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PS=ePM, thats just the definition of an ellipse, so you dont need to derive it.

dy/dt is max when glass is thinest and it is min when glass is widest becase it requires larger volume of water to fill when it is wide (hence the slower rate).

Can you assume the volume of a cylinder is (pi)(r^2)(h) when doing questions?

There are solutions to the book as well, not that you would need it. Link below: MEGAUPLOAD - The leading online storage and file delivery service What course are you doing, if you dont mind me asking?

link is here: http://www.megaupload.com/?d=UMCT2WI3 btw, whats so great about this book?

only t = 0, 1 satifies the inequality 1/(1+t<SUP>n</SUP>) <= 1/(1+t<SUP>n+1</SUP>) But as the question is asking about the integral (ie. area under the curve from 0 to 1), i dont think any value of n will satisfy the inequality(the one in the question).

True or false: \int_{0}^{1}\frac{dt}{1+t^n}\leq \int_{0}^{1}\frac{dt}{1+t^{n+1}} for n=1,2,3,... It's false cause they cant be equal right?

y=f(|x|) means: for x>0, y=f(x) for x<0, y=f(-x)

yeh i did, thanks for pointing it out.

\int_{e}^{e^4}\frac{dx}{x\ln x} for the above question you could easily use the substitution x=lnx but if we were to use parts, we get: 2\int_{e}^{e^4}\frac{dx}{x\ln x}=[\frac{\ln x}{\ln x}]^{e^4}_{e} by letting u=\frac{1}{\ln x} and v=\ln xdx Whats with that? How can you have 1...

\int \frac{du}{4+u^2} Substitute u=2tan@ du=(2sec^2@)d@ =\int \frac{2sec^2\Theta }{4sec^2\Theta }d\Theta =\frac{1}{2}\Theta =\frac{1}{2}\tan^{-1}\frac{u}{2}=\tan^{-1}\frac{e^x}{2} That takes ages though, just learn inverse trig.

how come?

The Fibanacci sequence is defined by t_{n+2}=t_{n+1}+t_{n}. Find the limit of \frac{t_{n+1}}{t_{n}} as n goes to infinite.

woops, thought it was an indefinite integral.

Substitue u=e^x du=(e^x)dx \int \frac{du}{4+u^2}=\frac{1}{2}\tan^{-1}\frac{u}{2} =\frac{1}{2}\tan^{-1}\frac{e^x}{2}

Heres a diagram

The volume is just a square pyramid height 4m and base 6m by 6m. Slicing the pyramid parallel to base gives square slices of side s and width dh with h being height of slice above base. By using similar triangles, \frac{s/2}{3}=\frac{4-h}{4} s=\frac{3}{2}(4-h) \delta...

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 y=\frac{b}{a}\sqrt{a^2-x^2} by considering top half only. A=4\frac{b}{a}\int_{0}^{a}\sqrt{a^2-x^2}dx due to symmetry =\pi ab by substitution of x=asin@ or by 1/4 area of circle radius a New Question: Prove that \frac{x_1+x_2+...+x_n}{n}\geq...

lol thats not much of a clue. Better clue: (sqrtx) = (sqrtx) -1 + 1 helps to get it in terms of In and In-1

The question was actually sin squared x not sin2x. actually dont worry, i think i got it now.