Calculate the average rate of change in number of student per computer in example one for the following items.

Year $=(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13)$

Student per computer $=(125, 75, 50, 37, 32, 25, 22, 20, 18, 16, 14, 10.5, 10)$

$$
text{color{#4257b2}(a) Years $2$ to $10$}
$$

Average rate of change $=dfrac{16-75}{10-2}$

$$
=dfrac{-59}{8} =-7.375
$$

The average rate of change in student per computer is decreased by $(7.375)$ students per computer.

$$
text{color{#4257b2}(b) Years $1$ to $5$}
$$

Average rate of change $=dfrac{32-125}{5-1}$

$$
=dfrac{-93}{4} =-23.25
$$

The average rate of change in student per computer is decreased by $(23.25)$ students per computer.

$$
text{color{#4257b2}(c) Years $10$ to $13$}
$$

Average rate of change $=dfrac{10-16}{13-10}$

$$
=dfrac{-6}{3} =-2
$$

$$
text{color{#4257b2}$therefore$ The average rate of change in student per computer is decreased by $(2)$ students per computer.}
$$

$$
text{color{Black}(a) The average rate of change in student per computer is \ decreased by $(7.375)$ students per computer.
\ \
(b) The average rate of change in student per computer is \ decreased by $(23.25)$ students per computer.
\ \
(c) The average rate of change in student per computer is \ decreased by $(2)$ students per computer.}
$$

$$
text{color{#4257b2}Predict the instantaneous rate of change in number of student per computer was the greatest.}
$$

The data show that there was a greater change in the number of student per computer in the first four years, so the decline was faster in these period.

$$
text{color{#c34632}In the first four years}
$$

We can start by entering the given data into the list. By pressing **STAT** then **ENTER** it will open a list where we can input our values. Into the list $L_1$ we should enter the data for years, pressing **ENTER** after each value to input the next one. Into column $L_2$ we should input the values of the number of students per computer, again pressing **ENTER** after each value.

Now, to make a scatter plot, press **2nd**, **Y=**, and **1** **ENTER**. Make sure that the plot is set to *ON* and choose the type and color of the graph you want to display. Press **ZOOM** and **9** to display the graph. It should look as depicted in *Figure 1*.

Figure 1.

To determine the equation that models this situation, i.e. the curve that best fits this data on the graph, we should press **STAT** and scroll to the right over **CALC**. By pressing **0 ExpReg** we will choose the exponential regression since the exponential function of a form $y=ab^x$ best fits the data in the table. By pressing **ENTER** the calculator gives
$$
a=96.3139313 text{and} b=0.8297552544$$
which gives the exponential equation of a form
$$ y=96.3139313(0.8297552544)^x.tag{1}$$
Now we can sketch this function and use the built-in function to set the tangent line to the curve at the points that correspond to the given years in each example.

*a)* The point that corresponds to the second year is the point $(2,75)$ on the graph. To sketch a tangent line to the curve at this point, press **2nd** and **PRGM** and from the list choose **5:Tangent(** and **ENTER**. Then press **2** for the $x$-coordinate of the point at which we need to sketch the tangent, and then **ENTER**. The tangent is drawn and its equation is displayed on the calculator screen. It yields
$$
y=-12.375x+91.06.$$
Hence, the slope of this tangent line is $-12.375$ and that is the instantaneous rate of change in the number of students per computer for the second year.

*b)* The point that corresponds to the seventh year is the point $(7,22)$ on the graph. To sketch a tangent line to the curve at this point, press **2nd** and **PRGM** and from the list choose **5:Tangent(** and **ENTER**. Then press **2** for the $x$-coordinate of the point at which we need to sketch the tangent, and then **ENTER**. The tangent is drawn and its equation is displayed on the calculator screen. It yields
$$
y=-4.868x+60.154.$$
Hence, the slope of this tangent line is $-4.868$ and that is the instantaneous rate of change in the number of students per computer for the second year.

