MA4001 Engineering Mathematics 1 Lecture 14 Derivatives of Trigonometric Functions Critical Points

Transcription

1 MA4001 Engineering Mathematics 1 Lecture 14 Derivatives of Trigonometric Functions Critical Points Dr. Sarah Mitchell Autumn 2014

2 An important limit To calculate the limits of basic trigonometric functions we nee sin x to know the limit lim x 0 x. From the graph it looks like the limit might be 1 (if the function is continuous).

3 sin x lim x 0 x As the function is even the two one-sie limits must be the sin x same, so we just nee to check lim x 0+ x. As x 0 we can assume that 0 < x < π 2 angle in the first quarant. an think of x as an

4 In this graph the point P has coorinates cos x, sin x an T has coorinates (1, tan x). The area of triangle OAP < the area of the sector OAP < the area of the triangle OAT.

5 Using the formula that the area of a triangle is 1 2 base perpenicular height: Area of triangle OAP = 1 1 sin x. 2 Area of triangle OAT = 1 1 tan x. 2 Also the area of a sector of a circle with angle θ is 1 2 θ raius2. (e.g., angle 2π, raius r, area=πr 2 ) Thus area of sector OAP is 1 2 x 12 = x 2.

6 We thus have the bouns: 1 2 sin x < x 2 < 1 2 tan x Thus or, inverting this: 1 < x sin x < 1 cos x cos x < sin x < 1 x Thus by the squeeze theorem, since we have lim x 0 sin x x 1 = lim x 0 cos x lim x 0 sin x x = 1 lim x 0 1 = 1

7 Other limits We can then obtain tan x lim x 0 x = lim x 0 sin x x cos x = lim x 0 sin x x lim x 0 1 cos x = 1 An also: lim x 0 1 cos x x 2 = lim x 0 2 sin 2 x/2 x 2 Let t = x, so that x = 2t an lim 2 lim t 0 x 0

8 Thus 1 cos x lim x 0 x 2 = lim t 0 2 sin 2 t (2t) 2 = 1 2 lim t 0 ( sin t t ) 2 = 1 ( lim 2 t 0 ) sin t 2 = 1 t 2 Remark These limits imply that for small values of x (in raians): sin x x; tan x x an 1 cos x x or cos 1 x 2 2

9 Derivative of sin x sin(x + h) sin x sin x = lim x h 0 h Using the formula sin a sin b = 2 sin a b 2 have cos a+b 2 we thus 2 sin h 2 sin x = lim cos(x + h 2 ) x h 0 h sin h 2 = lim lim h/2 0 h cos(x + h h 0 2 ) 2 = 1 cos x = cos x

10 Graphs

11 Derivative of cos x x cos x = ( π ) x sin 2 x = x sin u where u = π 2 x = u sin u by the chain rule u x = (cos u) ( 1) ( π ) = cos 2 x = sin x

12 Graphs

13 We can generalise these erivatives using the chain rule x f(kx + c) = f (kx + c) k an have an sin(kx + c) = k cos(kx + c) x cos(kx + c) = k sin(kx + c) x

14 Derivative of tan x an cot x x tan x = sin x x cos x Similarly one can show that = (cos x)(sin x) (sin x)(cos x) cos 2 x (cos x)(cos x) (sin x)( sin x) = cos 2 x = cos2 x + sin 2 x cos 2 x = 1 cos 2 x = sec2 x x cot x = csc2 x.

15 Derivative of sec x an csc x x sec x = 1 x cos x (cos x) = cos 2 (by the rule for reciprocals) x = sin x cos 2 x = (sec x)(tan x) Similarly one can show that csc x = (csc x)(cot x). x

16 Summary f(x) sin x cos x tan x cot x f (x) cos x sin x sec 2 x csc 2 x sec x (sec x)(tan x) csc x (csc x)(cot x)

17 Example: x ( x 2 sin x ) ( x 2 sin ) x x = (x 2 ) sin x + x 2 (sin x) (by the prouct rule) = 2x sin x + x 2 cos x x x (by the chain rule) = 2x sin x + x 2 cos x 1 2 x = 2x sin x x x cos x

18 Example Fin the tangent line to y = tan πx 4 at the point (1, 1). If the slope of the tangent line is m then the equation of the tangent line is y y 0 = m(x x 0 ) that is y tan π 4 = m(x 1) i.e., y 1 = m(x 1) The slope m is x tan πx 4 x=1 = π 4 Thus the tangent line is sec2 πx 4 = π π x=1 4 sec2 4 = π 4 cos 2 π 4 = π 2 y 1 = π (x 1) 2

19 Definition A critical point of the function f(x) is a point where f (x) = 0. Theorem If f(x) is ifferentiable in (a, b) an achieves a maximum or minimum at c (a, b), then f (c) = 0, i.e., c is a critical point.

20 Proof. If f(x) has a maximum at c then f(x) f(c) 0 for all x (a, b). Thus f f(c + h) f(c) (c) = lim 0, since h 0, an h 0+ h f f(c + h) f(c) (c) = lim 0, since h 0. h 0 h Thus f (c) = 0. The case of a minimum is similar. Remark f (c) = 0 oes not imply that there is a maximum or minimum at x = c.

21 Example y = x 3 y = 3x 2 = 0 at x = 0 Thus x = 0 is a critical point. But it is neither a maximum nor a minimum.