... Sections 9B,9C ...
Recall that a linear function is described by the formula
\(y=mx + b\)
\(y\) -- dependent variable
\(x\) -- independent variable
\(m\) -- slope aka rate of change
Dependent variable changes by the same amount \(m\) when independent variable changes by \(1\) unit.
A linear function describes growth when \(m>0\) and decline when \(m<0\).
The graph of a linear function is a straight line.
Charlie picks apples in the orchard at a constant rate. By 9:00am he has picked 150 apples, and by 11:00am he has picked 550 apples. If we use \(A\) for the number of apples and \(t\) for the time measured in hours after 9:00am, which function describes his harvesting?
Solution. The amount of apples picked grows linearly with time: \(A=mt+b\)
Here \(b=150\) is the amount of apples when \(t=0\), that is at 9:00am, and \(m=200\) is how much the amount of apples changes in one hour.
Answer: \(A=200t+150\)
A 1-degree change on the Celsius temperature scale is equivalent to a 9/5-degree change on the Fahrenheit scale. How much does the Fahrenheit temperature changes if the Celsius temperature increases by 5 degrees?
Solution. The Fahrenheit temperature changes with respect
to the Celsius temperature at a rate of 9/5
degrees F/ degree C.
An increase of 5 degrees Celsius results in an increase of 9 degrees Fahrenheit.
In 2012, Dana Vollmer set the women's world record in 100-meter butterfly (swimming) with a time of 55.98 seconds. Assume that the record falls at a constant rate of 0.05 seconds per year. Model the situation with a linear function, and predict the record in 2020.
Solution. We have linear model
\(R=-0.05t + 55.98\)
Where the independent variable \(t\) is time in years after 2012,
and dependent variable \(R\) is the record in seconds.
In particular, this model predicts in 2020 with \(t=8\)
\(R=-0.05\times 8 + 55.98 = 55.58\) seconds.
Suppose your pet dog weighted 3 pounds at birth and weighted 15 pounds one year later. Based on these two data points, find a liner function which describes how weight varies with age.
Solution. The variables are \(w\), the weight, and \(t\), the age.
The function should be
\(w=m\times t + b\)
We find \(b\) as the weight when \(t=0\): \(\hspace{3mm} b=3\)
We find \(m\) as the rate of change:
\(m = \frac{15-3}{1}=12\) if we measure the age in years, or
\(m = \frac{15-3}{12}=1\) if we measure the age in months
Suppose your pet dog weighted 3 pounds at birth and weighted 15 pounds one year later. Based on these two data points, find a liner function which describes how weight varies with age.
Answer. Both
\(w=12t+3\), where \(t\) is the age in years, and
\(w=t+3\), where \(t\) is the age in months
are correct.
Suppose your pet dog weighted 3 pounds at birth and weighted 15 pounds one year later. Based on these two data points, find a liner function which describes how weight varies with age. Use the function to predict your dog's weight at 6 months and 10 years of age.
Solution. We may use either formula. With
\(w=12t+3\), where \(t\) is the age in years,
we find for \(t=0.5\):
\(w=12\times 0.5+3 = 9\) pounds,
and for \(t=10\)
\(w=12\times 10+3 = 123\) pounds.
While the former prediction is reasonable, the latter is not.
\(a=\log_{10}x\) means nothing but \(10^a=x\)
Example. \(2=\log_{10}100 \) because \(10^2=100\)
Useful rules to apply
\(\log_{10}10^x=x\)
\(10^{\log_{10}x} =x \ \ \ (x>0)\)
\(\log_{10}(xy)=\log_{10}x+\log_{10}y \ \ \ (x>0 \ \text{and} \ y>0)\)
\(\log_{10}a^x=x\log_{10}a \ \ \ (a>0) \)
More Details in Brief Review on p488 and p542 of the textbook.