*c)* The point that corresponds to the twelth year is the point $(12,10.5)$ on the graph. To sketch a tangent line to the curve at this point, press **2nd** and **PRGM** and from the list choose **5:Tangent(** and **ENTER**. Then press **2** for the $x$-coordinate of the point at which we need to sketch the tangent, and then **ENTER**. The tangent is drawn and its equation is displayed on the calculator screen. It yields
$$
y=-1.915x+33.233.$$
Hence, the slope of this tangent line is $-1.915$ and that is the instantaneous rate of change in the number of students per computer for the second year.

a) $-12.375$
b) $-4.868$
c) $-1.915$

This investment is an exponential function of time. Note that the exponential function of time $t$ we can write in the form

$$y=a(1+r)^t,tag{1}
$$
where $a$ is the initial amount of the investment and $r$ is the growth rate of this function in decimal form.

*a)* Since the initial investment is given as $a=$6000$ and the rate of growth is $r=7.5%=0.075$ (to set the rate in decimal form, divide precentage amount by $100$), we can rewrite the Eq. $(1)$ as
$$y=$6000(1+0.075)^t=$6000(1.075)^t.tag{2}
$$

*b)* The instantaneous rate of change we can obtain as a slope of a tangent line at the point corresponding to the given year, using the tangent built-in function of the graphing calculator. Input the function set in Eq. $(2)$ and press **GRAPH** to draw it.

To sketch a tangent line to the curve at the point that corresponds to $10$ years, press **2nd** and **PRGM** and from the list choose **5:Tangent(** and **ENTER**. Then input $10$ for the $x$-coordinate of the point at which we need to sketch the tangent, and then press **ENTER**. The tangent is drawn and its equation is displayed on the calculator screen. It yields
$$y=864.33x+3422.88$$
Hence, the slope of this tangent line is $864.33$ and that is the instantaneous rate of change in the value for ten years.

*c)* If rate is compounded semi-annually, then for each compound period the rate is $r=0.075:2=0.0375$ and the function is
$$y=$6000(1+0.0375)^t=$6000(1.0375)^t.tag{3}
$$
Now if we sketch this function using the same built-in functions of the calculator, we can again set the tangent line at the desired point and read out the value of its slope. Note that in $10$ years 23 have a total of $10cdot2=20$ compound periods if the rate is compounded semi-annually. Hence, pressing **2nd** and **PRGM** and choosing option **5:Tangent(** from the list, we can set the tangent at the point $x=20$ and press enter. The tangent will be drawn and its equation displayed on the calculator. It yields:
$$y=461.24x+3304.13.$$
Hence, the slope of this tangent line is $461.24$ and that is the instantaneous rate of change in the value for ten years.

a) $y=$6000(1.075)^t$
b) $864.33$
c) $461.24$

Invest $(1000)$ dollar in saving account with compounded annually rate of $(6)$ hundred percent.

$$
text{color{#4257b2}(a) Calculate the rate at which amount growing over the firs for the following years. }
$$

$$
color{#4257b2}text{(I)} 2text{ years}
$$

Interest rate per year is $0.06cdot1000=60$ dollar per year

$a_{0}=1000$

$$
a_{1}=1000+60=1060=a_{0}+60
$$

$$
a_{2}=1060+60=1120=a_{1}+60=a_{0}+60+60=a_{0}+2(60)
$$

$$
color{#4257b2}a_{n}=1000+60n
$$

Year $=(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)$

Annually increase $=(1000, 1060, 1120, 1180, 1240, 1300, 1360, 1420, 1480, 1540, 1600)$

Average rate of change $=dfrac{1120-1060}{2-1}$

$$
=dfrac{60}{1} =60
$$

$$
text{color{#4257b2}$therefore$ The average rate of change in annually increase by $(60)$ dollar per year.}
$$

$$
color{#4257b2}text{(II)} 5text{ years}
$$

Average rate of change $=dfrac{1300-1060}{5-1}$

$$
=dfrac{240}{4} =60
$$

The average rate of change in annually increase by $(60)$ dollar per year.

