4U Revising Game (2 Viewers)
- Thread starter ronnknee
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harism
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my god lolokay.=ax^2+bx+c\\ f[f(x)] = a[f(x)]^2 + b[f(x)] + c \equiv [f(x)]^2\\ $equating coefficients gives a=1, b,c=0$\\ \therefore f(x)=x^2)
i was hearing about your answer from gurms.
we are both in definite belief you're our current era mathematician version of einstein.
(i) [maths]a=g-kv[/maths]
(ii) Terminal velocity: [maths]a\to 0,\, v\to \frac{g}{k}[/maths]
(iii)[maths]\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})[/maths]
(iv) [maths]x=\frac{g}{k}\int_{0}^{t}(1-e^{kt})dt\\=\frac{g}{k}(t+\frac{e^{-kt}}{k}-\frac{1}{k})[/maths]
(v)[maths]v\frac{dv}{dx}=g-kv\\\int_{0}^{x}-kdx=\int_{0}^{v}\frac{-kvdv}{g-kv}\\-kx=\int_{0}^{v}(1-\frac{g}{g-kv})dv\\-kx=v+\frac{g}{k}\ln(\frac{g-kv}{g})\\x=-\frac{v}{k}+\frac{g}{k^2}\ln(\frac{g}{g-kv})[/maths]
(ii) Terminal velocity: [maths]a\to 0,\, v\to \frac{g}{k}[/maths]
(iii)[maths]\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})[/maths]
(iv) [maths]x=\frac{g}{k}\int_{0}^{t}(1-e^{kt})dt\\=\frac{g}{k}(t+\frac{e^{-kt}}{k}-\frac{1}{k})[/maths]
(v)[maths]v\frac{dv}{dx}=g-kv\\\int_{0}^{x}-kdx=\int_{0}^{v}\frac{-kvdv}{g-kv}\\-kx=\int_{0}^{v}(1-\frac{g}{g-kv})dv\\-kx=v+\frac{g}{k}\ln(\frac{g-kv}{g})\\x=-\frac{v}{k}+\frac{g}{k^2}\ln(\frac{g}{g-kv})[/maths]
gurmies
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youngminii
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Oh I am so god damn annoyed. I had the solution and then I clicked copy to document on latex without pressing quick reply. Fuck me sideways.
yeah, i agree.
[maths]z_1z_2z_3=cis(\theta_1+\theta_2+\theta_3)=cis(2\pi)=1[/maths]
 \,\, Prove \,\, that \cos\theta_{k}=\frac{1}{2}(z_{k}+\frac{1}{z_{k}}) \,\, and \,\, \cos2\theta_{k}=\frac{1}{2}(z_{k}^{2}+\frac{1}{z_{k}^{2}}). \,\, Hence \,\, or \,\, otherwise, \,\, prove \,\, that \,\, \cos2\theta_{1}+\cos2\theta_{2}+\cos2\theta_{3}=4\cos\theta_{1}\cos\theta_{2}\cos\theta_{3}-1)
[maths]RHS=4\cos\theta_1\cos\theta_2\cos\theta_3-1\\=4(\frac{1}{2})^3(z_1+\frac{1}{z_1})(z_2+\frac{1}{z_2})(z_3+\frac{1}{z_3})-1\\=\frac{1}{2}(z_1z_2z_3+\frac{1}{z_1z_2z_3}+\frac{z_1z_2}{z_3}+\frac{z_1z_3}{z_2}+\frac{z_2z_3}{z_1}+\frac{z_1}{z_2z_3}+\frac{z_2}{z_1z_3}+\frac{z_3}{z_1z_2})-1\\=\frac{1}{2}(1+1+\frac{1}{z_3^2}+\frac{1}{z_2^2}+\frac{1}{z_1^2}+z_1^2+z_2^2+z_3^2-2)\\=\frac{1}{2}(z_1^2+\frac{1}{z_1^2})+\frac{1}{2}(z_2^2+\frac{1}{z_2^2})+\frac{1}{2}(z_3^2+\frac{1}{z_3^2})\\=\cos2\theta_1+\cos2\theta_2+\cos2\theta_3\\=LHS[/maths]
gurmies
Drover
Well done, nice method. More aesthetically pleasing to go from LHS to RHS xD
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.
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.
jet
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Now I hope this is right. I'm really tired, so there is a good chance I missed something.
(x+1)} \\ \text{Hence, the sum of the volumes of each cylinder gives the volume of the solid, as the value of } \delta x \to 0 \\ \therefore V = \lim_{\delta x \to 0}\sum_{0}^{4}\frac{2\pi x \delta x}{(2x+1)(x+1)} \\ = 2\pi\int^4_0 \frac{x \, dx}{(2x+1)(x+1)} \\ = 2\pi \int^4_0 \frac{1}{x+1} - \frac{1}{2x+1} \, dx \\ = 2\pi \left [ln \left |x + 1 \right | - \frac{1}{2}ln \left |2x + 1 \right | \right ]^4_0 \\ = 2\pi \left [ln \left |\frac{x+1}{\sqrt{2x + 1}} \right | \right ]^4_0 \\ = 2\pi \left [ln\left ( \frac{5}{3} \right ) - ln\left ( 1 \right ) \right ] \\ = 2\pi \, ln \left (\frac{5}{3} \right ) \\)
New question:
The tangent to x2 - y2 = c2 meets the lines y = x and y = -x at P and Q respectively. Find the equation of the tangent in cartesian coordinates (i.e NO PARAMETRICS) and hence prove that the area of the triangle OPQ is constant.
New question:
The tangent to x2 - y2 = c2 meets the lines y = x and y = -x at P and Q respectively. Find the equation of the tangent in cartesian coordinates (i.e NO PARAMETRICS) and hence prove that the area of the triangle OPQ is constant.
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harism
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for your question, jetblack, you mentioned "find the equation of the tangent".
where is the tangent produced?
did you mean line joining P and Q?
If you did, then:
P is (c/rt2 , c/rt2)
Q is (-c/rt2 , c/rt2)
line joining, PQ, has the equation: y=c/rt2
thus, for the traingle OPQ, you have height= c/rt2 and base = 2 . c/rt2 = c.rt2
finding area=hb/2
=(c/rt2).c.rt2/2
=c^2 / 2
which is a constant.
where is the tangent produced?
did you mean line joining P and Q?
If you did, then:
P is (c/rt2 , c/rt2)
Q is (-c/rt2 , c/rt2)
line joining, PQ, has the equation: y=c/rt2
thus, for the traingle OPQ, you have height= c/rt2 and base = 2 . c/rt2 = c.rt2
finding area=hb/2
=(c/rt2).c.rt2/2
=c^2 / 2
which is a constant.
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[maths]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 y=-x,$\,Q(\frac{c^2}{x_1+y_1},\frac{-c^2}{x_1+y_1})\\\\$Area$\\=\frac{1}{2}OP.OQ\\=\frac{1}{2}\sqrt{2(\frac{c^2}{x_1-y_1})^2}\sqrt{2(\frac{c^2}{x_1+y_1})^2}\\=(\frac{c^2}{x_1+y_1})(\frac{c^2}{x_1-y_1})\\=\frac{c^4}{x_1^2-y_1^2}\\=\frac{c^4}{c^2}\\=c^2[/maths]
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