Category Archives: MAC1105

Application of Quadratic Function: Maximize

Example:

A baseball player swings and hits a pop fly straight up in the air to the catcher.  The height of the baseball in meters t seconds after it is hit is given by the quadratic function h(t)=-4.9t^2+17.4t+1.  How long does it take the baseball to reach its maximum height?  What is the maximum height obtained by the baseball?

Solution:

h(t)=-4.9t^2+14.7t+1 is defined to be a quadratic function.  The graph of a quadratic function is a parabola.  This quadratic function has a leading coefficient of -4.9 which means that the parabola is opening down.

Find the vertex to find the maximum value.

The formula for the x coordinate of the vertex
t=-b/{2a}
For a quadratic function a is the coefficient of the square term, b is the coefficient of the linear term, and c is the constant
h(t)=-4.9t^2+14.7t+1
a=-4.9; b=14.7; c=1
Substitute the values of a and b into the formula
t=-b/{2a}
t={-(14.7)}/{2(-4.9)}
Simplify with a calculator
t={-(14.7)}/{2(-4.9)}
t=1.5

The ball reaches the maximum height 1.5 seconds after the ball was hit.

The maximum height can be found by substituting 1.5 seconds for time in the height function.

The height function
h(t)=-4.9t^2+14.7t+1
Substitute t=1.5 in the height function
h(t)=-4.9t^2+14.7t+1
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
Simplify using a calculator
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
h(1.5)=12.025

The maximum height of the ball is 12.025 meters.

 

Application: Exponential Growth and Decay (Half-life)

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution:  There is a two part process to this problem.  Part 1: Use some of the information to find the decay rate of radium.  Part 2: Answer the question using the rest of the given information.

Part 1:  Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

A=A _0 e^{kt}

A _0 is the initial amount

k is the decay rate

t is the time

A is the amount after t time has passed

Since radium has a half life of 1690 years, we know that whatever initial amount of radium is present after 1690 years there will be half of that initial amount left.  This allows me to identify A=1 /2 A _0 when the t=1690.

Substitute these values into the exponential decay formula and solve for k.

 

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for A and t since it is known that the half-life is 1690 years A=1 /2 A _0 and t=1690
A=A _0 e^{kt}
1 /2 A _0=A _0 e^{k(1690)}
Solve for the decay rate k:
Start by dividing both sides by the coefficient to isolate the exponential factor
1 /2 A _0=A _0 e^{k(1690)}
{1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0
1 /2= e^{k(1690)}
Solve for the decay rate k:
Take the natural log of both sides to get k out of the exponent
1 /2= e^{k(1690)}
ln(1 /2)= ln(e^{k(1690)})
Solve for the decay rate k:
Use the power rule for logarithms to get k out of the exponent
ln(1 /2)= ln(e^{k(1690)})
ln(1 /2)= k(1690)ln(e)
Solve for the decay rate k:
Simplify ln e = 1
ln(1 /2)= k(1690)ln(e)
ln(1 /2)= k(1690)(1)
ln(1 /2)= k(1690)
Solve for the decay rate k:
Solve for k by dividing by 1690 on both sides
ln(1 /2)= k(1690)
{ln(1 /2)}/1690= {k(1690)}/1690
{ln(1 /2)}/1690= k
-0.000410 approx k

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2:  Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years.  Symbolically that is A _0=50 when the t=630.  Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for initial amount (A_0), time (t), and the decay rate (k).
A _0=50 , t=630, and k={ln(1 /2)}/1690
A=A _0 e^{kt}
A=50 e^{630{ln(1 /2)}/1690}
Type the expression into your calculator and round to the thousandth place.
A=50 e^{630{ln(1 /2)}/1690}
A approx 38.615

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

Application of Quadratic Equation: Translation

Example:  The product of some negative number and 5 less than twice that number is 273.  Find the number.

Solution:  Translate the statement into an equation and then solve the equation.

“The product of some negative number…”

Let x be some negative number and multiply that number by the expression that comes next in the statement.

x  (expression that comes next)

“…and 5 less than twice that number…”

5 less than means take 5 away from what follows.  Twice that number refers to 2 times the negative number that was described before.  (2x-5)

x  (2x-5)

“…is 273”

This translates to equals 273.

x (2x-5)=273

Solve the equation.

 

The translated equation
x (2x-5)=273
Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation)
2x^2-5x=273
Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula)  I will demonstrate factoring so I need to move all terms to the same side to have zero on one side.
2x^2-5x=273
2x^2-5x-273=273-273
2x^2-5x-273=0
Factor
2x^2-5x-273=0
(       )(       )=0
(2x       )(x       )=0
(2x      21)(x     13)=0
(2x    +  21)(x  -   13)=0
Use the zero product property and set each factor equal to zero and solve.
(2x    +  21)(x  -   13)=0
2x    +  21=0 or x  -   13=0
2x    +  21-21=0-21 or x  -   13+13=0+13
2x =-21 or x =13
{2x}/2 ={-21}/2 or x =13
x ={-21}/2 or x =13

Since the statement says that the number is negative, the number is {-21}/2.

Application of Exponential Functions: Finding the Interest Rate

Example:

What is the interest rate necessary for an investment to quadruple after 7 year of continuous compound interest?

Solution:

Since this question involve continuous compound interest, we will use the associated formula.

A=Pe^{rt}

We are given that the invest quadruples in 7 years.  This tells me that when t=7 that A will be 4 times P.  I can write that in symbols A=4P.

Substitute these values into the continuous compound formula and solve for the interest rate.

