Solutions to Sample Problems for Calculus II | MATH 152, Assignments of Calculus

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di : 2 _ 24 ps 2 2(h? 6k 45) =0 & iy=k > y’ = 0,30 SF = y* — by? + 5y 4 O=kt-6k° 45k? ok? (k? — 6k +5) h(k-1)(k-5)=0 @& k=0,Lor5 dy dt wy di (b) y is increasing <> >0 & Yy-Dy-5)>0 = ye (—co,0)U (0,1) U (5,00) (c) yis decreasing © <0 & ye(l,5) 4. y' =y— x =O onthe line y = &, when x = 0 the slope is y, and when y = 0 the slope is —x. Direction field II satisfies these conditions. [Looking at the slope at the point (0, 2), LL looks more like it has a slope of 2 than does direction field I.] = x,y) = &— By. 4, (a) h = 0.2, to = 1, yo = 0, and F(x.) = 14. c wu, = 1.2andz2 = 1- made en veerees = (),392 & y{1.4). = 0.24 0.2[1.2 — (1:2)(0.2)| = 9. We need to find y2, tn = yo + RF (x0, yo) = 0+ 02m.) 0.2) wp =yt AF (21,91) = 0.2 + 0.2F(1.2, 0. (b) Now h = 0.1, so we need to find ya. wi =0+0.1f2 — (1)(0)] = 0.1, yo = 0.14 0.1[1.1 — (1.1)(0.1)] = 0.199, ys = 0.199 + 0.1[1.2 — (1.2)(0.199)] = 0.29512, and ya = 0.29512 + 0.1[1.3 — (1.3)(0.29512)] = 0.3867544 = y(1.4). / pH — pp gp? — 28/2 4¢, 4 VPLS dP/VP=Vidt = fPO?dP=fiedt = 3 dt 1/2 — 28/2 49,/2-2 = =2 C= 2/2-2,s02P? = Ft" + 2 a2 So 2V2=24+C > 3 P(1) 3 , 3/2 —_i VP =i8? 472-5 > p=(ie +v2 )