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age 23
visits member for 3 years, 10 months
seen Jun 30 '13 at 13:58

"An expert is a person who has made all the mistakes that can be made in a very narrow field." Niels Bohr


Mar
8
comment How to sketch the curve of parametric equations
sorry i had to finish editing my response
Mar
8
comment How to sketch the curve of parametric equations
i see now. so sin θ >= 0 and then sin θ <= 1. so x >= 0 and <=1?
Mar
8
comment How to sketch the curve of parametric equations
oh ok. so i could create a table of various points for t and then find the corresponding x and y values to determine the curve?
Mar
8
comment How to sketch the curve of parametric equations
oh ok. so essentially, you can just plot the initial point, the ending point to get the shape of the curve, and a point in between to determine the direction of the curve?
Mar
8
comment Find the area of the surface obtained by rotating the curve about the x axis
ahh now i see! thanks
Mar
7
comment Find the area of the surface obtained by rotating the curve about the x axis
After doing that I get: the integral from 0 to 1 of [(-2/pi) sqr(1 + u^2pi) du]. Would I have to do a trig substitution?
Mar
3
comment Determining a closed-form solution for the following sum
thanks! that cleared it up
Mar
3
comment Determining a closed-form solution for the following sum
but when evaluating the first summation, it does not need to be multiplied by (n+1) because you are plugging in for i directly?
Mar
3
comment Determining a closed-form solution for the following sum
in the second summation, are you adding n+1 because the index starts at 0 and not 1?
Mar
3
comment Determining a closed-form solution for the following sum
the second one.
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
great. thanks for your help!
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
Thanks. I finally understand, but shouldn't the evaluated integral be: -1/(10x^3) + x^5/6?
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
how did you get the second qauntity exactly?
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
is there supposed to be a sqr root?
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
@Tavares, yes it is
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
ok now i see. so then the integral would be sqr[(25x^8)/36 + 1/2 + 9/(100x^8)]
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
@Tavares where does the 1/2 come from?
Feb
27
comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$
It is 1/(10x^3)
Feb
25
comment Number Theory - Proof of divisibility by $3$
ahh i see now. makes perfect sense. thanks
Feb
25
comment Number Theory - Proof of divisibility by $3$
@yunone, how could you prove that 3|x by using x=y(mod 3) if 3|(x-y) only