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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: andrewloppol on April 26, 2010, 04:10:48 pm

Title: .
Post by: andrewloppol on April 26, 2010, 04:10:48 pm: .
Title: Re: Implicit Diff question
Post by: m@tty on April 26, 2010, 04:35:38 pm: After finding $\tan{y}$ you can construct a triangle. Opposite = 2 adjacent = 1 Therefore Hypotonuse = $\sqrt{5}$ So $\cos{y}=\frac{a}{h}=\frac{1}{\sqrt{5}}$ .

Then $\cos^2{y}=\frac{1}{5}$ .

So $\frac{dy}{dx}=-\frac{1}{5\left(\frac{1}{\sqrt{2}}\right)^2}=-\frac{2}{5}=-0.4$

Or you could just use the secant form you found initially:

$\sec{y}=\sqrt{5}$ ; $\sec{x}=\sqrt{2}$ .

Therefore $\frac{dy}{dx}=-\frac{\left(\sqrt{2}\right)^2}{\left(\sqrt{5}\right)^2}=-\frac{2}{5}=-0.4$ .

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