Estimating cot(46°) Using Method of Differentials

Sethka
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I am completely lost on these differentials! Can anyone help me make sense of them? Especially this question in particular:

cot(46(deg))

(Sorry, I don't know how to make that small little circle thing that denotes degree)

I'm supposed to use the method of differentials to estimate it to 4 decimal places.

Thanks!
 
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Show what you've tried. Do you know how to use this method?

By the way, I'd never heard of the "method of differentials" before, and I had to do a google search to figure out what it is. It seems to just be synonomous with "linear approximation", or, if you want a higher order approximation, "taylor series approximation", by which names I think the method is much more well known. Just another reason to always show your work if you want help.
 
45 degrees is \frac{\pi}{4} radians. 1 degree is \frac{\pi}{180} radians. It's better to use radians because that way the derivative is easier: if x is measured in radians then the derivative of y= cos(x) is y'= -sin(x) and so the differential is dy= -sin(x)dx. y+ dy= cos(x)- sin(x)dx.
To find cos(\frac{pi}{4}+ \frac{\pi}{180}, let x= \frac{\pi}{4} so that y= cos(\frac{\pi}{4})= \frac{\sqrt{2}}{2}, -sin(x)= -sin(\frac{\pi}{4})= -\frac{\sqrt{2}}{2} and dx= \frac{\pi}{180}.
 
Oh Thanks!

Thanks you guys, My textbooks are a little backwards it seems. Where one asks me to use method of differentials the other teaches linear aproximation, that was so confusing and now I see why. Thanks A bunch!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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