Last Big Questions After A Tiring Day Of Homework!?

[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.