# preTEST5A

## 3112 days ago by MATH4R2012

#1a) var('C,k') f(t)=C*e^(k*t);show(f) show(solve(f(0)==.5,C)) f(t)=(1/2)*e^(k*t);show(f)
 \newcommand{\Bold}{\mathbf{#1}}t \ {\mapsto}\ C e^{\left(k t\right)} \newcommand{\Bold}{\mathbf{#1}}\left[C = \left(\frac{1}{2}\right)\right] \newcommand{\Bold}{\mathbf{#1}}t \ {\mapsto}\ \frac{1}{2} \, e^{\left(k t\right)}
#1b) ans=solve(f(5)==5,k) show(len(ans)) show(ans.rhs()) f(t)=(1/2)*e^(log(10)*t/5);show(f)
 \newcommand{\Bold}{\mathbf{#1}}5 \newcommand{\Bold}{\mathbf{#1}}\frac{1}{5} \, \log\left(10\right) \newcommand{\Bold}{\mathbf{#1}}t \ {\mapsto}\ \frac{1}{2} \, e^{\left(\frac{1}{5} \, t \log\left(10\right)\right)}
#1c) ans=(solve(f(t)==1,t)) show(ans) show(ans.rhs().n())
 \newcommand{\Bold}{\mathbf{#1}}\left[t = \frac{5 \, \log\left(2\right)}{\log\left(10\right)}\right] \newcommand{\Bold}{\mathbf{#1}}1.50514997831991
#1d) print "t\tf(t)" for i in range(4): print 2^i,"\t",f(2^i).n()
 t f(t) 1 0.792446596230557 2 1.25594321575479 4 3.15478672240097 8 19.9053585276749 t f(t) 1 0.792446596230557 2 1.25594321575479 4 3.15478672240097 8 19.9053585276749
#2a) s(h)=25-13*log(h/12)/log(3);s show(s(2).n())
 \newcommand{\Bold}{\mathbf{#1}}46.2020867964289
#2b) show(s(15).n())
 \newcommand{\Bold}{\mathbf{#1}}22.3595178235248
#2c) show(solve(s(h)==50,h)) show((12*e^(-25*log(3)/13)).n())
 \newcommand{\Bold}{\mathbf{#1}}\left[h = 12 \, e^{\left(-\frac{25}{13} \, \log\left(3\right)\right)}\right] \newcommand{\Bold}{\mathbf{#1}}1.45090965795109

2d)

When there's 2" of snow on the ground, we can plow approximately 46.202 miles of road per hour!

When there's 15" of snow on the ground, we can plow approximately 22.356 miles of road per hour!

If we plow at a rate of  50 mph, then there's about 1.451" of snow on the road.

#2e) plot([s(h),50],0,15)  #3a) var('r,t') P(t)=750*e^(r*t);show(P) ans=solve(P(7.75)==1500,r) show(len(ans)) show(ans.rhs().n())
 \newcommand{\Bold}{\mathbf{#1}}t \ {\mapsto}\ 750 \, e^{\left(r t\right)} \newcommand{\Bold}{\mathbf{#1}}31 \newcommand{\Bold}{\mathbf{#1}}0.0894383458787026
#3b) P(t)=750*e^(4*log(2)*t/31);show(P) S(t)=750*(1+r)^t show(solve(P(1)==S(1),r)) show(100*4/31*log(2).n()) show(100*(e^(4/31*log(2))-1).n())
 \newcommand{\Bold}{\mathbf{#1}}t \ {\mapsto}\ 750 \, e^{\left(\frac{4}{31} \, t \log\left(2\right)\right)} \newcommand{\Bold}{\mathbf{#1}}\left[r = e^{\left(\frac{4}{31} \, \log\left(2\right)\right)} - 1\right] \newcommand{\Bold}{\mathbf{#1}}8.94383458787026 \newcommand{\Bold}{\mathbf{#1}}9.35599087586107
#3c) P(10).n()
 $\newcommand{\Bold}{\mathbf{#1}}1834.37052835120$
#3d) ans=solve(P(t)==3000,t) ans.rhs().n()
 $\newcommand{\Bold}{\mathbf{#1}}15.5000000000000$
#3e) ans=solve(P(t)==10000,t) ans.rhs().n()
 $\newcommand{\Bold}{\mathbf{#1}}28.9614833547881$
#4a linear) var('a,b') l=[[1,1.5],[2,7.4],[3,10.2],[4,13.4],[5,15.8],[6,16.3],[7,18.2],[8,18.3]] f(x)=a*x+b q=find_fit(l,f,solution_dict=True) show(q[a]) show(q[b]) points(l,color='blue')+plot(f(a=q[a],b=q[b]),(x,0,8),color='red')
 \newcommand{\Bold}{\mathbf{#1}}2.28928571429 \newcommand{\Bold}{\mathbf{#1}}2.33571428572  #4b logarithmic) var('a,b') #l=[[1,1.5],[2,7.4],[3,10.2],[4,13.4],[5,15.8],[6,16.3],[7,18.2],[8,18.3]] g(x)=a+b*log(x) q2=find_fit(l,g,solution_dict=True) show(q2[a]) show(q2[b]) points(l,color='blue')+plot(g(a=q2[a],b=q2[b]),(x,0,8),color='red')
 \newcommand{\Bold}{\mathbf{#1}}1.53794931133 \newcommand{\Bold}{\mathbf{#1}}8.373383353  y(t)=1.538+8.373*log(t);show(y(t)) show(y(10).n())
 \newcommand{\Bold}{\mathbf{#1}}8.37300000000000 \, \log\left(t\right) + 1.53800000000000 \newcommand{\Bold}{\mathbf{#1}}20.8175449836391
#4d) ans=solve(y(t)==20,t) ans.rhs().n()
 $\newcommand{\Bold}{\mathbf{#1}}9.06974785952418$

#4e)

After 10 minutes of reaction, the yield is about 20.818mg.

20mg yield occurs after approximately 9.070 minutes of reaction.