Tag Archives: differentiation

January 16, 2025 · 8:08 am

HSC Advanced 2024 Question 30

HSC Advanced 2024

Two circles have the same centre O. The smaller circle has a radius of 1 cm, while the larger has a radius of (x+1) cm. The circles enclose a region QRST, which is subtended by angle of {\theta} at O, as shaded.

The area of QRST is A cm2, where A is a constant and A>0

Let P cm be the perimeter of QRST

(a) By finding expressions for the area and perimeter of QRST show that P(x)=2x+\frac{2A}{x}

(b) Show that if the perimeter is minimised, then {\theta} must be less than 2.

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

P=\theta(1)+\theta(1+x)+2x
P=2\theta +\theta x +2x
P=\frac{2A}{x}+2x

I like it when the first part requires the student to show something and the second part has them use it (that way they can still do the second part even if they couldn’t do the first part).

(b) \frac{dP}{dx}=2-\frac{2A}{x^2}
0=2-\frac{2A}{x^2}
x=\sqrt{A}

\frac{d^2P}{dx^2}=\frac{4A}{x^3}
Both x and A are greater than zero, therefore \frac{d^2P}{dx^2}>0 and x=\sqrt{A} is a minimum.

Substitute x=\sqrt{A} into the Area formula
2A=A\theta+2\sqrt{A}\theta
\theta=\frac{2A}{A+2\sqrt{A}}
\theta=\frac{2A}{\sqrt{A}(\sqrt{A}+2)}
\theta=\frac{2\sqrt{A}}{\sqrt{A}+2}

Now \frac{2\sqrt{A}}{\sqrt{A}+2}<\frac{2\sqrt{A}}{\sqrt{A}}
Hence \theta<\frac{2\sqrt{A}}{\sqrt{A}}
and \theta<2

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January 9, 2025 · 9:40 am

Deriving the Chain Rule for Differentiation

How to differentiate something in the form y=[f(x)]^n

For example, y=(3x^2-2x+6)^5, we could expand the expression, but the Chain Rule provides a quick and easy method.

Differentiate y=[f(x)]^n

Let u=f(x), then y=u^n

We want to find \frac{dy}{dx}, but \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}

They’re not fractions, but limits of fractions, but they work like fractions.

\frac{du}{dx}=f'(x) and \frac{dy}{du}=nu^{n-1}

Therefore, \frac{dy}{dx}=f'(x)\times nu^{n-1}

Replace u with f(x)

(1)   \begin{equation*}\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)\end{equation*}

What about a function in the form y=f(g(x))?

We’re going to follow the same process.

Let u=g(x), then y=f(u)

\frac{du}{dx}=g'(x) and \frac{dy}{du}=f'(u)

Therefore \frac{dy}{dx}=f'(u)g'(x)

(2)   \begin{equation*}\frac{dy}{dx}=f'(g(x))g'(x) \end{equation*}

Equations 1 and 2 are versions of the Chain Rule.

Example

Find the derivative of y=(3x^2-2x+6)^5

    \begin{equation*}\frac{dy}{dx}=5(3x^2-2x+6)^4\times (6x-2)\end{equation}

    \begin{equation*}\frac{dy}{dx}=5(6x-2)(3x^2-2x+6)^4\end{equation}

    \begin{equation*}\frac{dy}{dx}=10(3x-1)(3x^2-2x+6)\end{equation}

Next time we are going to look at the Product Rule.

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September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

  • Find the function to optimise (in terms of one variable).
  • Find the stationary points.
  • Find the nature of the stationary points.
  • Find the maximum or minimum.
(a) Volume of the cylinder V=\pi r^2h
42=2r+h
h=42-2r
\therefore V_C=\pi r^2(42-2r)
Volume of spherical decorations V_S=\frac{4}{3}\pi( r_s)^3 where r_s=\frac{r}{3}
V_S=\frac{4\pi r^3}{81}
Volume unused space V=\pi r^2(42-2r)-20(\frac{4\pi r^3}{81})
V=2\pi (21r^2-r^3-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{81r^3}{81}-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{121r^3}{81})

(b) V=2\pi (21r^2-\frac{121r^3}{81})
\frac{dV}{dr}=2\pi (42r-\frac{121r^2}{27})
\frac{dV}{dr}=0
0=42r-\frac{121r^2}{27}
0=r(42-\frac{121r}{27})
r=0 or r=\frac{1134}{121}=9.372

\frac{d^2V}{dr^2}=2\pi (42-\frac{242r}{27})
(\frac{d^2V}{dr^2})_{|r=9.372}=-42
\therefore r=9.372 is a maximum.

Dimensions of the vase, internal diameter=18.7cm internal height=23.3cm

(c) Maximum volume of empty space =2\pi (21r^2-\frac{121r^3}{81})=3863.08cm^3
Volume of one sphere =\frac{4}{3}\pi r^3=3448.03cm^3

There is enough unused space for one extra decoration, but it would depend on how they are packed.

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