| Revision as of 22:14, 1 March 2009 edit24.40.202.246 (talk) →TingTing's Physics Problem: new section← Previous edit | Latest revision as of 11:03, 6 July 2012 edit undoHayson1991 (talk | contribs)242 edits ←Blanked the page | ||
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| <math>x = \tan\left(y\right)</math><br /><br /> | |||
| <math>1 = \sec^2\left(y\right)*\frac{dy}{dx}</math> (Chain rule, derivative of tan=sec^2)<br /><br /> | |||
| <math>\frac{1}{\sec^2\left(y\right)} = \frac{dy}{dx}</math><br /><br /> | |||
| <math>\cos^2\left(y\right) = \frac{dy}{dx}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \cos^2\left(y\right)</math><br /><br /> | |||
| == 9~ == | |||
| <math>x^{2}y + xy^2 = 6\,</math><br /><br /> | |||
| <math>\left(2x*y + x^{2}*\frac{dy}{dx}\right) + \left(1*y^2 + x*2y\frac{dy}{dx}\right) = 0</math><br /><br /> | |||
| <math>2xy + x^{2}\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0</math><br /><br /> | |||
| <math>x^{2}\frac{dy}{dx} + 2xy\frac{dy}{dx} = -2xy - y^2</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \frac{-2xy - y^2}{x^{2} + 2xy}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = -\frac{2xy + y^2}{x^{2} + 2xy}</math><br /><br /> | |||
| == Multiple u's == | |||
| To Find dy/dx for<br /> | |||
| <math>y = 2\cos\left(\left(5x\right)^2\right)</math><br /><br /> | |||
| ===The way she explains it=== | |||
| you'll make 3 u's<br /> | |||
| <math>\text{Let }u = 2\cos\left(u\right)</math><br /><br /> | |||
| <math>\text{Let }u = u^2\,</math><br /><br /> | |||
| <math>\text{Let }u = 5x\,</math><br /><br /> | |||
| == Gaaah, help~~ == | |||
| Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br /> | |||
| <math>x^2 + y^2 = 1\,</math><br /><br /> | |||
| <math>2x + 2y\frac{dy}{dx} = 0\,</math><br /><br /> | |||
| ===Find first derivative=== | |||
| <math>\frac{dy}{dx} = \frac{-2x}{2y}\,</math><br /><br /> | |||
| <math>\frac{dy}{dx} = -\frac{x}{y}\,</math><br /><br /> | |||
| ===Find second derivative=== | |||
| <math>2 + \left(2\frac{dy}{dx}*\frac{dy}{dx} + 2y*\frac{d^{2}y}{dx^2}\right) = 0\,</math><br /><br /> | |||
| <math>2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br /> | |||
| <math>2\left(-\frac{x}{y}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br /> | |||
| <math>2\frac{x^2}{y^2} + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br /> | |||
| <math>2y\frac{d^{2}y}{dx^2} = -2-2\frac{x^2}{y^2}\,</math><br /><br /> | |||
| <math>\frac{d^{2}y}{dx^2} = \frac{-2-2\frac{x^2}{y^2}}{2y}\,</math><br /><br /> | |||
| <math>\frac{d^{2}y}{dx^2} = -\frac{1}{y} - \frac{x^2}{y^3}\,</math><br /><br /> | |||
| <math>\frac{d^{2}y}{dx^2} = -\frac{y^2}{y^3} - \frac{x^2}{y^3}\,</math><br /><br /> | |||
| <math>\frac{d^{2}y}{dx^2} = -\frac{y^2+x^2}{y^3}\,</math><br /><br /> | |||
| <math>\frac{d^{2}y}{dx^2} = -\frac{1}{y^3}\,</math><br /><br /> | |||
| == Clock Problem ~ == | |||
| ===minute hand=== | |||
| <math>x=5\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{60}\right)</math><br /><br /> | |||
| <math>y=5\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{60}\right)</math><br /><br /> | |||
