4U Revising Game (1 Viewer)

[lolokay]

 

[harism]

my god lolokay.
i was hearing about your answer from gurms.

we are both in definite belief you're our current era mathematician version of einstein.

 

[azureus88]

(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]

 

[harism]

very nice way simon.

 

[gurmies]

 

[youngminii]

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.

 

[Drongoski]

You guys are all just too good. Just can't keep up ! Maybe I should just drop out.

 

[shaon0]

my god lolokay.
i was hearing about your answer from gurms.

we are both in definite belief you're our current era mathematician version of einstein.

yeah, i agree.

 

[azureus88]

[maths]z_1z_2z_3=cis(\theta_1+\theta_2+\theta_3)=cis(2\pi)=1[/maths]

[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]

Well done, nice method. More aesthetically pleasing to go from LHS to RHS xD

 

[azureus88]

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.

 

[jet]

Now I hope this is right. I'm really tired, so there is a good chance I missed something.



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.

 

[jet]

I already did. Gurmies told me you got the same.

 

[harism]

haha.
you know gurmies too?
seems like all of bos knows each other like family.
=)

 

[jet]

LOL. Gurmies approached me one day. We talk every now and then.

 

[harism]

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.

 

[jet]

Tangent to the curve at (x₁, y₁)

 

[harism]

oh i see...
hmm... i will have to rethink this then...

 

[azureus88]

[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]

 

[jet]

x^2 - y^2
Its a minus.