Continuous compound formula
A=Pe^{rt}
Substitute the values of t and A into the formula
A=Pe^{rt}
4P=Pe^{r*7}
4P=Pe^{7r}
Solve for r by dividing both sides by P and simplifying
4P=Pe^{7r}
{4P}/P={Pe^{7r}}/P
4=e^{7r}
Solve for r by taking the log of both sides.
4=e^{7r}
ln 4=ln e^{7r}
Solve for r by using the power rule and simplifying
ln 4=ln e^{7r}
ln 4=7r ln e
ln 4=7r (1)
ln 4=7r
Solve for r by dividing both sides by 7 and simplifying
ln 4=7r
{ln 4}/7={7r}/7
{ln 4}/7=r
Find the value in the calculator
{ln 4}/7=r
0.1980420516=r
Write the answer as a percentage rounded to two decimal places
r=0.1980420516
r=19.80420516%
r=19.80%

 

Solve an exponential equation: Take the log of both sides

Example:

(1.41)^x = (sqrt{2})^{1-4x}

Solution:

 

The exponential equation
 (1.41)^x = (sqrt{2})^{1-4x}
Since the bases cannot be easily written the same use the method of taking the log of both sides
(1.41)^x = (sqrt{2})^{1-4x}
ln (1.41)^x = ln (sqrt{2})^{1-4x}
Use the power rule for logarithms
ln (1.41)^x = ln (sqrt{2})^{1-4x}
x ln (1.41) = (1-4x) ln sqrt{2}
Use the distributive law
x ln (1.41) = (1-4x) ln sqrt{2}
x ln (1.41) = ln sqrt{2}-4x ln sqrt{2}
Collect the terms with x to one side and collect the terms without x on the other side
x ln (1.41) = ln sqrt{2}-4x ln sqrt{2}
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}-4x ln sqrt{2} +4x ln sqrt{2}
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}
Factor the common x
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}
x( ln (1.41) +4 ln sqrt{2})= ln sqrt{2}
Solve for x by dividing both sides by the factor in the parenthesis and simplify
x( ln (1.41) +4 ln sqrt{2})= ln sqrt{2}
{x( ln (1.41) +4 ln sqrt{2})}/{ ln (1.41) +4 ln sqrt{2}}= {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
x = {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
The solution
x = {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
x =0.2003

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator.  Here is an example of how you might enter it.

(ln (sqrt{2}))/(ln(1.41)+4 ln(sqrt{2}))

Solve the Logarithmic Equation by the one to one property

Example:

2 log_3(7-x)-log_3 2=log_3 18

Solution:

 

The logarithmic equation
2 log_3(7-x)-log_3 2=log_3 18
Use the power rule and the quotient rule to condense to a single logarithm
2 log_3(7-x)-log_3 2=log_3 18
log_3(7-x)^2-log_3 2=log_3 18
log_3((7-x)^2/ 2)=log_3 18
Since both sides of the equation have the same log base the expressions inside the logarithms must be equal
log_3((7-x)^2/ 2)=log_3 18
(7-x)^2/ 2= 18
Clear the denominator by multiplying by 2 on both sides and simplifying
(7-x)^2/ 2= 18
2*(7-x)^2/ 2= 2*18
(7-x)^2= 36
Get rid of the square by square rooting both sides and simplifying
(7-x)^2= 36
sqrt{(7-x)^2}= sqrt{36}
7-x= pm 6
Get x by itself by subtracting 7 on both sides
7-x= pm 6
7-7-x=-7 pm 6
-x=-7 pm 6
Get x by itself by dividing both sides by negative 1
-x=-7 pm 6
-x/-1={-7 pm 6}/-1
x=7 pm 6
x=7 + 6 or x=7 - 6
x=13 or x=1
Check x=13
2 log_3(7-13)-log_3 2=log_3 18
2 log_3(-6)-log_3 2=log_3 18
Log of a negative is undefined.  Exclude this solution.
Check x=1
2 log_3(7-1)-log_3 2=log_3 18
2 log_3(6)-log_3 2=log_3 18
log_3(6)^2-log_3 2=log_3 18
log_3 36-log_3 2=log_3 18
log_3 36/2=log_3 18
log_3 18=log_3 18
Keep this solution.

The solution to the equation is x=1.

Solve an Exponential Equation: Take the log of both sides

Problem:  Solve the exponential equation.

16^{3x-3}=3^{x-3}

Solution:

 

The exponential equation
16^{3x-3}=3^{x-3}
Since the bases cannot be easily written the same, use the method of taking the log of both sides
ln (16^{3x-3})=ln (3^{x-3})
Use the power rule for logarithms.
(3x-3)ln16=(x-3)ln3
Use the distributive law
3xln16-3ln16=xln3-3ln3
Collect the terms with x to one side and collect the terms without x on the other side
3xln16-3ln16=xln3-3ln3
3xln16-3ln16+3ln16=xln3-3ln3+3ln16
3xln16-xln3=xln3-xln3-3ln3+3ln16
3xln16-xln3=-3ln3+3ln16
3xln16-xln3=3ln16-3ln3
Factor the common x
3xln16-xln3=3ln16-3ln3
x(3ln16-ln3)=3ln16-3ln3
Solve for x by dividing both sides by the factor in the parenthesis and simplify
x(3ln16-ln3)=3ln16-3ln3
{x(3ln16-ln3)}/{3ln16-ln3}={3ln16-3ln3}/{3ln16-ln3}
x={3ln16-3ln3}/{3ln16-ln3}
The solution
x={3ln16-3ln3}/{3ln16-ln3}
x = 0.6956

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator.  Here is an example of how you might enter it.

(3ln(16)-3ln(3))/(3ln(16)-ln(3))