Peter the punter decided to place $10,000 into a high growth share portfolio. After 3 weeks his investment was looking rather sad, with the value of his portfolio falling $2,025. Unperturbed, Peter stuck with his original plan to hold the stock for at least 6 months. But after a further 3 weeks, things had not improved – his portfolio being worth only 64% of its original value at that point in time. Always a risk taker, Peter allowed his investment to continue until after a further 3 weeks, he found his portfolio to be worth only $5,275. Peter then called on you for advice. You know that the share market is full of ups and downs and that things should improve. You tell Peter you can use your tried and proven BiQuad financial modeling program to help forecast his future prospects. Peter agrees and asks for your help.
First you must develop the BiQuad model. To do this you are to use three of the data points available from observations. You will use the fourth data point to verify that your model is correct.
If the last digit of your Student ID No. is an integer power of 2, then you are to use the initial investment value and data from 3 weeks and 6 weeks to work out the model.
If the last digit of your Student ID No. is a prime number, then you are to use the initial investment value and the data from 3 weeks and 9 weeks to work out the model.
If the last digit of your Student ID No. is neither an integer power of 2 or a prime number, then you are to use the initial investment value and the data from 6 weeks and 9 weeks to work out the model.
(A) Write your Student ID No. ******54. Write the last digit here: 4. Write the three data points you are to use here:
Time (t) 0 weeks 3 weeks 6 weeks
Investment Value v(t) $10000 $2025 $6400
(B) Using the data you have written down in the table in (a) derive a quadratic function (you BiQuad model), 𝑣(𝑡), relating the value of Peter’s share portfolio to elapsed time. Show all working.
C) Verify that your function 𝑣(𝑡) is accurate by evaluating the function at the data point you did not use when deriving the function. Show substitutions and working
(D) By completing the square, determine the point in time when your model predicts Peter’s fortune will turn around. What is the lowest value that Peter’s investment will reach at this time? Using this knowledge and Peter’s original data, on a grid draw a graph of 𝑣(𝑡) up to this point in time
For all of the remaining questions you must use calculus to obtain the answers
(E) By finding the derivative of 𝑣(𝑡), determine how rapidly Peter’s fortune were varying after 5 weeks, and after 10 weeks.
(f) Use this derivative function 𝑣’(𝑡) to calculate the time when Peter should expect to see his investment turn the corner from its lowest value and start to improve. Compare your answer here with your answer to (d).
(g) If 𝑣(𝑡) was to be used to continue modelling Peter’s investment, determine at what future time his investment would reach the break-even point?
(H) However the quadratic function used in your BiQuad model changes once an investment has bottomed out. A second quadratic function takes over that predicts a more conservative recovery phase. In this case, recovery is given by the function 𝑤(𝑡) shown below. Use the product rule to determine 𝑤’(𝑡).
The new Function :
W(t) = 5(t-10) (t-20) + 4500, so
W’(t) =
(i) then provide calculations to prove that w(t)= v(t) and w’(t) = v’(t) at the turning point
(j) Using the new function 𝑤(𝑡), provide calculations to determine the predicted value of Peter’s investment at the end of the six month period.
(k) Using the new function 𝑤(𝑡), provide calculations to predict the time (nearest week) when Peter’s investment will have recovered to its initial value.
(l) Using data points created for the new function 𝑤(𝑡), show the graph of this function as an extension to the graph of v(t) in (d). Clearly label the six-month prediction point and breakeven point.
Quadratic Functions and Calculus - Steve, Tuesday, May 19, 2015 at 5:17am
Holy crap! Have you done anything on this?
Have you determined the three data points to use?
Have you decided what to do using the ID number?
Come back with some input, ok?
Quadratic Functions and Calculus - Anonymous, Wednesday, May 20, 2015 at 12:31am
I have used the three data point which is given in part A of the question.
the three inputs are $10000 in week 0
in week three it is $2025
and the 6th week it is 64% of the initial investment.
I have also put the last digit of my student ID which is an integer power of 2 which is 4
I HAVE DONE PART A!!!
Calculus and quadratic equations - Steve, Wednesday, May 20, 2015 at 11:56pm
The data points are:
(0,10000), (3,7075), (6,6400), (9,5275)
The instructions say:
If the last digit of your Student ID No. is an integer power of 2, then you are to use the initial investment value and data from 3 weeks and 6 weeks to work out the model.
So, now you know which three points to use for (B). If y = ax^2+bx+c, you have
0a+0b+c = 10000
9a+3b+c = 7975
36a+6b+c = 6400
Solve those to find your quadratic, and then start using it as instructed.
Calculus and quadratic equations - Reza, Thursday, May 21, 2015 at 2:26am
B)
step 1: substitute points to write to linear equations
y= ax^2+bx+c
the three points are (0,10000) (3,7595) (6,6400)
7975= a3^2+b x3+10000
7975=9a+3b+10000
-10000 -10000
-2025=9a+3b Equation (1)
6400=a(6^2)+6 xb+10000
6400=36a+6b+10000
-10000 -10000
-3600=36a+6b Equation 2
Step 2: Solve the equations
-2025=9a+3b (x4)
-3600=36a+6b (x2)
-8100=36+12b
-7200=72a+12b
-900=-36a
a= -900/-36=25
Sub a into Equation 2
-3600=36(25)+6b
-3600= 900+6b
-900 -900+6b
-4500=6b
b= -4500/6 = -750
so the quadratic equation is
y= 67.22x^2-1003.32x+10000
C)
25x^2- 750x+10000
25(0^2)-750(0)+10000
0 - 0 +10000=10000
Calculus and quadratic equations - Roney, Thursday, May 21, 2015 at 6:00am
Thanks,
I don't know how to do (e) of this question
(E) To find how rapidly Peter's fortune was varying after 5 weeks and 10 weeks, we need to find the derivative of the quadratic function v(t) = 67.22t^2 - 1003.32t + 10000.
First, let's find the derivative of the function using calculus:
v'(t) = d/dt (67.22t^2 - 1003.32t + 10000)
v'(t) = 2*67.22t - 1003.32
Now, substitute t = 5 to find the rate of change after 5 weeks:
v'(5) = 2*67.22(5) - 1003.32
v'(5) = 674.4 - 1003.32
v'(5) = -328.92
The rate of change after 5 weeks is -328.92.
Similarly, substitute t = 10 to find the rate of change after 10 weeks:
v'(10) = 2*67.22(10) - 1003.32
v'(10) = 1344.4 - 1003.32
v'(10) = 341.08
The rate of change after 10 weeks is 341.08.
To answer question (e), you need to find the derivative of the quadratic function 𝑣(𝑡) that you derived in part (B). The derivative of a quadratic function can be found by taking the derivative of each term separately.
The derivative of the first term, 25𝑡^2, is 50𝑡.
The derivative of the second term, -750𝑡, is -750.
The derivative of the third term, 10000, is 0.
So, the derivative of the function 𝑣(𝑡) is 𝑣'(𝑡) = 50𝑡 - 750.
To find how rapidly Peter's fortune varied after 5 weeks, substitute 𝑡=5 into the derivative function 𝑣'(𝑡).
𝑣'(5) = (50)(5) - 750 = 250 - 750 = -500
Therefore, after 5 weeks, Peter's fortune was varying at a rate of -$500 per week.
To find how rapidly Peter's fortune varied after 10 weeks, substitute 𝑡=10 into the derivative function 𝑣'(𝑡).
𝑣'(10) = (50)(10) - 750 = 500 - 750 = -250
Therefore, after 10 weeks, Peter's fortune was varying at a rate of -$250 per week.