Answers. *P*l=$\+**v') Igtri3, pages ) 72 * Cneprrn 2

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1 72 * Cneprrn 2 A calculator may be used for this question. 7. An isosceles triangle is inscribed in a semicircle, as shown in the diagram, and it continues to be inscribed as the semicircle changes size. The area of the semicircle is increasing at the rate of 1 cmzlsec when the radius of the semicircle is 3 cm. a. How fast is the radius of the semicircle increasing when the radius is 3 cm? Include units in your. b. How fast is the perimeter of the semicircle increasing when the radius is 3 cm? Include units in your. c. How fast is the area of the isosceles triangle increasing when the radius is 3 cm? Include units in your. d. How fast is the shaded region increasing when the radius is 3 cm? Include units in your. Answers Mumpu Cuorcr 1. (C) The given limit represents the alternative form of the definition of a derivative: f'(c) =,'* f(x) - f(c). x)c x-c If the function is defined as f(x) = In(x + 1), then substituting in c = E 1. (Catcutus Bth ed. pases / 9th ed. sives f'(1) = Igtri3, pages ) 4 (E) To flnd the derivative, change the form of the function: /=. to y =4x-3. Now find the derivative using the Power Rule and Constant Multiple Rule, {=a(-3)x 3' = -72x - = -+. (Calculus8th ed. pages 'dx,x* / 9th ed. pages ) 3. (D) To flnd the derivative, differentiate both sides of the equation implicitly with respect to x. *P*l=$\+**v') s(, *"EL) (" dx) =++2v!L "dx 3v " +3x!v = +*zrl. dx "dx

2 DlrrrRrrulRlolv r73 Now isolate the terms containing factor. z** -zvl = 4-3y dx "dx!rc*-2yr=4-3y dx dy dx and then factor out the common Now solve ro. {. dv - - n-'y = -,7,(lv - -4). = lu --n Note that the dx dx 3x-2y -1(2y-3x) 2y-3x 4. form for the could have been,n*'y. Be sure to watch for 3x -2y factoring out negative values on multiple-choice questions. tcalculus Bth ed. pages 747*145 / 9th ed. pages ) (A) Begin by differentiating both sides of the equation with respect to X. d,..., d. dr(""" )= *lvl -**, (,. dy) dy e " I I +- t=- I dxl dx -**, -*ru dy dy dx dx e,'y - =t--"*', L =l'r-",-, 14L dx dx t 'dx Solve for q-. o! =, "'"1,,, (Ca.lcu,lus Bth ed. pages / gth dx dx (1 -"',) ed. pages ) (D) To flnd the seventh derivative, differentiate until a pattern is observed. /=-sinx dv " - dx -cosx d'v -* = SInx d3v --* = CoSX y(a) = -sinx Because the fourth derivative is the same as the original function, the seventh derivative is the same as the third derivative. (Calculus Bth ed' page / 9ti;' ed. pages ) (E) To find the first derivative, differentiate using the Chain Rule as follows:! = +p*+ 3)3 (2) = B(Zx+ 3)'. Now differentiate again to find dx

3 71 * CuRprrn 2 the second derivative: {+ =B(3)(2x + 3)'(2) = 4B(2x +3)'. (Calculus Bth ed. pages / 9th ed. pages ) 7. (E) Differentiating y = 4sin2 (3x) with respect to x results in ll = +!lrinr3x)1'z. rlx dxl,r Using the Chain Rule, * = ta)2fsin (3x )lcos(3x)(3). AX Collecting like terms,!=z+"in(3x)cos(3x). (Calculus Bth ed. pages ax / 9th ed. pages ) D 1 -i 1^ v cm/sec v +**" B - 1.-/r"" 3 Let AC = y and BC - x. The area of LABC derivative, -l dk 1l dx dv =rl(v)+dr(x).]' = K =]-w. Takinu trre Bv substitution. dk = 1[[-1)',. *( -!lzo"]= 1,-u-10) = -7.5cm'/sec. " dr 2L\ 3) \ 2) ) 2' (Calculus Bth ed. pages / 9th ed. pages ) (B) Difflerentiatins both sides, 1[Et'l =zmtx)(!'l=i,,nx). sotvins y '( [dx.) x,j x r.,.9r. dy -r[f a)rnrxrl. dx dx "l\x) l (Calculus Bth ed. pages / Bth ed. pages ) 10. (E)Since 9(x) = f r(x) and e'(xl = =:. f 'lg( x)l' g'(-3) = =: * = -L= 1. wl,r, 9(-3) = z, the equation of the tangent " f'ts(-3)l f'(2) 3 " lineto g(x) is y - 2=1,r-S,, tcalculus Bth ed. pages / 9th ed. pages )

4 DrrreReurmoru * _1 (E) y'= _ " The tangent line has equation -1 1n(1 - x) -2 = ----:-(x-0) = x = x (Calculus Bth ed. pages / 9th ed. pages ) 72. (D) tim f(x) = lim f(x) = ax' +O =2r x-)ir- -c x-+z-. l2ax+cosx. x<tr f'(x)=1--- ^----'',+ 2, x > 7t x-+nt- x--'er Iim f'(x)= lim f'(x)=2ar-1=2. Solving for a,.> c=2r-j-(r'\=c=1 2ntt2 2an=3=a=+, and by substitution, 2n' (Calculus Bth ed. pages / 9th ed. pages ) 13. (C) If y = tan-1u,where u : f(x), then using the Chain Rule, dv 7 du 7.^ 2x+3 ---:-----: -t,1'- <l...adx 1+u2 dx (x' +3x)2 + 7 '-" -' (x' +3x)' +7 (Calculus Bth ed. pages / 9th ed. pages ) 14. (B) For p(x)=rf(x)* U(3x-Z)> p'(x)= f(x)+:tr'(x)- g'(3x-2)'3. Then p'(2) = f(2)+ 2.f'{2)- g(4x3) becomes p'(2) = 5 +2(7)- (-1X3) = fo. (Calculus Bth ed. pages / 9th ed. pages ) 15. (D) C=2zr-dC =zol. dctdrldh1, Sfnce _ 4' B 2 v = Erzh-dV =2orht* rr'ffi =_ CITVmln, 3 =_-= CITVITITIL = #lrf,- 2os1g4[+ n(4)'(, = 2on cm3/min (Calculus Bth ed. pages / 9th ed. pages )

5 76 * Cnaprrn 2 Fnre Rrsporusr Solution a. A= Lr' /tl - dr 1=3rt=d' =1= o.too cm./sec 3n Possible points 1,, 2:1 lr b. =, o nz\l - t +2 = ' 3n cmlsec,,{' Ir dp C. A =lnasex height =!1211, =,, 31 7T -t,,.b d. A = area of { of the circle minus the area of f of the triangle. A=-ftr-.7,7, 42 --r' io _,!4 =( l. -. ){ =:J-: - 3r - 6 = (2 ) 3tr 6tr o.1,zcm,/sec,.b -'t (.1 a,b, c, d(calculus Bth ed. pages / 9th ed. pages ) 1: {units for a, b, c, and d