“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.”

- S. Gudder

Kawan,

You've just re-derived Quotient rule.

When facing differentiation problem, you need to decide whether it is more efficient to use direct differentiation or implicit differentiation. Implicit differentiation in this case makes things complicated (for human) due to the nature of derivatives of trigonometric functions. For solving by hand, implicit differentiation is more suitable for polynomial functions.

My suggestions:

(1) Understanding Cultural Heritage: Learning from 1 MALAYSIA

(2) Extremism: A Threat to Cultural Diversity in Malaysia

(3) Contemporary Perspectives on Cultural Practices in Malaysia

(4) When Ways of Life Collide: Rethinking Cross-Cultural Approaches to Malaysian Harmony

This post has been edited by Critical_Fallacy: Jan 29 2015, 05:59 PM

Genius!

What kind of help do you require?

Can you at least get dy/dx?

So you also knew about Occam's Razor. It is the mentality thing.

Most Malaysian students are taught to work out the given problem directly, without recognizing the root of the problem and find a workaround to it. chocobo7779's method was commendable.

You can verify your answer at Wolfram Alpha.

Just copy and paste the LaTeX code.

As shown in (i), the function f(x) = 6x / (x² + x + 1) is bounded between –6 ≤ y ≤ 2 for all real values of x. The minimum y = –6 occurs at x = –1, and the maximum y = 2 occurs at x = 1.

The basic technique to sketch the graph of a rational function is to find the asymptotes and the intercepts. However, this method doesn't work well for the function f(x) in this question. Therefore, you need to employ the gradient method.

(1) Obtain the gradient:

dy/dx = f'(x) = –6(x² – 1) / (x² + x + 1)².

(2) Find the turning points (local extrema) of the gradient. These points mark the maximum steepness at certain x values.

x = –1.5321, –0.3473, 1.8794

(3) Calculate f'(x) and f(x) for a set of selected x values: [–2, –1.5321, –1, –0.5, –0.3473, 0, 0.5, 1, 1.8794, 2]

x	f'(x)	f(x)
–2.0000	–2.0000	–4.0000
–1.5321	–2.4534	–5.0642
–1.0000	0.0000	–6.0000
–0.5000	8.0000	–4.0000
–0.3473	8.8230	–2.6946
0.0000	6.0000	0.0000
0.5000	1.4694	1.7143
1.0000	0.0000	2.0000
1.8794	–0.3696	1.7588
2.0000	–0.3673	1.7143

(4) Plot the graph using the data and by tracing the slopes.

The way you interpreted Chocobo's challenge was misguided.

The challenge becomes crystal clear when it is rephrased:

Find the (positive) root of the Cubic equation h(x) = x³ – 6x – 1, such that it lies in the interval [2, 3].

So, by the IVT, you will just need to plug and play with the cubic equation h(x) for any value x within the interval [2, 3] until you get h(x) = 0. Technically, you can determine all the 3 coordinates (roots) if you are able to identify the correct intervals from the graph (not from hand sketch).

By the way, since x³ – 6x – 1 is already a transformed cubic equation (see Tschirnhausen Transformation), you can solve the cubic analytically using using Cardano's Formula.

This post has been edited by Critical_Fallacy: Feb 20 2015, 03:27 AM

Since Cardano's formula is not covered in STPM, you can simply use CASIO fx570 to get the roots of the cubic equation. Unless you are instructed to apply Newton-Raphson's method to find the root, I believe you can just use the built-in feature in the CASIO calculator. STPM Pure Math is still relatively straightforward.

Stick to STPM Math as what Just Visiting By advised. Cardano's formula is intended for those who are into the real "Pure Maths", by which a cubic equation is solved analytically.

In fact, I'm more worried about the way you interpret the math problems like the third (iii) problem. The techniques you apply to solve a problem reflect your understanding of what the question requires you to do.

While (iii) was designed to test your algebraic manipulation skill and your critical thinking skill to relate the cubic equation to the rational function, did you realize that (iii) also could be solved by obtaining the roots of x³ – 6x – 1 = 0?

This post has been edited by Critical_Fallacy: Feb 20 2015, 06:37 PM

It doesn't matter which equation you pick.



Because it is derived from the fundamental cubic equation f(x) = 0, they all have the same set of solutions.



Check the intersections of the graphs.



This post has been edited by Critical_Fallacy: Feb 20 2015, 09:45 PM

THe solution provided by chocobo7779 is similar to the Bisection method—simple but slow to converge.

From the practical side, because the cubic function is continuously differentiable, Newton–Raphson method (covered in STPM) is highly recommended.



To answer chocobo7779's challenge without using calculus, I have modified an optimization method (to find the extremum) to locate the root that is known to exist within a certain interval.



The results converge to 2.5289 (4 d.p).

This is the full version algorithm of chocobo7779's method. It requires at least 50 iterations and a series of tests to arrive at the accuracy of 5 decimal places. Compared to the Modified Golden Section Search, the rate of convergence is much slower. Of course, both cannot beat Newton–Raphson method.

This problem is related to the Nuclear Cooling Tower or Hyperboloid Structure.

If you are referring to 1(b), you can simplify the problem by shifting the y-axis.

Using the Disk Method, you should obtain something like this:



where h is the height of the water level.

Clue #1: Find a way to make this substitution:

Clue #2: Use Trigonometric Identities (after solving Clue #1).

For 2(b), using the Chain Rule, , find

(i).

(ii).

Since is given and can be obtained from the derivative of 1(b), the rest is Algebra and you can find the extrema on the closed interval .

For completeness, you should also determine the height of the water level at which the minimum and maximum of dh/dt occur.

scgoh123

For 2(b), I've got this. The max (1.22 m/hr) occurs around the bottleneck (8.36 m) of the Hyperboloid Container. Does it make sense to you?



When you get , it all goes back to the quadratics.



For your info, the dh/dt does not have a minimum in the calculus sense, but it contains an Infimum, which is the greatest lower bound on h ∈ [0, 14]. You can imagine that, as the water level is approaching 14 m, the flow rate dV/dt ≈ 0, causing the water level to increase extremely slow (to prevent overflow), and therefore dh/dt ≈ 0. It is inaccurate to put exactly dh/dt = 0, because that means the flow control valve is turned off completely (dV/dt = 0). Since dV/dh is non-zero, the Chain Rule becomes indeterminate. The argument of 2(b) is built on the assumption that the water is being pumped continuously into the cooling tower until it stops at the water level 14 m. Naturally, when you stop pumping water, the water level stops increasing.

Please note that in this case, the maximum is also the Supremum (the least upper bound) of the set.

This post has been edited by Critical_Fallacy: Apr 13 2015, 05:20 PM

Was that really a problem?

I believe that I've given an explanation in the previous post.

You could check your answer for h = 13.9 m and h = 13.99 m, to test the convergence of the result .

To make the problem more interesting, the parameters in part 2(b) should be chosen carefully k = 28 and n = 0.5, as both take roughly the same time (15.2 hours) to fill up the container up to the water level 13.9 m. But the nonlinear control mechanism in 2(b) will fill up the container 100% by 17 hours.



In reality, the process will be terminated (turn off the valve) if time taken to fill the container exceeds the expected time. The MRO (maintenance, repair, and overhaul) team will carry out the Fault Detection and Diagnosis manually on the system. Modern systems usually have the Fault-Tolerant Control System installed to minimize downtime of a system.

This post has been edited by Critical_Fallacy: Apr 13 2015, 09:39 PM