1. Entering the harbour

The depth y in metres above a horizontal mark on a harbour wall is given by

y=3sin(pi/12)

where t is the time in hours after midday on Monday.

(a) Plot the graph of y=3sin(pi/12) for 0_<t_<36

(b) The rate of change of depth of water in the harbour is given by dy/dt
Find dy/dt and calculate the rate of change of depth of water when t=19

(c) List the coordinates of all the stationary points on the graph.

(d) A luxury liner can only enter the harbour when the depth of water is at least 1.5 metres above the horizontal mark. At what times during the week can the liner enter the harbour?

2. Population growth

The population of a certain city was 30,000 in 1990 and 45,000 in 2000. The growth of this population can be described by the formula
...........kt
P=Po e

, where Po is the population size at a particular time t , and k is a constant. Use this information to find the value of k . Then make a prediction of the population size for the year 2020. Finally, apply a suitable transformation to the population growth formula so that you obtain an expression of the form y=mx=+c . How do x,y ,m and c relate to the variables P and t , and constants Po and k in the original formula?

2. Hello again, rom!

2. Population growth

The population of a certain city was 30,000 in 1990 and 45,000 in 2000.
The growth of this population can be described by the formula: $\displaystyle P \;=\;P_oe^{kt}$
. . where $\displaystyle P_o$ is the population size at a particular time $\displaystyle t$, and $\displaystyle k$ is a constant.

(a) Use this information to find the value of $\displaystyle k.$

In 1990 $\displaystyle (t = 0),\;P = 30,000$
. . So we have: .$\displaystyle 30,000 \:=\:P_oe^0\quad\Rightarrow\quad P_o = 30,000$
The function (so far) is: .$\displaystyle P \;=\;30,000e^{kt}$

In 2000 $\displaystyle (t=10),\;P = 45.000$
. . So we have: .$\displaystyle 45,000 \:=\:30,000e^{10k}\quad\Rightarrow\quad e^{10k} \:=\:1.5$

Take logs: .$\displaystyle \ln\left(e^{10k}\right) \:=\:\ln(1.5)\quad\Rightarrow\quad 10k(\ln e) \:=\:\ln(1.5)$

. . Hence: .$\displaystyle k \:=\:\frac{\ln(1.5)}{10}\:\approx\:0.04$

Therefore, the function is: .$\displaystyle P \;= \;30,000e^{0.04t}$

(b) Then make a prediction of the population size for the year 2020.

In 2020 $\displaystyle (t = 30)\!:\;\;P \:=\:30,000e^{(0.04)(30)} \:\approx\:99,604$

(c) Apply a suitable transformation to the population growth formula
so that you obtain an expression of the form: $\displaystyle y \:=\:mx=+c$

We have: .$\displaystyle P \:=\:30,000e^{0.04t}$

Take logs:
. . $\displaystyle \ln(P) \:=\:\ln\left(30,000e^{0.04t}\right) \:= \:\ln(30,000) + \ln\left(e^{0.04t}\right) \:=\:\ln(30,000) + 0.4t(\ln e)$

And we have: . . . . $\displaystyle \underbrace{\ln(P)}_\downarrow \;= \;\underbrace{0.4t}_\downarrow + \underbrace{\ln(30,000)}_\downarrow$
which is of the form: .$\displaystyle y \;\;\;= \;\;\;mx \quad + \quad c$

(d) How do $\displaystyle x,\,y,\,m$ and $\displaystyle c$ relate to the variables $\displaystyle P$ and $\displaystyle t$,
and the constants $\displaystyle P_o$ and $\displaystyle k$ in the original formula?

. . $\displaystyle \begin{array}{cccc}x \\ y \\ m \\ c\end{array} \begin{array}{cccc} = \\ = \\ = \\ = \end{array} \begin{array}{cccc}t \\ \ln(P) \\ k \\ \ln(P_o)\end{array}$