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27.21 Let f (x) = sin x + sin |x|. (a) Sketch the graph of f . (b) Is f continuous at 0 (c) Is f differentiable at x = 0 27.22 Find the slope of the tangent line to the graph of: (a) y = x + cos (xy) at (0, 1); (b) sin (xy) = y at ( /2, 1). [Hint: Use implicit differentiation.] 27.23 Use implicit differentiation to nd y : (a) cos y = x (b) sin (xy) = y2 27.24 (a) A ladder 26 feet long is leaning against a vertical wall (see Fig. 27-9). If the bottom of the ladder A is slipping away from the base of the wall at the rate of 3 feet per second, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 10 feet from the base of the wall (b) An airplane is ascending at a speed of 400 kilometers per hour along a line making an angle of 60 with the ground. How fast is the altitude of the plane changing 27.25 (a) A man at a point P on the shore of a circular lake of radius 1 mile (see Fig. 27-10) wants to reach the point Q on the shore diametrically opposite P. He can row 1.5 miles per hour and walk 3 miles per hour. At what angle (0 /2) to the diameter PQ should he row in order to minimize the time required to reach Q (When = 0, he rows all the way; when = /2, he walks all the way.) (b) Rework part (a) if, instead of rowing, the man can paddle a canoe at 4 miles per hour.

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In particular, the position vector or radius vector r from O to the point x; y; z is written p and has magnitude r jrj x2 y2 z2 : r xi yj zk 3

Fig. 27-9

DOT OR SCALAR PRODUCT The dot or scalar product of two vectors A and B, denoted by A B (read A dot B) is de ned as the product of the magnitudes of A and B and the cosine of the angle between them. In symbols, A B AB cos ; 0@ @ 4

27.26 (a) Find the absolute extrema of f (x) = x sin x on [0, /2]. (b) From part (a), infer that sin x < x holds for all positive x. (c) GC Verify on a graphing calculator that cos x

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You can look at the available printers using the Options dialog box inside AutoCAD, as shown in Figure 15-2. The Plotting tab displays all the available printer device names. Unfortunately, AutoCAD doesn t provide a means to see this list programmatically.

Assuming that neither A nor B is the zero vector, an immediate consequence of the de nition is that A B 0 if and only if A and B are perpendicular. Note that A B is a scalar and not a vector. The following laws are valid: 1. A B B A Commutative Law for Dot Products

Fig. 27-10

2. A B C A B A C Distributive Law 3. m A B mA B A mB A B m, where m is a scalar. 4. i i j j k k 1; 5. i j j k k i 0 A B A1 B1 A2 B2 A3 B3 The equivalence of this component form the dot product with the geometric de nition 4 follows from the law of cosines. (See Fig. 7-8.) If A A1 i A2 j A3 k and B B1 i B2 j B3 k, then

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opposite side adjacent side adjacent side cot opposite side hypotenuse sec adjacent side hypotenuse csc opposite side

In particular, jCj2 jAj2 jBj2 2jAjjBj cos Since C B A its components are B1 A1 ; B2 A2 ; B3 A3 and the square of its magnitude is B2 B2 B2 A2 A2 A2 2 A1 B1 A2 B2 A3 B3 1 2 3 1 2 3 or jBj2 jAj2 2 A1 B1 A2 B2 A3 B3

sin x cos x (cos x)Dx (sin x) (sin x)Dx (cos x) = (cos x)2 (cos x)(cos x) (sin x)( sin x) = cos2 x 2 x + sin2 x 1 cos = = 2x cos cos2 x 2x = sec 1 Dx (cot x) = Dx ((tan x) 1 ) = Dx (tan x) (tan x)2 1 1 1 = = 2 (cos x)2 (tan x) (tan x cos x)2 1 = (sin x)2 Dx (tan x) = Dx = csc2 x Differentiating the rst identity of Theorem 28.3, 2(tan x)(sec2 x) = 2(sec x)Dx (sec x) and dividing through by 2(sec x), which is never zero, gives Dx (sec x) = tan x sec x Similarly, differentiation of the second identity of Theorem 28.3 gives Dx (csc x) = cot x csc x

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