Category Archives: MAC1140

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.

 

Form a Polynomial given the Degree and Zeros

Example: Form a polynomial f(x) with real coefficients having the given degree and zeros.

Degree 4; Zeros -2-3i; 5 multiplicity 2

Solution:

By the Fundamental Theorem of Algebra, since the degree of the polynomial is 4 the polynomial has 4 zeros if you count multiplicity.

There are three given zeros of -2-3i, 5, 5.

The remaining zero can be found using the Conjugate Pairs Theorem.  f(x) is a polynomial with real coefficients.  Since -2-3i is a complex zero of f(x) the conjugate pair of -2+3i is also a zero of f(x).

Now that all the zeros of f(x) are known the polynomial can be formed with the factors that are associated with each zero.

Since f(x) has a zero of 5, f(x) has a factor of x-5

Since f(x) has a second zero of 5, f(x) has a second factor of x-5

Since f(x) has a factor of -2-3i, f(x) has a factor of x-(-2-3i)

Since f(x) has a factor of -2+3i, f(x) has a factor of x-(-2+3i)

Form the polynomial using all of the factors.  The leading coefficient will remain unknown.
f(x)=a(x-5)(x-5)(x-(-2-3i))(x-(-2+3i))
Multiply the factors with complex numbers.  Doing so will cancel the complex numbers from the expression

  • Distribute the minus
  • Multiply each term in one factor by each term in the other factor
  • simplify i^2=-1
  • combine like terms
f(x)=a(x-5)(x-5)(x-(-2-3i))(x-(-2+3i))
f(x)=a(x-5)^2(x+2+3i)(x+2-3i)
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i-9i^2)
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i-9(-1))
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i+9)
f(x)=a(x-5)^2(x^2+4x+13)
Multiply the other pair of factors
f(x)=a(x-5)^2(x^2+4x+13)
f(x)=a(x-5)(x-5)(x^2+4x+13)
f(x)=a(x^2-5x-5x+25)(x^2+4x+13)
f(x)=a(x^2-10x+25)(x^2+4x+13)
Multiply the two trinomials by multiplying each term in the first trinomial by each term in the other trinomial and then combine like terms
f(x)=a(x^2-10x+25)(x^2+4x+13)
f(x)=a(x^4+4x^3+13x^2-10x^3-40x^2-130x+25x^2+100x+325)
f(x)=a(x^4-6x^3-2x^2-30x+325)

The polynomial with degree 4 and zeros of -2-3i and 5 wiht multiplicity 2 is f(x)=a(x^4-6x^3-2x^2-30x+325)

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.

Partial Fraction Decomposition

Example:  Find the partial fraction decomposition for the rational expression.

11/{x(x^2+11)}

Solution:

The rational expression
 11/{x(x^2+11)}
Factor the denominator completely
11/{x(x^2+11)}
Write the expression with each factor as a separate fraction.
11/{x(x^2+11)}=?/x+?/{x^2+11}
Decide what type of expression to put in the numerator

x is linear so we put a constant in the numerator

x^2+11 is a quadratic so we put a linear expression in the numerator

11/{x(x^2+11)}=?/x+?/{x^2+11}
11/{x(x^2+11)}=A/x+?/{x^2+11}
11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
Now that the set up is done we need to solve for the unknowns A, B and C
11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
Multiply both sides of the equation by the common denominator {x(x^2+11)} and simplify by canceling out common factors
11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
{x(x^2+11)}11/{x(x^2+11)}={x(x^2+11)}(A/x+{Bx+C}/{x^2+11})
11={x(x^2+11)}{A/x}+{x(x^2+11)}{Bx+C}/{x^2+11}
11={(x^2+11)}{A}+x(Bx+C)
Use the distributive property and collect the like terms with the x’s
11={(x^2+11)}{A}+x(Bx+C)
11=Ax^2+11A+Bx^2+Cx
11=Ax^2+Bx^2+Cx+11A
11=(A+B)x^2+Cx+11A
Equate coefficients to create a system of equations to solve
11=(A+B)x^2+Cx+11A
0x^2+0x+11=(A+B)x^2+Cx+11A
On the left 11 is the constant term.  There is no linear term so the coefficient is zero and there is no quadratic term so the coefficient is zero.

On the right the constant (no x’s) is 11A, the linear coefficient is C and the quadratic coefficient is A+B.

System of equations from equating the coefficients
matrix{3}{1}{{11=11A} {0=C} {0=A+B}}

Now solve the system of equations.

The first equation is  11=11A.  Since it only has one variable, I can solve for A by hand.

11=11A
11/11={11A}/11
1=A

The second equation is 0=C .  It is already solved.

The third equation is 0=A+B.  Since I know A=1, I can substitute and solve for B.

0=A+B
0=1+B
0-1=1-1+B
-1=0+B
-1=B

Using the original set up of 11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}, substitute the values of A, B, and C.

11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
11/{x(x^2+11)}=1/x+{-1x+0}/{x^2+11}
11/{x(x^2+11)}=1/x+{-x}/{x^2+11}

The partial fraction decomposition of 11/{x(x^2+11)} is 1/x+{-x}/{x^2+11}

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%