Practice TEST 4 The actual TEST 4 is Friday, December 11, 2009 During Class 7:30 to 9:00 AM 1. (a) Give the definition of y = ln(x). ......................................................... ln(a) is the area under the curve y = 1/x from x=1 to x=a. Area to the right of 1 is positive. Area between 0 and 1 is negative. ln(1) = 0 the ln(x) is not defined for x <= 0. ......................................................... x (b) Give the definition of y = e x y = e is the inverse of the function y = ln(x). ......................................................... x (c) If (a,b) is on the curve, y = e , what the derivative of y = ln(x) when x = b. x If (a,b) is on y = e, then (b,a) is on y = ln(x). The derivative of ln(x) is 1/x, so the derivative of ln(x) at x=b is 1/b. ......................................................... 2. Express the following as rational numbers. 2 ln(3/4) (a) e 9/16 3 (b) ln( Sqrt[ e ] ) 3/2 (c) Log (27/8) 3 3/2 3 (d) Log Sqrt[a ] 3/4 2 a ......................................................... 3. Find the following derivatives. / (a) y = ln(x) y = 1/x x / x (b) y = e y = e / (c) y = Sinh[x] y = Cosh[x] / (d) y = Cosh[x] y = Sinh[x] / 2 (e) y = Tanh[x] y = Sech [x] / 1 (f) y = ArcSin[x] y = ------------- _ _ | 2 | Sqrt|1-x | |_ _| / -1 (g) y = ArcCos[x] y = ------------ _ _ | 2 | Sqrt|1-x | |_ _| / 1 (h) y = ArcTan[x] y = --------- 2 1+x 1 (g) y = ArcSec[x] ----------------- _ _ | 2 | |x| Sqrt| x - 1 | |_ _| ......................................................... / 4. Derive the solution of y = -a y + b (a >= 0) dy/dt = -a y + b dy ------- = dt -ay+b -ady ------- = -a dt -ay+b ady ------- = -a dt ay-b ln(ay-b) = -a t + c -a t ay - b = C e -a t b + C e y = -------------- <=========== a / -at -a t Check y = 1/a C e (-a) = - C e / -a t y = -( ay-b) = - a y + b = -b - C e +b ......................................................... 5. A population started with 64 individuals and every 13 years the population doubles. Write the equation t/13 A(t) = 64 2 ......................................................... 6. Room temperature is 20 degrees centigrade. A hot brick starts at 200 degrees centigrade and cools to 160 degrees centigrade in 5 minutes. (a) How hot will it be after a total of 30 minutes? dT/dt = k(T-20) dT -------- = k dt T-20 ln(T-20) = k t + c kt T = 20 + C e at t = 0, T = 200, so C = 180. kt T = 20 + 180 e at t = 5 T = 160 so k5 160 = 20 + 180 e 140 k5 ---- = e 180 k = ln(140/180)/5 ln(140/180)t/5 T = 20 + 180 e _ _ t/5 | 140 | T = 20 + 180 |-----| |_180_| _ _ 30/5 | 140 | T(30) = 20 + 180 |-----| = 59.8479 degrees Centigrade. |_180_| (b) When will it be 25 degrees centigrade? _ _ t/5 | 140 | 25 = 20 + 180 |-----| |_180_| t/5 | 140 | 5/180 = |-----| | 180 | ln(5/180) t = 5 -------------- = 71.2955 minutes ln(140/180) ------------------------------------------------ Get["font.math"]; T[t_] := 20 + 180 (140/180)^(t/5); p1 = Plot[T[t],{t,0,80}]; p2 = Plot[20,{t,0,80},PlotStyle->{RGBColor[1,0,0]}]; p3 = ListPlot[{{0,200},{5,160},{71.2955,25}}, PlotStyle->{PointSize[0.02],RGBColor[1,0,0]}]; p4 = ListPlot[{{71.2944, 25},{30,T[30]}}, PlotStyle->{RGBColor[0,1,0],PointSize[0.02]}]; p5 = Plot[T[30],{x,0,30},PlotStyle->{RGBColor[0,0,1]}]; p6 = Plot[25,{x,0,5 Log[5/180]/Log[140/180]}, PlotStyle->{RGBColor[0,0,1]}]; p7 = ParametricPlot[{30,y},{y,0,T[30]},PlotStyle->{RGBColor[0,0,1]}]; p8 = ParametricPlot[{5 Log[5/180]/Log[140/180],y},{y,0,25}, PlotStyle->{RGBColor[0,0,1]}]; p9 = Show[p1,p2,p3,p4,p5,p6,p7,p8, PlotLabel->"P6: T[t] := 20 + 180 (140/180)^(t/5)", PlotRange->All, Axes->True, AxesLabel->{Minutes,Temperature}, AxesOrigin->{0,0}, AxesStyle->Directive[Orange, 8], TicksStyle->Directive[Red,12]]; Export["/math/www/hentzel/class.165.09/dec09.p6.pdf",p9]; ......................................................... 7. A brick starts at 300 degrees at t = 0. At t = 1 it is 210 degrees. At t = 2 it is 192 degrees. What is the ambient temperature? dT/dt = k(T-S) dT ------ = k dt T-S ln(T-S) = k t + c kt T-S = C e kt T = S + C e at t = 0 300 = S + C k at t = 1 210 = S + C e 2k at t = 2 192 = S + C e _ _ 2 | 210-S | | 192-S | |------ | = |-------| |_ C _| | C | _ _ 2 | 210-S | | 192-S | |------ | = |-------| |_300-S_| | 300-S | 2 (210-S) = (192-S)(300-S) 2 2 2 210 -420 S + S = 192 300 - 492 S + S 44100 -420 S = 57600 - 492 S 72 S = 13500 S = 187.5 300 = S + C = 187.5 + C C = 112.5 k 210 - S 210-187.5 1 e = --------- = --------- = ----- C 112.5 5 t T(t) = 187.5 + 112.5 (1/5) t = 0 gives 187.5 + 112.5 = 300 t = 1 gives 187.5 + 22.5 = 210 t = 2 gives 187.5 + 4.5 = 192 ----------------------------------------------------------------------- Get["font.math"]; T[t_] = 187.5 + 112.5 (1/5)^t; p1 = Plot[T[t],{t,0,4},PlotRange->All]; p2 = ListPlot[{{0,300},{1,210},{2,192}}, PlotStyle->{PointSize[0.02],RGBColor[1,0,0]}]; p3 = Plot[187.5,{t,0,4},PlotStyle->{RGBColor[0,1,0]}]; p4 = Plot[0,{t,0,4}]; p5 = Plot[210,{t,0,1},PlotStyle->{RGBColor[0,0,1]}]; p6 = Plot[192,{t,0,2},PlotStyle->{RGBColor[0,0,1]}]; p7 = ParametricPlot[{1,y},{y,0,210},PlotStyle->{RGBColor[0,0,1]}]; p8 = ParametricPlot[{2,y},{y,0,192},PlotStyle->{RGBColor[0,0,1]}]; p9 = Show[p1,p2,p3,p4,p5,p6,p7,p8, PlotLabel->"P7: T[t_] = 187.5 + 112.5 (1/5)^t", PlotRange->All, Axes->True, AxesLabel->{Time,Temperature}, AxesOrigin->{0,0}, AxesStyle->Directive[Orange,8], TicksStyle->Directive[Red,12]]; Export["/math/www/hentzel/class.165.09/dec09.p7.pdf",p9]; ---------------------------------------------------------------------- 8. Find the derivative of _ _ | 2 | 13 15 | x | (x+3) (x-5) |_ _| y = --------------- e 4 (x+29) 2 ln(y) = 13 ln(x+3) + 15 ln(x-5) + x - 4 ln(x+29) / 13 15 -4 1/y y = ------- + -------- + 2x + ------ x+3 x-5 x+29 _ _ | 2 | _ _ 13 15 | x || | / (x+3) (x-5) |_ _|| 13 15 -4 | y = --------------- e |------ + --------- + 2x + -------| 4 | x+3 x-5 x+29 | (x+29) |_ _| ......................................................... Get["font.math"]; f[x_] := (x+3)^13 (x-5)^15 E^(x^2) (x+29)^(-4); p1 = Plot[f[x],{x,-3.7,5.4},PlotRange->All]; p2 = Show[p1, PlotLabel->"P8: f[x] = (x+3)^13 (x-5)^15 e^(x^2)/(x+29(^4)", PlotRange->All, Axes->True, AxesLabel->{Time,Temperature}, AxesOrigin->{0,0}, AxesStyle->Directive[Orange,8], TicksStyle->Directive[Red,12]]; Export["/math/www/hentzel/class.165.09/dec09.p8x.pdf",p2]; fp[x_] := f[x](13/(x+3)+15/(x-5) + 2 x -4/(x+29)); p1 = Plot[fp[x],{x,-3.6,5.35},PlotRange->All]; p2 = Show[p1, PlotLabel->"P8: fp[x] = f[x](13/(x+3)+15/(x-5)+2x-4/(x+29)", PlotRange->All, Axes->True, AxesLabel->{Time,Temperature}, AxesOrigin->{0,0}, AxesStyle->Directive[Orange,8], TicksStyle->Directive[Red,12]]; Export["/math/www/hentzel/class.165.09/dec09.p8y.pdf",p2]; ......................................................... 