Task 2
Video : Do you believe
Presenting by : “Dalton Sherman”,
I believe in me
Do you believe in me?
Do you believe that I can stand up here fearless and talk over 21 thousand of you?
No matter where we come from, what is sunny salt Dallas, hat is placing brown, what is … or Nort Dallas or West Dallas or whenever, you aren’t better nice give up o us?
No, you are better nice because as you know in some cases you ae all we have got.
Everyone need you, We need you now, wanted ever!
Believe in yourself, finally do you believe that you can do anything, be anything, make anything.
Kamis, 15 Januari 2009
LESSON WITH A FRIEND
MIXTURE PROBLEM
We now consider rate problem involving mixtures. A substances S is allowed fo flow into a certain mixture in a container at a certain rate, and the mixture is kept uniform by stirring. Further. In one such situation, this uniform mixture simultaneously flows out of the container at another (generally different) rate; in another situation this may not be the case. In either case we seek to determine the quantity of the substance S present in the mixture at time t.
Let :
x denote the amount of S present at time t,
dx/dt denotes the rate of changer of x at time t.
INPUT denotes rate S which enter the mixture.
OUTPUT denotes rate S which leaves the mixture.
Then we have at one the basic equation
dx/dt =INPUT-OUTPUT
We have a problem
A tank initially contains 50 gal of pure water. Starting at time t=0, a brine containing 2 lb of dissolved salt per gallon flow into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the function of salt in the tank!
First we must make the mathematical formula. We apply the basic equation
dx/dt=INPUT-OUTPUT
The brine flows in at the rate of 3 gal/min and each gallon contain 2 lb of salt.
We formulating the INPUT as the weight of salt which flows in tank at time t and OUTPUT as the weight of salt which leaves tank at time t
In Physics, we have formula that
Concentration = weight / volume
Debit =volume/time
Find weight of salt at time t (weight/ time)
Weight/time = (concentration) (debit) = ( weight/ volume ) (volume/time)= weight/time
We apply at the problem to INPUT and OUTPUT
INPUT = (concentration which flows in) (debit which flows in)
= (2 lb/gal) (3 gal/min) = 6 lb/min
To find the rate of salt that leaves the tank at time t, we must find concentration of salt. The tank contains 50 gallon of the mixture at any time t. this 50 gallon contain x lb salt at time t, so the concentration is (x/50) lb/gal and the debit of mixture which leaves tank is 3 gal/min.
OUTPUT = (concentration which leaves tank) (debit which leaves tank)
= (x/50)lb/gal) (3 gal/min) = (3x/50)lb/min
We apply it in basic equation
dx/dt = 6 - (3x/50)
and we have initial condition at t=0 , x=0
x(0)=0
Then we determine this problem.
The equation is both linear and separable.
dx/dt = 6 - (3x/50)
We equaling the denominator, then we have
dx/dt = (300-3x)/50)
We group x with dx and t with dt. We change dt with (100-x) and (100-x) with dt.
dx/(100-x)=3 dt /50
We change this form into
dx/(x-100)= - 3dt/50 with multiply by -1
Then integrating both of side, then we get
ln (x-100) = -3t/50+c1
We use exponential to get the function of x
exp(ln (x-100)) = exp(-3t/50+c1)
Because of exp(ln x) = x, so
x-100=exp(-3t/50+c1)
Then move -100 to other side
x=100+exp(-3t/50+c1)
exp(c1)=c so
x=100+c.exp(-3t/50) ………(1)
We have intial value that x(0)=0 , at t=0 x =0,
then we substitute t=0 and x=0 into (1)
0= 100 + c exp(-3(0)/50)>>> exp(0)=1
0= 100 + c
c= -100
Then c = -100, substitute c into (1), we have
x=100-100.exp(-3t/50), we simply it become
x=100(1-exp(-3t/50)
x=100(1-exp(-3t/50) , is function of how much salt in the tank at t time.
