Hard HSC question (1 Viewer)

BlueGas · Aug 22, 2015

I was looking at the solutions and I dont understand the 4th step where x turned to 2x. Also how do you differentiate 2x and you get 2x^2 in part ii?

BreathingMaths · Aug 22, 2015

They multiplied the 1/2 inside the big brackets.

rand_althor · Aug 22, 2015

For i) they multiplied everything in the brackets by 1/2.
$\begin{align*}&=\frac{1}{2}s\sin{a}(2x-2l+\frac{l^2}{x}) \\&=s\sin{a}(\frac{2x}{2}-\frac{2l}{2}+\frac{1}{2}\times\frac{l^2}{x}) \\&=s\sin{a}(x-l+\frac{l^2}{2x})\end{align*}$

For ii) you aren't differentiating x, rather x^{-1}
$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x}\frac{l^2}{2x}&=\frac{\mathrm{d}}{\mathrm{d}x}\frac{l^2}{2}x^{-1} \\&=\frac{l^2}{2}\times -1 \times x^{-2} \\&=-\frac{l^2}{2x^2}\end{align*}$

Drsoccerball · Aug 22, 2015

They brought half into the brackets

Drsoccerball · Aug 22, 2015

rand_althor said:
For i) they multiplied everything in the brackets by 1/2. For ii) you aren't differentiating x^2, rather x^{-2}

dat timing

SlimiJay · Aug 22, 2015

what hsc question is this?

rand_althor · Aug 22, 2015

SlimiJay said:
what hsc question is this?

2008 2U Question 10 (b)

Drsoccerball · Aug 23, 2015

BreathingMaths said:
They multiplied the 1/2 inside the big brackets.

Edits his post haha

Hard HSC question (1 Viewer)

BlueGas

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BreathingMaths

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rand_althor

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Drsoccerball

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rand_althor

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