Recent content by chewy

ahh, i see, i was taking it too literaly, thks very much

ok i see, thks, but I am still confused on how 1 + (dy/dx)^2 is a perfect square, doesn't there need to be a third term??

dont worry about it i figured it out

ok, i found the derivative first which was y^2 - 1/(4y^2), i then squared this which gave me y^4 + 1/(16y^4) - 1/2, i then stuck this in the formula and added the one, which gave y^4 +1/(16y^4) + 1/2, i then put the terms together (16y^8 + 8y^4 + 1)/16y^4, i then seen the numerator is the a...

sorry that was meant to be y is equal or more than -pi/4 and y is equal or less that pi/4

Homework Statement definate integral to find length of line x = int [sqrt( (sec t)^4 - 1) integrate from 0 to y with -pi/4 < y > pi/4 it is actually y is equal or more/less than pi/4 The Attempt at a Solution Ive worked out that the length of the line will be the same...

Find the length of which line ? The equation you have posted will graph out to be a complex curve! the line will be continuous from y =1 to y= 3, not quite sure what u mean, there is only one line x = (y^3/3) + 1/(4y).

The formula for length of a line y = f(x) Length = int [sqrt(1+ (dx/dy)^2)] dx in this case the variable is y,

ok, i found the derivative first which was y^2 - 1/(4y^2), i then squared this which gave me y^4 + 1/(16y^4) - 1/2, i then stuck this in the formula and added the one, which gave y^4 +1/(16y^4) + 1/2, i then put the terms together (16y^8 + 8y^4 + 1)/16y^4, i then seen the numerator is the a...

Homework Statement finding the lengths of a line Homework Equations x = (y^3/3) + 1/(4y) from y =1 to y=3 hint:: 1 + (dx/dy)^2 is a perfect square The Attempt at a Solution I found the solution, i did this by just finding the derivative and then putting it into the equation for...

thks very much, I've learned more of these two questions, than the previous easier 30.

Homework Statement prove by substitution that definite integral int (1/t)dt from [x to x*y] = int (1/t)dt from [1 to y]. Homework Equations The Attempt at a Solution i can do this problem if i integrate and use the log laws, no probs, but the question says to use a substitution...

forget that last reply, i figured it out this morning when i woke with a fresh head. the graphs have the same area but the new one comes from the other side, when f(x) = 0, then f(a-x)= a, when f(x)= a the f(a-x) = 0. painfully obvious now. this is website is a great idea, thanks again for...

thanks very much for the reply, i followed your advice and the integrand i had was 1, which in turn leads to a when integrated, I am a little confused over the dummy variable, is u=x because the limits of the integrand are still [0-a] when reversed? also the answer in the book is a/2, does this...

Homework Statement if f is a continuous function, find the value of the integral I = definte integral int [ f(x) / f(x) + f(a-x) ] dx from 0 to a. by making the substitution u = a - x and adding the resulting integral to I. this is one of the last questions in the thomas international...