The trouble at I explaining this problem to my friend Bibid are :
1.Sometimes, I forgot some word that I must say to Bibid, like exponent, power, etc.
2.I forgot how I explaining integral.
3.Sometimes I must read the source book to make Bibid say “I see”
We now consider rate problem involving mixtures. A substances S is allowed fo flow into a certain mixture in a container at a certain rate, and the mixture is kept uniform by stirring. Further. In one such situation, this uniform mixture simultaneously flows out of the container at another (generally different) rate; in another situation this may not be the case. In either case we seek to determine the quantity of the substance S present in the mixture at time t.
Let :
x denote the amount of S present at time t,
dx/dt denotes the rate of changer of x at time t.
INPUT denotes rate S which enter the mixture.
OUTPUT denotes rate S which leaves the mixture.
Then we have at one the basic equation
dx/dt =INPUT-OUTPUT
We have a problem
A tank initially contains 50 gal of pure water. Starting at time t=0, a brine containing 2 lb of dissolved salt per gallon flow into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the function of salt in the tank!
First we must make the mathematical formula. We apply the basic equation
dx/dt=INPUT-OUTPUT
The brine flows in at the rate of 3 gal/min and each gallon contain 2 lb of salt.
We formulating the INPUT as the weight of salt which flows in tank at time t and OUTPUT as the weight of salt which leaves tank at time t
In Physics, we have formula that
Concentration = weight / volume
Debit =volume/time
Find weight of salt at time t (weight/ time)
Weight/time = (concentration) (debit) = ( weight/ volume ) (volume/time)= weight/time
We apply at the problem to INPUT and OUTPUT
INPUT = (concentration which flows in) (debit which flows in)
= (2 lb/gal) (3 gal/min) = 6 lb/min
To find the rate of salt that leaves the tank at time t, we must find concentration of salt. The tank contains 50 gallon of the mixture at any time t. this 50 gallon contain x lb salt at time t, so the concentration is (x/50) lb/gal and the debit of mixture which leaves tank is 3 gal/min.
OUTPUT = (concentration which leaves tank) (debit which leaves tank)
= (x/50)lb/gal) (3 gal/min) = (3x/50)lb/min
We apply it in basic equation
dx/dt = 6 - (3x/50)
and we have initial condition at t=0 , x=0
x(0)=0
Then we determine this problem.
The equation is both linear and separable.
dx/dt = 6 - (3x/50)
We equaling the denominator, then we have
dx/dt = (300-3x)/50)
We group x with dx and t with dt. We change dt with (100-x) and (100-x) with dt.
dx/(100-x)=3 dt /50
We change this form into
dx/(x-100)= - 3dt/50 with multiply by -1
Then integrating both of side, then we get
ln (x-100) = -3t/50+c1
We use exponential to get the function of x
exp(ln (x-100)) = exp(-3t/50+c1)
Because of exp(ln x) = x, so
x-100=exp(-3t/50+c1)
Then move -100 to other side
x=100+exp(-3t/50+c1)
exp(c1)=c so
x=100+c.exp(-3t/50) ………(1)
We have intial value that x(0)=0 , at t=0 x =0,
then we substitute t=0 and x=0 into (1)
0= 100 + c exp(-3(0)/50)>>> exp(0)=1
0= 100 + c
c= -100
Then c = -100, substitute c into (1), we have
x=100-100.exp(-3t/50), we simply it become
x=100(1-exp(-3t/50)
x=100(1-exp(-3t/50) , is function of how much salt in the tank at t time.
The trouble at I explaining this problem to my friend Bibid are :
1.Sometimes, I forgot some word that I must say to Bibid, like exponent, power, etc.
2.I forgot how I explaining integral.
3.Sometimes I must read the source book to make Bibid say “I see”
Representing The Video of Learning Mathematics
Video 4
INVERSE FUNCTION
INVERSE FUNTION
Before we start study the Inverse Function, better we review the definition of a function.
