Quiz on GCE Further Mathematics

[SowWillClimbTree]

Question:

The orthodox method of using First Principles to find a derivative is to deploy Taylor's Theorem first. When the power series of a function is computed by the said theorem, divide every term with h such that the terms with higher degrees will become zero.

Using the above information, compute the first derivative for tan^{-1}(x).

[SowWillClimbTree]

Another question:

Using radian measure of \theta, prove the the area of a sector is equal to r^2\theta/2 and the arc length of the sector is equal to r\theta

[SowWillClimbTree]

Solution on first question without using Taylor's theorem:

Some identities are even NOT taught in HKAL Pure Mathematics curriculum, so the above solution is just for our reference: don't replicate it in HKDSE!

[SowWillClimbTree]

Hint:

Just plug the right hand side to the definition of derivative. I will show the workout later.

[SowWillClimbTree]

Solution:

Denote tan^{-1}(x) = x - x^3/3 + o(x^3),

then:

\frac {d tan^{-1}(x)}{dx} = \lim\limits_{h \to 0} \frac{tan^{-1}(x+h) - tan^{-1}(x)}{h}

= 1 - x^2 + o(x^2)

Note that |x|<1, then the result can be expressed as sum of geometric series of common ratio -x^2, i.e.

\frac {1}{1 - (-x^2)} = \frac {1}{1 +x^2}

Q.E.D.

[SowWillClimbTree]

Solution to the second question:

Note this triangle:

Now we want infinite copies of sliced triangle such the areas of triangles will tend to be the area of a sector. We shall set up the following limit:

\lim\limits_{n \to \infty} n r^2 sin(\frac{\theta}{n})/2

By letting mn = 1, then the limit becomes:

\lim\limits_{m \to 0} \frac{r^2 sin(m\theta)}{2m}

By small value approximation that sin(\theta) ~ \theta, the limit becomes:

\frac{r^2 m\theta}{2m} = r^2\theta /2

[SowWillClimbTree]

For the second part, note that the above triangle is an isosceles triangle, where the 2 equal angles other than A is given by:

\frac{\pi - A}{2}

Then:

sin(\frac{\pi - A}{2}) = cos (A/2)

Now by sine law, denote the infinite sum of infinitesimal arc length as l, we have:

\frac{l/n}{sin(A/n)} = \frac{r}{cos(A/2n)}

Note that by double angle formula:

sinA = 2 sin(A/2) cos(A/2)

Rearranging, we have:

l = \lim\limits_{n \to \infty} 2rn sin(A/2n)

By denoting mn=1 again, the limit becomes:

l = \lim\limits_{m \to 0} \frac{2rsin(Am/2)}{m}

By small value approximation again:

l = \frac{2rAm}{2m} = rA

Change A to \theta, done.

[SowWillClimbTree]

Note on Mathjax:

To google the LaTeX form of any math symbol, use the convention below:

theta latex

then the result will give:

\theta

which will display:

\theta

Google supports MathML from the AI result, so you can directly copy the result to this forum without further formatting.