#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)
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#1b)
ans=solve(f(5)==5,k)
show(len(ans))
show(ans[4].rhs())
f(t)=(1/2)*e^(log(10)*t/5);show(f)
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#1c)
ans=(solve(f(t)==1,t))
show(ans)
show(ans[0].rhs().n())
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#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())
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#2b)
show(s(15).n())
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#2c)
show(solve(s(h)==50,h))
show((12*e^(-25*log(3)/13)).n())
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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)
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#3a)
var('r,t')
P(t)=750*e^(r*t);show(P)
ans=solve(P(7.75)==1500,r)
show(len(ans))
show(ans[30].rhs().n())
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#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())
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#3c)
P(10).n()
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#3d)
ans=solve(P(t)==3000,t)
ans[3].rhs().n()
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#3e)
ans=solve(P(t)==10000,t)
ans[3].rhs().n()
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#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')
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#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')
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y(t)=1.538+8.373*log(t);show(y(t))
show(y(10).n())
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#4d)
ans=solve(y(t)==20,t)
ans[0].rhs().n()
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#4e)
After 10 minutes of reaction, the yield is about 20.818mg.
20mg yield occurs after approximately 9.070 minutes of reaction.
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