## Review for the final

... Sections 9B,9C ...

#### Linear modelling

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.

Exercise 10 p536

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: $$A=200t+150$$

Exercise 19 p537

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.

Exercise 24 p538

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.

Exercise 29 p538 ... modified

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

Exercise 29 p538 ... continued

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.

$$w=12t+3$$, where $$t$$ is the age in years, and
$$w=t+3$$, where $$t$$ is the age in months are correct.

Exercise 29 p538 ... continued

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.

Review of logarithms for Exercises 11-26

$$a=\log_{10}x$$ 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.

Logarithms. Exercise 13 on p551

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 $$x<5$$

Logarithms. Exercise 25 on p551

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 $$1$$

Exponential growth and decay. Exercise 33 p551

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 $$r=0.05$$

$$Q= 2000 \times (1.05)^t$$

Exponential growth and decay. Exercise 33 p551

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

Exponential growth and decay. Exercise 39 p551

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?

$$Q=1500 \times 0.92^t$$
We now want to find $$t$$ such that $$Q=30$$:
$$30=1500 \times 0.92^t$$

We need to solve the equation for $$t$$.

Exponential growth and decay. Exercise 39 p551 ... continued

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

Exponential growth and decay. Exercise 39 p551 ... continued

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
$$T_{\text{half}} = -\frac{ \log_{10}2}{\log_{10}(0.92)} \approx 8.31$$

Exponential growth and decay. Exercise 39 p551 ... continued

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.

Exponential growth and decay. Exercise 44 p551

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.