Last Big Questions After A Tiring Day Of Homework!?

[mAX9]

Hi I'm up to the last part of the long homework we have to do. If i complete this as well I will be able to pass everything at school.

Modelling Water In A Reservoir

In the year 2001/2002 measurements taken of the volume of water in the Delmaine reservoir ('00 Megalitres) varied with time thus:

t (# months after October 1'st).:1 l 3 l 4 7
V (100's ML)..........................:7 l 6 l 5 l3

The minimum volume was found to be at t = 7

a. Find a quadratic equation of the form y = a(x-h)*squared* + K that approximately fits the data explaining carefully how you obtained the values of a, h and k.

b. Sketch the graph of the function you have obtained in part (a), over the domain [0,12].

c. According to this model, what will be the volume of water 9 months after October 1'st?

It was a long day and I need sleep, please help me out here, I really need answers to these, Thank you.

[NickThissen]

You are given the general form of the equation, so you can enter the values given to find the unknown constants.

The equation will be of the form
V(t) = a ( t - h )² + K

where V and t are the volume and time as given in the table, and a, h and K are, at this time, unknown constants. This is simply given in the question, except that I replaced y by V and x by t as appropriate (the volume is a function of time, just like y is a function of x).

You have four combinations of V and t that should work for this formula, so you can simply enter them.
You know that these equations must all be true:

7 = a ( 1 - h )² + K
6 = a ( 3 - h )² + K
5 = a ( 4 - h )² + K
3 = a ( 7 - h )² + K

There are three unknowns, and four equations, so (if this system has a solution, and it should otherwise the question is wrong), you are able to solve it. In fact, you only need three of the four equations.

How do you do it? Well, you can start by extracting one unknown (let's say K) from one equation. For exmaple, you can rewrite the first equation to
K = 7 - a (1 - h)²

You still can't solve this for K because you don't know a and h yet. But you've got two (actually three even) more equations. So, you can enter this equation for K into the others. Let's do that for the second one:
6 = a ( 3 - h )² + K
6 = a ( 3 - h )² + 7 - a (1 - h)²

Now, you only have two unknowns left! You can repeat this until you have only one unknown. You can then solve that, and work your way back to the top, until you know all unknowns.

That gives you your answer on question a. The rest should be simple

[mAX9]

You are given the general form of the equation, so you can enter the values given to find the unknown constants.

The equation will be of the form
V(t) = a ( t - h )² + K

where V and t are the volume and time as given in the table, and a, h and K are, at this time, unknown constants. This is simply given in the question, except that I replaced y by V and x by t as appropriate (the volume is a function of time, just like y is a function of x).

You have four combinations of V and t that should work for this formula, so you can simply enter them.
You know that these equations must all be true:

7 = a ( 1 - h )² + K
6 = a ( 3 - h )² + K
5 = a ( 4 - h )² + K
3 = a ( 7 - h )² + K

There are three unknowns, and four equations, so (if this system has a solution, and it should otherwise the question is wrong), you are able to solve it. In fact, you only need three of the four equations.

How do you do it? Well, you can start by extracting one unknown (let's say K) from one equation. For exmaple, you can rewrite the first equation to
K = 7 - a (1 - h)²

You still can't solve this for K because you don't know a and h yet. But you've got two (actually three even) more equations. So, you can enter this equation for K into the others. Let's do that for the second one:
6 = a ( 3 - h )² + K
6 = a ( 3 - h )² + 7 - a (1 - h)²

Now, you only have two unknowns left! You can repeat this until you have only one unknown. You can then solve that, and work your way back to the top, until you know all unknowns.

That gives you your answer on question a. The rest should be simple

Oh my God, This is what I really needed, THank You!! The rest should be peace of cake. I'll do it in the morning Thanks again!

[jemidiah]

Actually, the question can't be solved exactly. Nick's approach is correct if the given equation would fit the data exactly. For instance, using his approach and the first three data points gives {K -> 169/24, a -> -(1/6), h -> 1/2} (solved with Mathematica), while using the last three data points gives {K -> 193/24, a -> -(1/24), h -> -4}. Clearly they can't both be right.

The question asks for an "approximate fit" of the data. Most likely, the question wants you to minimize the squared error of the fit, that is, the sum of the squares of the difference between your fit's prediction and the true values. However, trying that right now would be an annoying problem in minimizing with respect to three variables. The question seems to come to the rescue: it says "The minimum volume was found to be at t = 7". Clearly the fit will be a parabola pointing up; the minimum value of a parabola is at its vertex. In the form of the equation given, the vertex occurs at h, so h=7. We can also find K, since

V(7) = a(7-7)^2 + K = K = 3

so h=7, K=3. Now we just need to find a, where a minimizes the error between the fit and the remaining points. This can be done many ways, but again assuming it wants you to minimize the sum of the squared error, first calculate the sum of the squared error:

(V(1)-7)^2 + (V(3)-6)^2 + (V(4)-5)^2 + (V(7)-3)^2
= (a(1-7)^2+3-7)^2 + (a(3-7)^2+3-6)^2 + (a(4-7)^2+3-5)^2 + (a(7-7)^2+3-3)^2
= (36 a - 4)^2 + (16a - 3)^2 + (9a - 2)^2
= 1296 a^2 - 288 a + 16 + 256 a^2 - 96 a + 9 + 81 a^2 - 36 a + 4
= 1633 a^2 - 420 a + 29

Taking the derivative with respect to a to minimize this quantity gives
3266 a - 420 = 0
=> a = 420/3266 ~= 0.1286

If you don't know calculus, you could also find the vertex of this parabola. For a general parabola

c1 x^2 + c2 x + c3

the vertex occurs at -c2/c1, giving the same answer as the above.

Evaluating at the given points gives

V(1) ~= 7.63
V(3) ~= 5.06
V(4) ~= 4.16
V(7) = 3

This is quite close to the original data, somewhat validating the original fit function. You can of course graph it decently well from the calculated quantities. After 9 months, it's easy to see the predicted volume is ~3.51.