| ===hour hand=== | |||
| <math>x=4\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br /> | |||
| <math>y=4\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br /> | |||
| == Piston speed ~ == | |||
| <math>x_w = r\cos\left(\theta\right)</math><br /><br /> | |||
| <math>y_w = r\sin\left(\theta\right)</math><br /><br /> | |||
| <math>D_p = \sqrt{L^2-x_w^2}+y_w</math><br /><br /> | |||
| <math>\frac{dD_p}{dt} = \frac{1}{2}\left(L^2-x_w^2\right)^{-\frac{1}{2}}\cdot\left(-2x_w\frac{dx_w}{dt}\right)+\frac{dy_w}{dt}</math><br /><br /> | |||
| <math>\frac{dx_w}{dt} = -r\sin\left(\theta\right)\cdot\omega</math><br /><br /> | |||
| <math>\frac{dy_w}{dt} = r\cos\left(\theta\right)\cdot\omega</math><br /><br /> | |||
| <math>\theta = t\cdot\omega</math><br /><br /> | |||
| == Feon's Question 1~ == | |||
| === Solution 1 === | |||
| <math>y = \tan\left(\arcsin\left(x\right)\right)</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \frac{\sec^2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br /> | |||
| === Solution 2 === | |||
| <math>y = \tan\left(\arcsin\left(x\right)\right)</math><br /><br /> | |||
| <math>y = \frac{\sin\left(\arcsin\left(x\right)\right)}{\cos\left(\arcsin\left(x\right)\right)}</math><br /><br /> | |||
| <math>y = \frac{x}{\cos\left(\arcsin\left(x\right)\right)}</math><br /><br /> | |||
| <math>y = x\sec\left(\arcsin\left(x\right)\right)</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + x\cdot\frac{\sec\left(\arcsin\left(x\right)\right)\tan\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + x\cdot\frac{\sin\left(\arcsin\left(x\right)\right)}{\cos^2\left(\arcsin\left(x\right)\right)\sqrt{1-x^2}}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + \frac{x^2}{\cos^2\left(\arcsin\left(x\right)\right)\sqrt{1-x^2}}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + \frac{x^2\sec^2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br /> | |||
| ==Feon's Question 2== | |||
| <math>y = x\cdot\arcsec\left(x\right)</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \arcsec\left(x\right) + \frac{x}{\left|x\right|\sqrt{x^2-1}}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \begin{cases} | |||
| \arcsec\left(x\right) + \frac{1}{\sqrt{x^2-1}}& \mbox{if }x>0 \\ | |||
| \arcsec\left(x\right) - \frac{1}{\sqrt{x^2-1}}& \mbox{if }x<0 | |||
| \end{cases}</math><br /><br /> | |||
| ==Feon's Question 3~~== | |||
| <math>y = x\left(\arcsin\left(x\right)\right)^2-2x+2\sqrt{1-x^2}\arcsin\left(x\right)</math><br /><br /> | |||
| <math>y = x\left(\arcsin\left(x\right)\right)^2-2x+2\left(1-x^2\right)^\frac{1}{2}\arcsin\left(x\right)</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \left(\arcsin\left(x\right)\right)^2+x \cdot \frac{2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}-2+2\cdot \frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}} \cdot -2x \cdot \arcsin\left(x\right) + \frac{2\sqrt{1-x^2}}{\sqrt{1-x^2}}</math><br /><br /><math>\frac{dy}{dx} = \left(\arcsin\left(x\right)\right)^2+x \cdot \frac{2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}-2+2\cdot \frac{1}{2\sqrt{1-x^2}} \cdot -2x \cdot \arcsin\left(x\right) + \frac{2\sqrt{1-x^2}}{\sqrt{1-x^2}}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \arcsin^2\left(x\right)+\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}}-\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}}-2+2</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \arcsin^2\left(x\right)</math><br /><br /> | |||
