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Since air is leaking out at the rate of 3 cubic inches per minute, the volume V of the balloon is decreasing at the rate 4 of dV/dt = 3. But the volume of a sphere of radius r is V = 3 r 3. Hence, 3 = So, d 4 3 dV = r dt dt 3 dr 3 = dt 4 r 2 = 4 d(r 3 ) 4 dr = 3r 2 3 dt 3 dt = 4 r 2 dr dt

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The number of objects in the polar array. A nonzero value representing the angle, measured counterclockwise in radians, over which the array path will extend. A three-element array of doubles specifying the 3-D WCS coordinates of the arc s center.

Since this is an identity, 5A 3B 6, 2A B 1 and A 3=11, B 17=11. Then 6 x 3=11 17=11 3 17 dx dx dx ln jx 3j ln j2x 5j c x 3 2x 5 x 3 2x 5 11 22 Method 2: Substitute suitable values for x in the identity (1). For example, letting x 3 and x 5=2 in (1), we nd at once A 3=11, B 17=11.

CHAP. 20]

1 +

Substituting r = 5, 3 3 dr = 0.009 55 dt 100 314 Thus, when the radius is 5 inches, the radius is decreasing at about 0.01 inch per minute.

Fig. 5-7

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20.2 A 13-foot ladder leans against a vertical wall (see Fig. 20-3). If the bottom of the ladder is slipping away from the base of the wall at the rate of 2 feet per second, how fast is the top of the ladder moving down the wall when the bottom of the ladder is 5 feet from the base

Always set the NumberofObjects parameter greater than 1. The original object is counted as an element of the polar array, even though this method doesn t return it in the object array. Consequently, the size of the object array is NumberofObjects-1. If you set NumberofObjects to 1, the method returns an empty array, but doesn t raise an error. Figure 10-8 shows an ellipse arranged along an arc using the ArrayPolar method.

Let x y. Then x sin x y sin y sin y y sin y I dx dy dy dy 2x 2y 2y 2 0 1 cos 0 1 cos 0 1 cos 0 1 cos y d cos y I tan 1 cos y j I 2 =2 I 0 2 0 1 cos y i.e.; I 2 =2 I or I 2 =4:

Fig. 20-3

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Letting x =2 y, we have p p =2 =2 =2 p cos y cos x sin x p p dx I p p dx p p dy sin x cos x cos x sin x cos y sin y 0 0 0 Then I I p p =2 cos x sin x p p dx p p dx cos x sin x sin x cos x 0 0 =2 p p =2 sin x cos x p p dx dx 2 sin x cos x 0 0 =2

Let x be the distance of the bottom of the ladder from the base of the wall, and let y be the distance of the top of the ladder from the base of the wall. Since the bottom of the ladder is moving away from the base of the wall at 2 feet per second, dx/dt = 2. We wish to compute dy/dt when x = 5 feet. Now, by the Pythagorean theorem, (13)2 = x 2 + y2 Differentiation of this, as in example (c), gives 0=x But when x = 5, (1) gives y= so that (2) becomes dy dt 2(5) 5 dy = = dt 12 6 0 = 2(5) + 12 Hence, the top of the ladder is moving down the wall (dy/dt < 0) at 5 feet per second when the bottom of the ladder is 6 5 feet from the wall. (13)2 (5)2 = 169 25 = 144 = 12 dx dy dy +y = 2x + y dt dt dt (2) (1)

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