$$
color{#4257b2}text{(III)} 10text{ years}
$$

Average rate of change $=dfrac{1600-1060}{10-1}$

$$
=dfrac{540}{9} =60
$$

$$
text{color{#4257b2}$therefore$ The average rate of change in annually increase by $(60)$ dollar per year.}
$$

$$
text{color{#4257b2}(b) Why the change rate is constant?}
$$

Because the annually compounded is increasing at a constant rate, so the instantaneous change in the rate is constant at any period of time.

$$
text{color{#c34632}(a)
\ \
(I) The average rate of change in annually increase by $(60)$ dollar per year.
\ \
(II) The average rate of change in annually increase by $(60)$ dollar per year.
\ \
(III) The average rate of change in annually increase by $(60)$ dollar per year.
\ \
(b) Because the annually compounded is increasing at a constant rate, so the instantaneous change in the rate is constant at any period of time.}
$$

The amount remaining is given by $left(M(t)=500(0.5)^{tfrac{t}{5.2}}right)$.

Answer for the following questions.

$$
{color{#4257b2}text{(a)} Calculate the amount remaining after one day.}
$$

One day=$24$ hours

$$
M(t)=500(0.5)^{tfrac{24}{5.2}}
$$

$$
M(t)=500(0.5)^{4.615}
$$

$$
M(t)=500cdot0.040808
$$

$$
M(t)=20.4040text{ gm}
$$

$$
{color{#4257b2}text{(b)} Calculate the instantaneous rate of change in mass at one day.}
$$

Rate of change in mass at one hour, can be calculated as follows:

$$
M(t)=500(0.5)^{tfrac{1}{5.2}}
$$

$$
M(t)=500(0.5)^{0.1923}
$$

$$
M(t)=500cdot0.8752
$$

$$
M(t)=437.6046text{ gm}
$$

Average rate of change $=dfrac{20.4040-437.6046}{24-1}$

$$
=dfrac{-417.2006}{23} =-18.1392
$$

$$
text{color{#4257b2}$therefore$ The average rate of change in one day is decreased by $(18.1392)$ gm/hr.}
$$

{$text{color{#c34632}(a) $M(t)=20.4040$ gm
\ \
(b) The average rate of change in one day is decreased by $(18.1392)$ gm/hr.}$

The population in town is decline by rate of $(1.8)$ hundred percent per year and the population today is $(12000)$

$$
text{color{#4257b2}(a) Write an equation that represent the population in the town}
$$

Interest rate per year is $.018cdot12000=216$ population per year

$a_{0}=12000$ dollar

$$
a_{1}=12000-216=11784=a_{0}-216
$$

$$
a_{2}=11784-216=11568=a_{1}-216=a_{0}-216-216=a_{0}-2(216)
$$

$$
color{#4257b2}a_{n}=12000-216n
$$

$$
text{color{#4257b2}(b) Calculate the instantaneous rate of change in the population $(10)$ years from now}
$$

$$
a_{n}=12000-(216cdot10)
$$

$$
a_{n}=12000-2160
$$

$$
a_{n}=9840
$$

Instantaneous rate of change $=dfrac{9840-12000}{10-0}=dfrac{2160}{10}$

Instantaneous rate of change $=216$ population.

$$
text{color{#4257b2}(c) Calculate the instantaneous rate of change when the population is half current now }
$$

Population current now $0.5*12000=6000$

Interest rate per year is $.018cdot6000=108$ population per year

$a_{0}=6000$ dollar

$$
a_{1}=6000-108=5892=a_{0}-108
$$

$$
a_{2}=5892-108=5784=a_{1}-108=a_{0}-108-108=a_{0}-2(108)
$$

$$
color{#4257b2}a_{n}=6000-108n
$$

$$
a_{n}=6000-(108cdot10)
$$

$$
a_{n}=6000-1080
$$

$$
a_{n}=4920
$$

Instantaneous rate of change $=dfrac{4920-6000}{10-0}=dfrac{1080}{10}$

Instantaneous rate of change $=108$ population.