9. Find the derivative. Do not simplify. 2 2 (x + 6 x + 2 ) / (x + 6 x + 2 ) (a) e y = e (2x+6) 2 / 14 + 6x (b) ln( 14 x + 3 x ) y = ------------ 2 14 x + 3 x -3 ----- 2 / x (c) ArcTan[ 3/x ] y = ------------- _ _ 2 | 3 | 1 + |---- | |_ x _| -1/2 (d Sinh[ Sqrt[x] ] y' = Cosh[Sqrt[x]] 1/2 x ......................................................... 10. Evaluate these integrals. x/2 e x/2 (a) INT -------- dx 2 ArcTan[e ] + C x 1 + e Sinh[x] (b) INT ----------- dx ln(Cosh[x]) + C Cosh[x] 1 (c) INT --------------- dx 3 ArcSin[x/3] + C _ _ | x 2 | Sqrt| 1-(---) | |_ 3 _| Sqrt[x] e Sqrt[x] (d) INT -------------- dx 2 e + C Sqrt[x] ......................................................... 11. Give the formula for these compounding times. n (a) Yearly (1+r/100) 12 n (b) Monthly (1+r/1200) 365 n (c) Daily (1+r/36500) n r/100 (d) Instantaneously e ......................................................... x -x e - e 12. (a) Find the inverse function for y = -------- 2 x -x 2y = e - e _ _ 2 x | x | 2 y e = |e | - 1 |_ _| _ _ 2 | x | x 0 = |e | -2 y e -1 |_ _| _ _ | 2 | 2y +/- Sqrt|4 y + 4 | x |_ _| e = ------------------------ 2 _ _ x | 2 | e = y +/- Sqrt| y + 1 | |_ _| _ _ | _ _ | | | 2 | | x = ln| y + Sqrt| y +1 | | <=========== | |_ _| | |_ _| x -x e - e (b) Find the derivative of the inverse function of y = -------- 2 _ _ | _ _ | | | 2 | | y = ln| x + Sqrt| x +1 | | | |_ _| | |_ _| _ _ -1/2 | 2 | 1 + 1/2 |x + 1 | (2x) |_ _| dy/dx = ---------------------------- _ _ | 2 | x + Sqrt| x + 1 | |_ _| _ _ | 2 | Sqrt| x + 1 | + x |_ _| dy/dx = ----------------------------------------- _ _ | _ _ | _ _ | | 2 | | | 2 | |x + Sqrt| x + 1 | | Sqrt| x + 1 | | |_ _| | |_ _| |_ _| 1 dy/dx = ---------------- _ _ | 2 | Sqrt| x + 1 | |_ _| 13. A tank has 100 gallons of water and contains 5 pounds of disolved salt. Water with 1 pound of salt per gallon flows into the tank at 2 gal per minute and the well mixed solution drains out of the tank at 2 gallons per minute. (a) Write the differential equation for the amount of salt in the tank at time t. (b) How much salt will be in the tank after 30 minutes. S(0) = 5 dS/dt = 2 - 2/100 S dS/dt + 2/100 S = 2 2/100 t 2/100 t e ( dS/dt + 2/100 S ) = 2 e t/50 t/50 d( e S ) = 2 e t/50 t/50 e S = 2 e /(1/50) + C S(0) = 5 5 = 100 + C So C = -95 t/50 t/50 e S = 100 e - 95 -t/50 S = 100 - 95 e <================= / \/ S[0] = 5 -t/50 dS/dt = 95/50 e -t/50 -t/50 dS/dt +2/100 S = 95/50 e + 2/100 (100-95 e ) = 2 It checks. -30/50 S[30] = 100 - 95 e = 47.8629 pounds ============================ f[t_] := 100 - 95 E^(-t/50); f[30] = 47.8629 ==================================================== Get["font.math"]; f[t_] := 100 - 95 E^(-t/50); p1 = Plot[f[t],{t,0,200}]; p2 = Plot[100,{t,0,200},PlotStyle->{RGBColor[0,1,0]}]; p3 = Plot[f[30],{t,0,30},PlotStyle->{RGBColor[0,0,1]}]; p4 = ParametricPlot[{30,y},{y,0,f[30]},PlotStyle->{RGBColor[0,0,1]}]; p5 = ListPlot[ { {0,5}, {30,f[30]} },PlotStyle->{RGBColor[1,0,0]}]; p6 = Show[p1,p2,p3,p4,p5,PlotLabel->"P13 S[t] = 100 - 95 e^(-t/50)"]; Export["/math/www/hentzel/class.165.09/dec09.p13.pdf",p6]; ====================================================