Question

Question #e8dd9

Answer 1

See below.

Explanation:

Given a model

`y(a,t)=a_1t+a_2+a_3sin(omega_0t)`

and a set of points

`P=((t_1,y_1),(t_2,y_2),cdots,(t_n,y_n))`

We will try to adjust the model to the `P` table, by choosing `a_i` such that is minimum the error defined as

`E(a)=sum_(i=1)^n(y(a,t_k)-y_k)^2`

The determination of `a=(a_1,a_2,a_3)` is done by solving the linear system obtained through the stationary conditions

`(partial E(a))/(partial a_i) = 0` for `i=1,2,3`

so we have to solve the linear system

`M a = b`

where

#M = ( (sum t_k^2, sum t_k,sum sin(omega_0t_k)t_k), ( sum t_k,n,sum sin(omega_0t_k)), (sum sin(omega_0t_k)t_k,sum sin(omega_0t_k),sum sin(omega_0t_k)^2))#

and

`b=((sum t_k y_k),(sum y_k),(sum sin(omega_0 t_k)y_k))`

Putting numeric values into the formulas we get

`M=((540, 60, -12),(60, 10, 0),(-12, 0, 4))`
`b=((2430.), (350.), (10.))`

obtaining `{a_1 = 2.5, a_2 = 20, a_3 = 10`

Attached a plot showing the results.

Answering the requested items:

a) Using an square minimum error criteria, the parameters are:

`m = a_1 = 2.5`
`b = a_2 = 20`
`A = a_3 = 10`

b) The stock appreciation is linked to a positive rate so

`(df)/(dt) = m + A omega_0 cos(omega_0 t) > 0`

This is attained for

`0 le t le 3.95` and `8.049 le t le 12`

c) The stock loses value for `3.95 le t le 8.049`

Answer 2

Given that `f(t)` the value of a stock is modeled as

`f(t)=mt+b+Asin((pit)/6)` where t is in months since Jan1

Let us consider

`"At Jan1 "t= 0 and f(0)=$20.00`

`"At Apr1 "t= 3and f(3)=$37.50`

`"At Jul1 "t= 6 and f(6)=$32.50`

`"At Oct1 "t= 9 and f(6)=$35.00`

`"At Jan1 "t= 12 and f(6)=$50.00`

so
`"At Jan1 "t= 0 and f(0)=$20.00`

`=>mxx0+b+Asin(0)=20=>b=20.......(1)`

`"At Apr1 "t= 3and f(3)=$37.50`

`=>mxx3+20+Asin((pi*3)/6)=37.50`

`=>3m+20+A=37.50`

`=>3m+A=37.50-20.00`

`=>3m+A=17.50............................(2)`

`"At Jul1 "t= 6 and f(6)=$32.50`

`=>6xxm+20+Asin((pi*6)/6)=35.00`

`=>6xxm+20+0=35`

`=>m=15/6=2.5..........................(3)`

From (2) and (3) we get

`=>3xx2.5+A=17.50`

`=>A=10.00`

Hence
(a)
`color(red)(m=2.50" "b=20 and A=10)`
(b)

Now the given equation becomes

`f(t)=2.5t+20+10sin((pit)/6)`

Differentiating this equation w,r to t we get the rate of change of the price of stock as

`f'(t)=d/(dt)(2.5t)+d/(dt)(20)+d/(dt)(10sin((pit)/6))`

`=>f'(t)=2.5+(10pi)/6cos((pit)/6)`
In this equation the value of `cos((pit)/6)` varies with time.
`cos((pit)/6)` has got maximum positive value at t=0 and t =12 as `cos 0 and cos (2pi) = 1`
So rate of positive change of price of stock will occur in this period i.e. Jan every year this means the stock appreciate s most in this period.

(c)

Taking `f'(t)=0` we have

`2.5+(10pi)/6cos((pit)/6)=0`

`=>cos((pit)/6)=-1.5/pi`

`=>t=6/picos^-1(-1.5/pi)~~3.95~~4->color(red)"month of May"`

when t =4

`f''(4)=-(10pi^2)/36sin((pi*4)/6)<0->"indicates maximum"`

Again `t = 6/pi(2pi-cos^-1(-1.5/pi))=12-6/picos^-1(-1.5/pi)`

`=12-3.95=8.05->" indicates month of September"`

when t =8

`f''(8)=-(10pi^2)/36sin((pi*8)/6)>0->color(red)"indicates Minimum"`

So during the period May -September growth of price gets diminished i.e the price of the stock is actually losing in this period.

Answer 3

Alternate solution for part (c)

Explanation:

For Solution posted by @dk_ch
The modelling equation becomes

`f(t)=2.5t+20+10sin((pit)/6)`

Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for `t=3.951` and `8.049` respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.

--.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-

For Solution posted by @Cesareo R

`f(t)=2.25t+18.49+7.29sin((pit)/6)`, values rounded to two decimal places

`m = a_1 = 2.25439`, `b = a_2 = 18.49`, `A = a_3 = 7.29285`

Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for `t=4.204` and `7.796` respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.