$$
text{color{#c34632}(a) $a_{n}=12000-216n$
\ \
(b) Instantaneous rate of change $=216$ population.
\ \
(c) Instantaneous rate of change $=108$ population.}
$$

$$
text{color{#4257b2}Discus how the instantaneous rate of changes for the following function}
$$

$$
f(x)=left(dfrac{1}{2}right)^x f(x)=log_{tfrac{1}{2}}(x)
$$

$$
text{color{#4257b2}For $f(x)=left(dfrac{1}{2}right)^x$}
$$

The instantaneous rate of change in the $y$ values is close to $(0)$ for small value of $x$. As $x$ increase, the instantaneous rate of change gets large very quickly. As $xrightarrowinfty$, the instantaneous rate of change$rightarrowinfty$

$$
text{color{#4257b2}For $f(x)=log_{tfrac{1}{2}}(x)$}
$$

The instantaneous rate of change in the $y$ values is very large for small value of $x$. As $x$ gets larger, the instantaneous rate of change gets smaller very quickly. As $xrightarrowinfty$, the instantaneous rate of change$rightarrow0$

$$
color{#c34632}{color{#4257b2}For $f(x)=left(dfrac{1}{2}right)^x$}
$$

The instantaneous rate of change in the $y$ values is close to $(0)$ for small value of $x$. As $x$ increase, the instantaneous rate of change gets large very quickly. As $xrightarrowinfty$, the instantaneous rate of change$rightarrowinfty$

$$
text{color{#4257b2}For $f(x)=log_{tfrac{1}{2}}(x)$}
$$

The instantaneous rate of change in the $y$ values is very large for small value of $x$. As $x$ gets larger, the instantaneous rate of change gets smaller very quickly. As $xrightarrowinfty$, the instantaneous rate of change$rightarrow0$

*(a)* Entering the given function as a function of $x$ in the graphing calculator and pressing GRAPH, we obtain the graph as depicted in *Figure 1*:

Figure 1.

*(b)* The average rate of change over the distance interval $[d_1,d_2]$ is
$$
r_a=dfrac{S(d_2)-S(d_1)}{d_2-d_1}$$

Hence, for $d_1=10$ and $d_2=100$ the velocities of a wind at these distances are
$$
begin{align*}
S(d_1)&=93log10+65=158\
S(d_2)&=93log100+65=251\
end{align*}$$
Hence, the average rate of change in a velocity of a wind over the distance interval $[10,100]$ is
$$
r_a=dfrac{251-158}{100-10}=1.03 mathrm{dfrac{miles}{hour}}
$$

*(c)* The instantaneous rate of change in a velocity of a wind we can obtain by choosing a small distance interval $[d_1,d_2]$ around the given time distance in miles and using the formula

$$r_i=dfrac{S(d_2)-S(d_1)}{d_2-d_1}.$$

Hence, if we choose distances $d_1=9.99$ and $d_2=10.01$ around the $10$th mile, the velocities of wind at these distances are
$$
begin{align*}
S(d_1)&=93log9.99+65=157.96\
S(d_2)&=93log10.01+65=158.04\
end{align*}$$
Hence, the instantaneous rate of change around the $10$th mile is
$$
r_i=dfrac{158.04-157.96}{10.01-9.99}=4 mathrm{dfrac{miles}{hour}}
$$

Hence, if we choose distances $d_1=99.99$ and and $d_2=100.01$ around the $100$th mile, the velocities of a wind at these distances are
$$
begin{align*}
S(d_1)&=93log99.99+65=250.9959609\
S(d_2)&=93log100.01+65=251.0040387\
end{align*}$$
Hence, the instantaneous rate of change around the $100$th mile is
$$
r_i=dfrac{251.0040387-250.9959609}{100.01-99.99}=0.403 mathrm{dfrac{miles}{hour}}
$$

a) See the graph.
b) $r_a=1.03$
c) $4$ and $0.403$ miles per hour
d) See the explanation.

The instantaneous rate of change we can obtain numerically by choosing a small interval $[x_1,x_2]$ around the certain point and using the formula

$$r_i=dfrac{f(x_2)-f(x_1)}{x_2-x_1},$$

where $f(x_1)$ and $f(x_2)$ are the values of the given exponential function at points $x_1$ and $x_2$.