Apr 23, Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 1 / and 19pa
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1 Calculus with Algebra and Trigonometry II Lecture 23 Final Review: Curve sketching and parametric equations Apr 23, 2015 Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 1 / and 19pa
2 Curve sketching To sketch the graph of a function, y = f (x), follow these steps Zeros Find the values of x where f (x) = 0 Asymptotes The function will have horizontal asymptotes if lim x ± f (x) exist. The function will have a vertical asymptote at x = a if lim x a f (x) doesn t exist. The function will have a slant asymptote if lim x ± f (x) x exists First derivative Use the first derivative to determine the critical points and the intervals where the function is increasing or decreasing Second derivative Use the econd derivative to find the points of inflection and the intervals where the function is convex or concave. Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 2 / and 19pa
3 Example 1 Sketch the function f (x) = x 3 + 3x 2 2 Zeros One zero is -1. This implies (x + 1) divides evenly into x 3 + 3x 2 2 x 2 + 2x 2 x + 1 ) x 3 + 3x 2 2 x 3 x 2 2x 2 2x 2 2x 2x 2 2x Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 3 / and 19pa
4 The other zeros are solutions of x 2 + 2x 2 = 0. Using the quadratic formula gives x = 2 ± = 1 ± 3 2 Asymptotes lim f (x) x lim f (x) x So there are no horizontal asymptotes. so there are no vertical asymptotes. so there is no slant asymptote either. lim f (x) exists for all a R x a f (x) lim x x Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 4 / and 19pa
5 First derivative The critical points are x = 0, 2 f (x) = 3x 2 + 5x = 3x(x + 2) The function is increasing on (, 2) and (0, ) and decreasing on ( 2, 0) Thus x = 2 is a maximum and x = 0 is a minimum. Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 5 / and 19pa
6 Second derivative f (x) = 6x + 6 f (x) has a point of inflection at x = 1. It is convex on ( 1, ) and concave on (, 1). Putting it all together gives Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 6 / and 19pa
7 Example 2 Sketch g(x) = Zeros There is only one at x = 0. Asymptotes lim x 2x 1 + x 2 2x 1 + x 2 = 0 lim x 2x 1 + x 2 = 0 So the x axis is a horizontal asymptote. Since 1 + x 2 is always positive there are no vertical asymptotes. First derivative f (x) = 2(1 + x 2 ) 2x(2x) (1 + x 2 ) 2 = 2(1 x 2 ) (1 + x 2 ) 2 Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 7 / and 19pa
8 The critical points are x = 1 a minimum and x = 1 a maximum. The function is increasing on ( 1, 1) and decreasing everywhere else Second derivative f (x) = 2(1 x 2 )(4x(1 + x 2 ) + 4x(1 + x 2 ) 2 (1 + x 2 ) 4 = 4x(3 x 2 ) (1 + x 2 ) 3 The points of inflection are 0, ± 3. The function is convex on ( 3, 0) and ( 3, ) and concave on (, 3 and (0, 3). Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 8 / and 19pa
9 The graph of the function is Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr Curve 23, 2015 sketching 9 / and 19pa
10 Example 3 Sketch h(x) = x 2 6x + 11 x 2 Zeros Using the quadratic formula x = 6 ± so the function has no real zeros Asymptotes lim h(x) x ± so it doesn t have a horizontal asymptote. = 3 ± 2i do not exist lim h(x) x 2 doesn t exist so it has a vertical asymptote at x = 2. Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 10 / and 19pa
11 h(x) lim = 1 x x so it has a slant asymptote. Using long division so So the slant asymptote is x 4 x 2 ) x 2 6x + 1 x 2 + 2x 4x + 1 4x 8 7 h(x) = x x 2 y = x 4 Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 11 / and 19pa
12 First derivative h (x) = 1 3 (x 2) 2 The function has crical points at x = 2 3 a maximum and x = a minimum. Second derivative h (x) = 3 (x 2) 3 so the function is convex for x > 2 and concave for x < 2. Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 12 / and 19pa
13 The graph is Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 13 / and 19pa
14 Parametric equations Curves given in parametric form x = x(t) y = y(t) have critical points when x (t) = 0, corresponding to a vertical tangent line, and y (t) = 0, corresponding to a horizontal tangent. in general the slope of the tangent line is dy dx = y (t) x (t) and the arc length is given by the expression s = x 2 + y 2 dt Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 14 / and 19pa
15 Example 4 For the curve x = t y = t 1 t t > 0 (a) Find the cartesian equation, (y = f (x)), for the curve. (b) Find the critical points. (c)find the equation of the tangent line to the curve at t = 4. (a) x 2 = t y = x 2 1 x (b) so there are no vertical tangents. x (t) = 1 2 t 1/2 0 y (t) = t 3/2 0 so there are no horizontal tangents either Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching / and 19pa
16 (c) x(4) = 2 y(4) = 7 2 x (4) = 1 y (4) = dy dx = y x = = 17 4 So the equation of the line is y 7 2 = 17 (x 2) 4 Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 16 / and 19pa
17 Example 5 For the curve (a) Find the critical points (b) Find the length of the curve x = 3 2 t2 y = 4 3 t3 0 t 1 x = 3t y = 4t 2 so the only possible critical point is at t = 0. The slope there is dy dx = 4t2 3t = 4 3 t = 0 so the tangent line is horizontal. 1 1 s = 9t t 4 dt = t t 2 dt (1) 0 = u du = [ 2 3 u3/2 ] 25 = (2) 9 9 Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 17 / and 19pa
18 Example 6 For the curve x = 2 cos t y = 2 sin t cos t 0 t π 2 (a) Find the critical points (b) Find the length of the curve Vertical tangents Horizontal tangents x = 4 sin t cos t y = 2(cos 2 t sin 2 t) x = 0 t = 0, π 2 y = 0 t = π 4 Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 18 / and 19pa
19 = Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 19 / and 19pa
20 The arc length is s = = 2 = 2 = 2 π 2 0 π 2 0 π 2 0 π sin 2 t cos 2 t + 4(sin 2 t cos 2 t) 2 dt (3) 4 sin 2 t cos 2 t + sin 4 t 2 sin 2 t cos 2 t cos 4 t dt (4) (sin 2 t + cos 2 t) 2 dt (5) dtπ (6) Calculus with Algebra and Trigonometry II Lecture 23Final Review: Apr 23, Curve 2015sketching 19 / and 19pa
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