Author Topic: . (Read 520 times) Tweet Share

m@tty · « **Reply #1 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 $\frac{dy}{dx}=-\frac{\left(\sqrt{2}\right)^2}{\left(\sqrt{5}\right)^2}=-\frac{2}{5}=-0.4$ .

Search the forums now!

We have moved!

Author Topic: . (Read 520 times) Tweet Share

andrewloppol

.

m@tty

Re: Implicit Diff question

Recent Posts

User:
Password: