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For x immediately to the left of 1 , x 1 < 0 and x 3 < 0, and so, f (x) > 0. For x immediately to the right of 2 2 2 1 , x 1 > 0 and x 3 < 0 and so, f (x) < 0. Thus, we have the case {+, }, and f has a relative maximum at x = 1 . 2 2 2 2 (The second-derivative test could have been used instead.) The function f has no absolute minimum on [0, 1). Its graph has 4x 2 3 = (see Fig. 24-3).1 the line x = 1 as a vertical asymptote, since lim f (x) = lim x 1 x 1 x 1

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By further di erentiation, using the notation f 00 u d 2 f =du2 , g 00 v d 2 g=dv2 , we have 1 @2 Y @Yt @Yt @u @Yt @v @ @ fa f 0 u a g 0 v g a fa f 0 u a g 0 v g a @t @u @t @v @t @u @v @t2 a2 f 00 u a2 g 00 v @2 Y @Yx @Yx @u @Yx @v @ @ f f 0 u g 0 v g f f 0 u g 0 v g @x @u @x @v @x @u @v @x2 f 00 u g 00 v Then from (1) and (2), @2 Y=@t2 a2 @2 Y=@x2 .

Fig. 24-3

Although the BlockReference object has identical geometry, each block reference s AttributeReference objects may have different values. You can access these attribute references using the GetAttributes method.

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CHAP. 6]

Fig. 26-5 Theorem 26.5 (Law of Cosines): In any triangle ABC (see Fig. 26-7), c2 = a2 + b2 2ab cos

Method 1: Di erentiate the given equations with respect to x, considering u and v as functions of x and y. Then 1 @u 1 12v ; @x 1 8uv 2u @u @v 3 @x @x 2 @u @v 4v 1 @x @x

Fig. 26-6

Solving,

For a proof, see Problem 26.3. Note that the Pythagorean theorem is obtained as the special case = /2.

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Use the GetAttributes method of a BlockReference or MInsertBlock object to retrieve any associated AttributeReference objects. It returns an array, varAttributeRefs, that contains all the nonconstant attributes. This method has the following syntax: varAttributeRefs = Object.GetAttributes() The following example uses the Debug window to display information about each AttributeRef object associated with a block reference. Try it on an attributed block, such as the one created in the previous example. Public Sub TestGetAttributes() Dim varPick As Variant Dim objEnt As AcadEntity Dim objBRef As AcadBlockReference Dim varAttribs As Variant Dim strAttribs As String Dim intI As Integer On Error Resume Next With ThisDrawing.Utility '' get an entity from user .GetEntity objEnt, varPick, vbCr & "Pick a block with attributes: " If Err Then Exit Sub

Solving,

Fig. 26-7 EXAMPLE Find angle in the triangle of Fig. 26-8.

We have, of course, assumed that 1 8uv 6 0. Method 2: The given equations are F u2 v 3x y 0, G u 2v2 x 2y 0. Then by Problem 6.31,

Solving the law of cosines for cos , cos = Then, from Table 26-1, = /3.

CHAP. 6]

Fig. 26-8

'' cast it to a blockref Set objBRef = objEnt '' exit if not a block If objBRef Is Nothing Then .Prompt vbCr & "That wasn't a block." Exit Sub End If '' exit if it has no attributes If Not objBRef.HasAttributes Then .Prompt vbCr & "That block doesn't have attributes." Exit Sub End If '' get the attributerefs varAttribs = objBRef.GetAttributes '' show some information about each strAttribs = "Block Name: " & objBRef.Name & vbCrLf For intI = LBound(varAttribs) To UBound(varAttribs) strAttribs = strAttribs & " Tag(" & intI & "): " & _ varAttribs(intI).TagString & vbTab & " Value(" & intI & "): " & _ varAttribs(intI).TextString & vbCrLf Next End With MsgBox strAttribs End Sub

PARTIAL DERIVATIVES Fx @ F; G Gx @u @ x; v Fu @ F; G @x Gu @ u; v 3 1 Fv 1 4v 1 12v Gv 2u 1 1 8uv Fv 1 4v Gv

a 2 + b2 c 2 (5)2 + (2)2 ( 19)2 25 + 4 19 1 = = = 2ab 2(5)(2) 20 2

6.33. If F u; v; w; x; y 0, G u; v; w; x; y 0, H u; v; w; x; y 0, nd @v ; b @x ; c @w : a @y x @v w @u y

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