Solve the equation for \(x\)
\(3^x=99\)
Solution. Take logarithms of both sides:
\(x \log_{10}3 = \log_{10}99 \)
\(x= \frac{\log_{10}99}{\log_{10}3}\) \(\approx 4.18 \) <---- Answer
Check that roughly:
since \(3^4=81\), we know \(x>4\)
since \(3^5=243\), we know \(x<5\)
Solve the equation for \(x\)
\(\log_{10}(4+x)=1.1\)
Solution. Raise \(10\) to the power of both sides:
\(10^{ \log_{10}(4+x)} = 10^{1.1} \)
\(4+x=10^{1.1}\)
\(x=10^{1.1}-4 \) \(\approx 8.59 \) <---- Answer
Check that roughly:
\(\log_{10}(4+ 8.59)=\log_{10}(12.59)\) is slightly bigger than \(1\)
Your starting salary at a new job is \(\$2000 \) per month, and you get annual raise of \(5\%\) per year. Create an exponential function which models the situation.
Solution. The function is
\( Q= Q_0 \times (1+r)^t \),
where \(Q\) is your salary \(t\) years after the beginning.
Here the initial value \(Q_0=2000\), and the rate of change \(r=0.05\)
Answer:
\( Q= 2000 \times (1.05)^t \)
Your starting salary at a new job is \(\$2000 \) per month, and you get annual raise of \(5\%\) per year. What is your projected salary \(10\) years later?
Solution. We plug in \(t=10\) into our exponential model \( Q= 2000 \times (1.05)^t \)
\( Q= 2000 \times (1.05)^{10} \approx \$ 3,258 \) <---- Answer
Suppose that poaching reduces the population of an endangered animal by \(8\%\) per year. Further suppose that when the population of this animal falls below \(30\), its extinction is inevitable. If the current population of the animal is \(1500\), when will it face extinction?
Solution.
We start with an exponential model
\(Q=1500 \times 0.92^t\)
We now want to find \(t\) such that \(Q=30\):
\(30=1500 \times 0.92^t\)
We need to solve the equation for \(t\).
Solving for \(t\):
\(30=1500 \times 0.92^t\)
\(\frac{1}{50}=0.92^t\)
\(\log_{10}(1/50)=t \times \log_{10}(0.92)\)
\(t=\frac{\log_{10}(0.02)}{\log_{10}(0.92)} \approx 47\) years
Suppose that poaching reduces the population of an endangered animal by \(8\%\) per year. Further suppose that when the population of this animal falls below \(30\), its extinction is inevitable. The current population of the animal is \(1500\). What is the half-life time of the population?
Solution. We use the half-life time formula
\(T_{\text{half}} = -\frac{ \log_{10}2}{\log_{10}(1+r)}\)
with \(r=-0.08\) to find
\(T_{\text{half}} = -\frac{ \log_{10}2}{\log_{10}(0.92)} \approx 8.31\)
Suppose that poaching reduces the population of an endangered animal by \(8\%\) per year. Further suppose that when the population of this animal falls below \(30\), its extinction is inevitable. The current population of the animal is \(1500\). What is the half-life time of the population?
Remarks.
We found \(T_{\text{half}} \approx 8.31\) while the approximate half-life formula
\(T_{\text{half}} \approx \frac{70}{8} \approx 8.75 \) gives us a good approximation
Roughly, a bit more than \(8\) years later, there will be only \(750\) animals left.
The half-life of carbon-14 is about \(5700\) years. A well-preserved piece of wood found at an archaeological site has \(12.3 \%\) of the carbon-14 that it must have had when it was alive. Estimate when the wood was cut.
Solution. We model with the exponential function
\(Q=Q_0 \times \left( \frac{1}{2} \right) ^{t/T_{\text{half}}} = Q_0 \times \left( \frac{1}{2} \right) ^{t/5700}\)
We want \(t\) such that \(Q/Q_0 = 12.3 \% = 0.123 \)
We solve for \(t\) the equation \(0.123= \left( \frac{1}{2} \right) ^{t/5700}\):
\(t=5700 \times \frac{\log_{10}0.123}{\log_{10}0.5} \approx 17,000 \) years.