NAME____________________________________ SCHOOL__________________________________ TEST 2 Friday, October 10, 2008 ___________________________________________ //-----------------------------------------\\ // \\ || FAX your answers to: || || || || Irvin Roy Hentzel at Fax 515-294-5454 || \\ // \\_________________________________________// \\_______________________________________// ___________________________________________ /-------------------------------------------\ // \\ || Snail Mail your answers to: || || || || Irvin Roy Hentzel || || Department of Mathematics || || 432 Carver Hall || || Iowa State University || || Ames, Iowa 50011-2064 || \\____________________________________________// \\__________________________________________// 1. Find the Oblique asymptote to 4 3 x + 17 x + 17 x + 5 f(x) = ------------------- 3 x + x + 1 x + 17 ___________________________ | 3 | 4 3 x + x + 1 | x + 17 x + 17 x + 5 4 2 x + x + x ------------------------- 3 2 17 x - x + 16 x + 5 3 17 x + 17 x + 17 ---------------------------- 2 - x - x - 12 4 2 2 x + 4 x + 17 x + 5 -x - x - 12 f(x) = ------------------------ = x + 17 + ---------------- 3 3 x + x + 1 x + x + 1 The oblique asymptote is y = x+17. <============ Answer / \/ x+17+(-x^2-x-12)/(x^3+x+1) 350 + 870 t 2. Find Limit ------------------------------------- t->Infinity __________________________________ / / 2 2 \/ ( 165 + 51 t) + (432 t ) 350 + 870 t Limit ------------------------------------- t->Infinity __________________________________ / / 2 2 \/ ( 165 + 51 t) + (432 t ) 350 --- + 870 t Limit ------------------------------------- t->Infinity __________________________________ / / 2 / / 165 \ 2 / ( ---- + 51 ) + ( 432 ) \/ \ t / 870 = ---------------------------- __________________________ / 2 2 \/ 51 + 432 870 = ---------------- = 2 <============ Answer 435 ___________ / \/ 51^2+432^2 = 435 Limit[ (350 + 870 t)/Sqrt[ ( 165 + 51 t)^2 + (432 t )^2 ],{t->Infinity}] = 2 3. Find the equation of the tangent line to 4 3 3 4 2 2 y x + y x + 4 x y = 40 at the point (2,1). 3 / 3 4 2 2 / 4 3 3 2 2 / 4 y y x + y 3 x + 3 y y x + y 4 x + 8 x y + 4 x 2 y y = 0 / / / 32 y + 12 + 48 y + 32 + 16 + 32 y = 0 / 112 y = -60 / -60 -15 y = --------- = ------ 112 28 y-1 = -15/28(x-2) <========Answer f[x_,y_] := y^4 x^3 + y^3 x^4 + 4 x^2 y^2 - 40; n[x_, y_] := y - f[x,y]/(4 y^3 x^3 + 3 y^2 x^4 + 8 x^2 y); q[x_,y_] := (a = y; Do[ (a = n[x,a]),{i,1,1000}];Return[a]); p1 = Plot[ q[x,1.0],{x,0,5}] p2 = Plot[1-(15/28)(x-2),{x,0,5},PlotStyle->{RGBColor[1,0,0]}]; p3 = ListPlot[{{2,1}},PlotStyle->{RGBColor[1,0,1],PointSize[0.01]}]; p4 = Show[p1,p2,p3]; p5 = Show[p4,PlotRange->{{1,3},{0,2}}] 4. A 25 foot ladder is leaning against a wall. The bottom is 7 feet from the wall and moving away from the wall at 2 ft/sec. How fast is the top of the ladder moving down the wall. |\ | \ | \ | \ | \ | \ 25 y | \ | \ | \ | \ | \ |___________\ x y = (25^2 - x^2)^(1/2) dy/dt = (1/2) (25^2 - x^2)^(-1/2) (-2) x dx/dt -x dx/dt dy/dt = --------------------- Sqrt[ 25^2 - x^2 ] -14 -14 -7 dy/dt = --------------- = ---- = ----- = -0.583333 ft/sec Sqrt[25^2-49] 24 12 f[x_] := Sqrt[25^2 - x^2]; Limit[ (f[7+2*t]-f[7])/t ,{t->0}] ////////////////TOOOOOOOOOOOOOOO HEREEEEEEEEEEEEEEEEEE/////////////// 2 2 m s 5. The equation for horsepower is hp = ----------------- 3 550 t Where m = weight/32; s = distance in feet, and t = time in seconds. If weight = 160 +/- 1 lbs. s = 363 ft +/- 1 ft. t = 11 +/- 0.01 sec. (a) Find the horsepower. (b) how does the horsepower change due to the error in weight? (c) how does the horsepower change due to the error in distance? (d) How does the horsepower change due to the error in time? 2 160/32 363 363 hp = ================== = 9/5 horsepower. 550 11 11 11 9/5 dhp/dw /\w = -------- 1/32 = 0.01125 error from weight 160/32 2 (9/5) dhp/ds /\s = --------- 1 = 0.00991736 error from distance 363 (-3)(9/5) dhp/dt /\t = -------- 0.01 = -0.00490909 error from time 11 / 6. Suppose f(13) = 32 and f (13) = 2.73 Use differentials to approximate f(11.4). / f(11.4) = f(13) + f (13)(-1.6) = 32 + (2.73)(-1.6) = 27.632 7. Your balloon is falling at a constant rate of 22 ft/sec. When the balloon is at 210 feet, you untie 50 lbs of sand. (a) How long till the sand hits the ground. (b) What is the velocity of the sand when it hits the ground. a = -32 v = -32 t -22 s = -16 t^2 -22 t + 210 0 = -2 (-3 + t) (35 + 8 t) t = 3 seconds or t = -35/8 seconds. t = 3 seconds <======================= v (3) = -32 3 - 22 = -118 ft/sec. <================ 8. Gravel is being dumped from a conveyor belt at the rate of 20 ft^3/min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 feet high. V = 1/3 B H 2 3 V = 1/3 Pi r h = 1/3 Pi h /4 2 dV/dt = Pi/4 h dh/dt 20 = Pi/4 14 14 dh/dt 80 ---------- = dh/dt Pi 14 14 20 dr/dt = ----- = 0.129922 ft/min 49 Pi In problems 9 through 15, Find the following derivatives. 2 9. f[x] = x Tan[ x ] / 2 2 2 f [x] = Tan[x ] + x Sec [x ] 2x 2 10. f[x] = x Tan [x] / 2 2 f [x] = Tan [x] + 2 x Tan[x] Sec [x] 2 Pi 11. f[x] = (3 + x ) / 2 Pi-1 f [x] = Pi (3 + x ) 2 x 12. f[x] = Tan[ x Sec[x] ] / 2 f [x] = Sec [ x Sec[x] ] ( Sec[x] + x Sec[x] Tan[x] ) 13. f[x] = Cos[ x/2 ] / f [x] = -Sin[x/2] (1/2) 14. f[x] = Sin[2x] Cos[3x] / f [x] = Cos[2x] 2 Cos[3x] + Sin[2x](-Sin[3x] 3) 3 x + Sin[x] 15. f[x] = -------------- 2 x + Cos[x] 2 2 3 / ( x + Cos[x])(3 x + Cos[x]) - (x + Sin[x])(2 x - Sin[x] ) f [x] = ----------------------------------------------------------------- 2 2 (x + Cos[x] ) _ _ | 2 | Sqrt| 4 + 2 x | |_ _| 16. f[x] = -------------------- 2 Pi _ _ -1/2 | 2 | 1/2 | 4 + 2 x | 4 x / |_ _| f [x] = ------------------------------------ 2 Pi 14 17. f[x] = Sin [x] / 13 f [x] = 14 Sin [x] Cos[x]