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utoCAD gives you plenty of methods and properties so you can edit drawing objects, which you d normally have to do in the AutoCAD interface. By employing these methods and properties, you can adapt and combine editing tasks in macros and VBA applications as needed. This chapter considers editing objects, first through their methods and then through their properties. Specifically, this chapter covers the following: Copying, deleting, exploding, highlighting, mirroring, moving, offsetting, rotating, and scaling objects Working with polar and rectangular arrays Changing the color, layer, linetype, and visibility of objects In versions prior to 2006, whenever you modify an object in your code, the changes to the drawing don t appear until you call the object s Update method, the Application object s Update method, or the Document object s Regen method. In some cases, AutoCAD updates the display when your macro or program is complete, but it is safest for your code to force the update. This chapter s examples use the modified object s Update method to update the drawing display: DrawingObject.Update

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y 1 1

CHAP. 17]

Then the required value 2=3 10=3 8:3. (c) Since t 0 at 0; 1 and t 1 at 1; 2 , the line integral equals 1 1 ft2 t2 1 g dt f t2 1 2 tg 2t dt 2t5 4t3 2t2 2t 1 dt 2

Fig. 17-5

t 0 0

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It is dif cult to compute a value of x in (1, 3) for which f (x) = 12x 5 40x 4 + 24x 3 3x 2 + 12x 11 = 0 However, f (x) is itself a continuous function such that f (1) = 12 40 + 24 3 + 12 11 = 6 < 0 f (3) = 2916 3240 + 648 27 + 36 11 = 322 > 0 Hence, the intermediate-value theorem assures us that there must be some number c between 1 and 3 for which f (c) = 0.

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The following sample rotates an object based on user input. Public Sub RotateObject() Dim objDrawingObject As AcadEntity Dim varEntityPickedPoint As Variant Dim varBasePoint As Variant Dim dblRotationAngle As Double On Error Resume Next ThisDrawing.Utility.GetEntity objDrawingObject, varEntityPickedPoint, _ "Please pick an entity to rotate: " If objDrawingObject Is Nothing Then MsgBox "You did not choose an object." Exit Sub End If varBasePoint = ThisDrawing.Utility.GetPoint(, _ "Enter a base point for the rotation.") dblRotationAngle = ThisDrawing.Utility.GetReal( _ "Enter the rotation angle in degrees: ") 'convert to radians dblRotationAngle = ThisDrawing.Utility. _ AngleToReal(CStr(dblRotationAngle), acDegrees) 'Rotate the object objDrawingObject.Rotate varBasePoint, dblRotationAngle objDrawingObject.Update End Sub Use the Rotate3D method to perform a three-dimensional rotation of an object around an axis. This method has the following syntax: DrawingObject.Rotate3D AxisPoint1, AxisPoint2, RotationAngle Table 10-6 explains this method s parameters. Table 10-6. Rotate3D Method Parameters

10.2. If A 3x2 6yz i 2y 3xz j 1 4xyz2 k, evaluate A dr from 0; 0; 0 to 1; 1; 1 along C the following paths C: 2 3 a x t; y t ; z t b The straight lines from 0; 0; 0 to 0; 0; 1 , then to 0; 1; 1 , and then to 1; 1; 1 c The straight line joining 0; 0; 0 and 1; 1; 1

Since f (0) = 4 and f (2) = 16 + 2 4 = 14, the intermediate-value theorem guarantees that f has a zero between 0 and 2; call it x0 . Because f (x) = 6x 2 + 1 > 0, f (x) is increasing everywhere (Theorem 17.3). Therefore, when x > x0 , f (x) > 0; and when x < x0 , f (x) < 0. In other words, there is no zero other than x0 .

f 3x2 6yz i 2y 3xz j 1 4xyz2 kg dxi dyj dzk C 3x2 6yz dx 2y 3xz dy 1 4xyz2 dz

17.6 Prove Rolle s theorem (Theorem 17.1).

(a) If x t; y t2 ; z t3 , points 0; 0; 0 and 1; 1; 1 correspond to t 0 and t 1, respectively.

Case 1: Case 2: f (x) = 0 for all x in [a, b]. Then f (x) = 0 for all x in (a, b), since the derivative of a constant function is 0. f (x) > 0 for some x in (a, b). Then, by the extreme-value theorem (Theorem 14.2), an absolute maximum of f on [a, b] exists, and must be positive [since f (x) > 0 for some x in (a, b)]. Because f (a) = f (b) = 0, the maximum is achieved at some point c in the open interval (a, b). Thus, the absolute maximum is also a relative maximum and, by Theorem 14.1, f (c) = 0. f (x) < 0 for some x in (a, b). Let g(x) = f (x). Then, by Case 2, g (c) = 0 for some c in (a, b). Consequently, f (c) = g (c) = 0.

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