Unformatted text preview: SECTION 4.2 THE MEAN VALUE THEOREM 275 4. In the primary rainbow. the rainbow angle gets smaller as k gets larger. as we found in Problem 2. so the colors
appear from top to bottom in order of increasing k. But in the secondary rainbow. the rainbow angle gets larger as It
gets larger. To see this. we ﬁnd the minimum deviations for red light and for violet light in the secondary rainbow. /1.. 1 2 —
For k: R: 1.3318 (red light) the minimum occurs at m % arccos E? m 1.255 radians. and so the rainbow angle is D(a1) — 71' m 50.60. For k 2 1.3435 (violet light) the minimum occurs at 2
. — 1
cm s arccos [I % x 1.248 radians. and so the rainbow angle is D(a2) — 7r 2 53.60. Consequently. the rainbow angle is larger for colors with higher indices of refraction. and the colors appear from bottom to top in order
of increasing k, the reverse of their order in the primary rainbow. Note that our calculations above also explain why the secondary rainbow is more spread out than the primary
rainbow: in the primary rainbow, the difference between rainbow angles for red and violet light is about 1.70. whereas in the secondary rainbow it is about 30. 4.2 The Mean Value Theorem 1. f(r) 2 $2 2 4m + 1. [0. 4]. Since f is a polynomial. it is continuous and differentiable on R. so it is continuous on
[0.4] and differentiable on (0,4). Also. f(O) 2 12 f(4). f’(c) 2 0 (2 2c — 4 2 0 (2 c 2 2. which is in the open interval (0. 4). so 6 2 2 satisﬁes the conclusion of Rolle”s Theorem. 2. f(z) 2 m3 — 3w2 + 230 + 5, [0. 2]. f is continuous on [0. 2] and differentiable on (0. 2). Also. f(O) 2 5 2 f(2). f’(c)—0 a 3c2 60:2 0 2 c 6:Vii—24—1__l\/§.bothin(0.2). 3. f(at) 2 sin 27m. [—1, 1]. f. being the composite of the sine function and the polynomial 27m. is continuous and differentiable on R. so it is continuous on [—1.1] and differentiable on (21.1). Also. f(—1) 2 0 2 f(l). f’(c)_0 < 2ncos2nc—0 z cosZHC—O > 2/ic2_g+27m (2 c::§+n.1fn:00r i1. then c : :41. ii is in (21, 1). 4. f(az) 2 mx/a: + 6. [26. 0]. f is continuous on its domain. [26. 00). and differentiable on (—6. 00). so it is
3c l— 12 *20
2x/c+6 continuous on [26.0] and differentiable on (26. 0). Also. f(26) 2 0 2 f(0). f’(c) 2 0 (2 c 2 —4. which is in (—6. 0). 5. f(m) 2 1 2 $2/3. f(21) 2 1 2 (21)”3 2 1 ~12 0 2 f(1). f/(ac) 2 2§$_1/3. so f’(c) 2 Ohas no
solution. This does not contradict Rolle’s Theorem. since f’(0) does not exist. and so f is not differentiable 011(21.1). 5 f(JI) = (w — 1)‘2 f(0) = (0 — 1)‘2 =1:(2 1)E2 I f(2)‘ f’($) = 22($ — 1)_3 => ﬂat?) is never 0. This does not contradict Rolle’s Theorem since f’(1) does not exist. ...
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 Spring '10
 Ban
 Calculus

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