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\delta V=\pi\left \{ (4-x+\delta x)^2-(4-x)^2 \right \}(2y)\\=4\pi y(4-x)\delta x\\=2\sqrt{3}\pi(4-x)\sqrt{4-x^2}\delta x\\\\V=\lim_{\delta x\to 0}\sum_{x=-2}^{2}2\sqrt{3}\pi(4-x)\sqrt{4-x^2}\delta x\\=8\sqrt{3}\pi\int_{-2}^{2}\sqrt{4-x^2}\delta x-2\sqrt{3}\pi\int_{-2}^{2}x\sqrt{4-x^2}\delta...

It doesnt matter whether its pi or n. Its assumed that they're positive constants anyway so it should work either way. Anyway, new question: Find all solutions to z^4 = (z-1)^4

f'(x)=-\frac{1}{x^2}+\frac{1}{n}\frac{2nx^2\sin(nx)\cos(nx)-2x\sin^2(nx)}{x^4}\\=-\frac{1}{x^2}+\frac{1}{n}\frac{nx\sin(2nx)-2\sin^2(nx)}{x^3}\\=\frac{-nx+nx\sin(2nx)+\cos(2nx)-1}{nx^3}\\=\frac{nx(\sin(2nx)-1)+(\cos(2nx)-1)}{nx^3}\\<0\,\,$for$\,\,x>0\\$since$\,\,-1\leq \sin(2nx)\leq...

x_1=Ut\cos\alpha \\y_1=-\frac{g}{2}t^2+Ut\sin\alpha \\x_2=Vt\cos\alpha \\y_2=-\frac{g}{2}t^2+Vt\sin\alpha \\\\m=\frac{y_2-y_1}{x_2-x_1}=\frac{t\sin\alpha (V-U)}{t\cos\alpha (V-U)}=\tan\alpha \\\therefore $Angle of inclination$\,=\alpha \\\\$Let T be time of flight for particle...

You got x^2/25 + y^2/16 =1 Sub in y=mx+b and you get a quadratic in x. Make the discriminant equal to 0. You now have the general tangent. It should be something like y=mx+-sqrt(16m^2 + 25). Since m=1, y=x+-sqrt(41)

wow, thats a good solution. Nice work.

oh yeh, dam, i see my mistake now. Im missing two terms, -2/2001 and -2/2000. cbb editing.

What did you get as an answer?

lol gurmies, its kinda hard to type out a solution for that question. If you write it out in columns, a lot of stuff cancels out and you get...

Yeh, that was silly of me. It's been edited. Gurms, you ask a question.

x^2-y^2=c^2\\2x-2y\frac{dy}{dx}=0\\\frac{dy}{dx}=\frac{x}{y}=\frac{x_1}{y_1}$at$\,(x_1,y_1)\\y-y_1=\frac{x_1}{y_1}(x-x_1)\\xx_1-yy_1=x_1^2-y_1^2=c^2\\\therefore Tangent:\,xx_1-yy_1=c^2\\\\$Sub y=x,$\,P(\frac{c^2}{x_1-y_1},\frac{c^2}{x_1-y_1})\\$Sub...

New Question: The region bounded by the curve y=1/[(2x+1)(x+1)], the coordinate axis and the line x=4 is rotated through one revolution about the y axis. Use the method of cyclindrical shells to find volume of solid generated.

z_1z_2z_3=cis(\theta_1+\theta_2+\theta_3)=cis(2\pi)=1...

(i) a=g-kv (ii) Terminal velocity: a\to 0,\, v\to \frac{g}{k} (iii)\frac{dv}{dt}=g-kv\\e^{kt}\frac{dv}{dt}+kve^{kt}=ge^{kt}\\\frac{d}{dt}(ve^{kt})=ge^{kt}\\ve^{kt}=\int_{0}^{t}ge^{kt}dt=\frac{g}{k}(e^{kt}-1)\\\frac{dx}{dt}=v=\frac{g}{k}(1-e^{-kt}) (iv)...

sub x=sin@

\int_{0}^{\pi/4}(\cos x+\sec x)^2dx\\=\int_{0}^{\pi/4}(\cos^2x+sec^2x+2)dx\\=\int_{0}^{\pi/4}(\frac{5}{2}+\frac{1}{2}\cos2x+\sec^2x)dx\\=[\frac{5}{2}x+\frac{1}{4}\sin2x+\tan x]^{\pi/4}_0\\=\frac{5\pi}{8}+\frac{5}{4} New Question: Find @ if the quadratic y=ax(x-1) is tangent to circle x^2...

r\binom{n}{r}=\frac{rn!}{r!(n-r)!}=n\frac{(n-1)!}{(r-1)!(n-r)!}=n\binom{n-1}{r-1} P_1+P_2+P_3+...+P_n\\=\sum_{r=1}^{n}\binom{n}{r}x^r(1-x)^{n-r}\\=(1-x+x)^n\\=1...

\int_{0}^{4}(4-x)dx\\=[4x-\frac{x^2}{2}]^4_0\\=8 V=\pi\int_{0}^{4}(4-y)dy\\=\pi\int_{0}^{4}(4-x)dx\\=8\pi New Question: Find dy/dx for: y=\ln(\frac{x^{99}\sqrt[5]{5x+1}}{(x+1)^{21}\sqrt[3]{x+1}}) Do not expand it out lol

In answer to Trebla's question: x=\sum_{i=1}^{k}\sin(nt+\alpha _i)+\sum_{j=1}^{r}\cos(nt+\alpha_j) \frac{dx}{dt}=n\sum_{i=1}^{k}\cos(nt+\alpha _i)-n\sum_{j=1}^{r}\sin(nt+\alpha_j) \frac{d^2x}{dt^2}=-n^2(\sum_{i=1}^{k}\sin(nt+\alpha _i)+\sum_{j=1}^{r}\cos(nt+\alpha_j)) =-n^2x

\ln x-\frac{3}{\ln x}=2\\(\ln x)^2-2\ln x-3=0\\(\ln x-3)(\ln x+1)=0\\\ln x=-1,3\\x=\frac{1}{e},e^3 They're both valid?? Or is there something i missed.