| ===Last Part=== | |||
| <math>u \cdot v=2\sqrt{1-x^2}\cdot\arcsin\left(x\right)</math><br /><br /> | |||
| <math>u = 2\sqrt{1-x^2}</math><br /><br /> | |||
| <math>v = \arcsin\left(x\right)</math><br /><br /> | |||
| <math>u' = 2\cdot\frac{1}{2\sqrt{1-x^2}}\cdot-2x</math><br /><br /> | |||
| <math>u' = \frac{-2x}{\sqrt{1-x^2}}</math><br /><br /> | |||
| <math>v' = \frac{1}{\sqrt{1-x^2}}</math><br /><br /> | |||
| <math>u'v + v'u = \frac{-2x}{\sqrt{1-x^2}} \cdot \arcsin\left(x\right) + \frac{1}{\sqrt{1-x^2}} \cdot 2\sqrt{1-x^2}</math><br /><br /> | |||
| <math>u'v + v'u = -\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}} + 2</math><br /><br /> | |||
| == ln derivative == | |||
| <math>y = \ln{\sqrt{\frac{x-1}{x+1}}}</math><br /> | |||
| <math>\frac{dy}{dx} = \frac{1}{\sqrt{\frac{x-1}{x+1}}}\cdot\frac{1}{2\sqrt{\frac{x-1}{x+1}}}\cdot\frac{1\left(x+1\right)-1\left(x-1\right)}{\left(x+1\right)^2}</math><br /> | |||
| <math>\frac{dy}{dx} = \sqrt{\frac{x+1}{x-1}}\cdot\left(\frac{1}{2}\cdot\sqrt{\frac{x+1}{x-1}}\right)\cdot\frac{2}{\left(x+1\right)^2}</math><br /> | |||
| <math>\frac{dy}{dx} = \frac{x+1}{2x-2}\cdot\frac{2}{\left(x+1\right)^2}</math><br /> | |||
| <math>\frac{dy}{dx} = \frac{1}{\left(x-1\right)\left(x+1\right)}</math><br /> | |||
| <math>\frac{dy}{dx} = \frac{1}{x^2-1}</math><br /> | |||
| == ln derivative 2~ == | |||
| <math>y = \ln{\left(x+\sqrt{4+x^2}\right)}</math><br /> | |||
| <math>\frac{dy}{dx} = \frac{1}{\left(x+\sqrt{4+x^2}\right)}\cdot\left(1+\frac{1}{2\sqrt{4+x^2}}\cdot 2x\right)</math><br /> | |||
| == Difference == | |||
| For the original function f: | |||
| ===1=== | |||
| <math>\frac{1}{b-a}\int_a^b{fdx}</math> | |||
| ===2=== | |||
| <math>\frac{f(b)-f(a)}{b-a}</math> | |||
| == TingTing's Physics Problem == | |||
| {| class="wikitable" border="1" | |||
| |- | |||
| ! <math>\,x\,</math> | |||
| ! <math>\,y\,</math> | |||
| |- | |||
| | <math>x = \mbox{given} \,</math> | |||
| | <math>y = \mbox{given} \,</math> | |||
| |- | |||
| | <math>t = \,</math> | |||
| | <math>t = \,</math> | |||
| |- | |||
| | <math>v = v_0 \cos\left(\theta\right) \,</math> | |||
| | <math>v = v_0 \sin\left(\theta\right) \,</math> | |||
| |- | |||
| | <math>a = 0 \,</math> | |||
| | <math>a = -9.81 \,</math> | |||
| |} | |||
| <math>y = v_0 \sin\left(\theta\right) \frac{1}{2} a t^2\,</math><br /><br /> | |||
| <math>x = v_0 \cos\left(\theta\right) t\,</math><br /> | |||
| <math>\frac{x}{v_0 \cos\left(\theta\right)} = t\,</math><br /> | |||
| <math>t = \frac{x}{v_0 \cos\left(\theta\right)}\,</math><br /><br /> | |||
| <math>y = v_0 \sin\left(\theta\right) \frac{1}{2} a \left(\frac{x}{v_0 \cos\left(\theta\right)}\right)^2\,</math><br /> | |||