expected, Theorem 5 gives: 9.2

  22. Example 4 • On the other hand, suppose we use the representation given in Example 3 in Section 9.1:x = sin 2t y = cos 2t 0 ≤ t ≤ 2p • Then, dx/dt = 2cos 2t and dy/dt = –2sin 2t 9.2

  23. Example 4 • Then, the integral in Theorem 5 gives: 9.2

  24. Example 4 • Notice that the integral gives twice the arc length of the circle. • This is because, as t increases from 0 to 2π, the point (sin 2t, cos 2t) traverses the circle twice. • In general, when finding the length of a curve C from a parametric representation, we have to be careful to ensure that C is traversed only once as t increases from a to b. 9.2

  25. Example 5 • Find the length of one arch of the cycloid x = r(– sin) y = r(1 – cos) • SOLUTION • From Example 3, we see that one arch is described by the parameter interval 0 ≤ ≤ 2p. • Since 9.2

  26. Example 5 SOLUTION • We have 9.2

  27. Example 5 SOLUTION • To evaluate this integral, we use the identity with  = 2x. • Thisgives 1 – cos  = 2sin2(/2). • Since 0 ≤  ≤ 2π, we have 0 ≤ /2 ≤ π, and so sin(/2) ≥ 0. • Therefore, 9.2

  28. Example 5 SOLUTION • Hence, 9.2

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