We have relation of F(x,y), this represent to function y=f(x) if we can solve for y uniquely in term of x as 1-1 function if also we can solve x uniquely in term of y.
If we see y=x2, it is a function but not 1-1 function, because if we choose a value of y, we have two value of x. 1-1 function only mapping a x into a y.
So lets start at y = 2x-1
Let’s look at the graph function
y intercepts -1 , x intersects ½, the slope =m= 2
We have a point (1/2,0) and (0, -1) . To draw this line, find (1/2,0) and (0, -1) first, then draw a line that across the points.
Then attend it
y = x intersects y = 2x-1
x = 2x-1 (add 1 in both of side)
1+x = 2x (move the x into right side)
1 = x
Substitute x = 1 in the equation y = x and y = 2x-1, we get y =1 (x =1 satisfying the both equations because x is intersect of y= x and y = 2x-1
Of course, the point of intersection of y= x and y = 2x-1 is (1,1)
We want to solve for x in term of y
2x-1 = y
2x = y+1
x = ½ (y+1)
x = ½ y + ½
Then in this equation, I want to give exchange rule of x and y, so I’m going to do now is rewrite x into y, and y into x,
We get y = ½ x + ½ ……………..(**)
The x intercepts in -1, and y intercepts in ½
We have point (0,1/2) and (-1,0)
So the inverse function to given function passes through that point (1,1)
We have f(x) = 2x-1
We have g(x) = ½ x + ½
Then take the value of g and plug in for f
We want to compute f(g(x))
f(g(x)) = 2 […] – 1 , then fill the bracket with ½ x + ½
= 2 [½ x + ½] – 1= x + 1 -1= x
in the other hand, we take g(f(x))
g(f(x)) = ½ […..] + ½ = ½ [2x-1] +½ = x-½ +½ = x
Then we write
g = f -1
f(g(x)) = f(f -1 (x)) = x and
g(f(x)) = f -1 (f(x)) = x
Other example. Let’s take the function y =(x-1)/(x+2). This function has a vertical asymptote as line x = -2 and also has horizontal asymptote as a line y = 1
x intercept is approximately 1and y intercept is approximately -½
The function hyperbole
Then let’s take the equation y =(x-1)/(x+2) and solve for x in term of y
y(x+2) = (x-1)
yx-2y = x-1 (take 2y into the right side, move x to the left side)
yx-x =-1-2y
(y-1)x =-1-2y
X = (-1-2y)/(y-1)
we are going to exchange the rule of y and x
y= (-1-2x)/(x-1)
at x = 0 y = -1
at y = 0 -1-2x = 0
-1 = 2x or x =-½
y= (-1-2x)/(x-1) has vertical asymptote as a line x = 1 and also horizontal asymptote as a line y = -2
We have
f(x) = (x-1)/(x+2)
f -1(x) = (-1-2x)/(x-1)
Then
f(f -1 (x)) = (((-1-2x)/(x-1))-1)/(( (-1-2x)/(x-1))+2)
= ((((-1-2x)-(x-1))/(x-1)))/((((-1-2x)+2(x-1)/(x-1)))
= (-1-2x-x+1)/(-1-2x+2x-2)
= -3x/-3
= x
Then find f -1(f(x))
f -1(f(x)) =((-1-2((x-1)/(x+2)))/((x-1)/(x+2) – 1))
= (-1(x+2)-2((x-1)/(x+2))/((x-1)- 1(x+2)/(x+2))
= ((-1x-2-2x+2)/(x+2))/((x-1- x-2)/(x+2)))
= (-1x-2-2x+2)/(x-1- x-2)
= -3x/-3
= x
We find that
f(g(x)) = x
g(f(x)) = x
Range of g is a domain of f, and the range of f is a domain of g
The two function are inverses function.
we just write g = f -1
INVERSE FUNCTION
INVERSE FUNTION
Before we start study the Inverse Function, better we review the definition of a function.
