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 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