We have relation of F(x,y), this represent to function y=f(x) if we can solve for y uniquely in term of x as 1-1 function if also we can solve x uniquely in term of y.
If we see y=x2, it is a function but not 1-1 function, because if we choose a value of y, we have two value of x. 1-1 function only mapping a x into a y.
So lets start at y = 2x-1
Let’s look at the graph function
y intercepts -1 , x intersects ½, the slope =m= 2
We have a point (1/2,0) and (0, -1) . To draw this line, find (1/2,0) and (0, -1) first, then draw a line that across the points.
Then attend it
y = x intersects y = 2x-1
x = 2x-1 (add 1 in both of side)
1+x = 2x (move the x into right side)
1 = x
Substitute x = 1 in the equation y = x and y = 2x-1, we get y =1 (x =1 satisfying the both equations because x is intersect of y= x and y = 2x-1
Of course, the point of intersection of y= x and y = 2x-1 is (1,1)
We want to solve for x in term of y
2x-1 = y
2x = y+1
x = ½ (y+1)
x = ½ y + ½
Then in this equation, I want to give exchange rule of x and y, so I’m going to do now is rewrite x into y, and y into x,
We get y = ½ x + ½ ……………..(**)
The x intercepts in -1, and y intercepts in ½
We have point (0,1/2) and (-1,0)
So the inverse function to given function passes through that point (1,1)
We have f(x) = 2x-1
We have g(x) = ½ x + ½
Then take the value of g and plug in for f
We want to compute f(g(x))
f(g(x)) = 2 […] – 1 , then fill the bracket with ½ x + ½
= 2 [½ x + ½] – 1= x + 1 -1= x
in the other hand, we take g(f(x))
g(f(x)) = ½ […..] + ½ = ½ [2x-1] +½ = x-½ +½ = x
Then we write
g = f -1
f(g(x)) = f(f -1 (x)) = x and
g(f(x)) = f -1 (f(x)) = x
Other example. Let’s take the function y =(x-1)/(x+2). This function has a vertical asymptote as line x = -2 and also has horizontal asymptote as a line y = 1
x intercept is approximately 1and y intercept is approximately -½
The function hyperbole
Then let’s take the equation y =(x-1)/(x+2) and solve for x in term of y
y(x+2) = (x-1)
yx-2y = x-1 (take 2y into the right side, move x to the left side)
yx-x =-1-2y
(y-1)x =-1-2y
X = (-1-2y)/(y-1)
we are going to exchange the rule of y and x
y= (-1-2x)/(x-1)
at x = 0 y = -1
at y = 0 -1-2x = 0
-1 = 2x or x =-½
y= (-1-2x)/(x-1) has vertical asymptote as a line x = 1 and also horizontal asymptote as a line y = -2
We have
f(x) = (x-1)/(x+2)
f -1(x) = (-1-2x)/(x-1)
Then
f(f -1 (x)) = (((-1-2x)/(x-1))-1)/(( (-1-2x)/(x-1))+2)
= ((((-1-2x)-(x-1))/(x-1)))/((((-1-2x)+2(x-1)/(x-1)))
= (-1-2x-x+1)/(-1-2x+2x-2)
= -3x/-3
= x
Then find f -1(f(x))
f -1(f(x)) =((-1-2((x-1)/(x+2)))/((x-1)/(x+2) – 1))
= (-1(x+2)-2((x-1)/(x+2))/((x-1)- 1(x+2)/(x+2))
= ((-1x-2-2x+2)/(x+2))/((x-1- x-2)/(x+2)))
= (-1x-2-2x+2)/(x-1- x-2)
= -3x/-3
= x
We find that
f(g(x)) = x
g(f(x)) = x
Range of g is a domain of f, and the range of f is a domain of g
The two function are inverses function.
we just write g = f -1
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