Question

I am studying differentiation and integration on the circle and have difficulty with the slope which appears to simplify to the Cotangent function. By implication Integration on Cot(x) should indicate the circle as its area function however this is not working out - any assistance or clarification is greatly appreciated

Answer 1

can you type out the question

Answer 2

Based upon the circle equation x^2 + y^2 = 1 my attempt at
dy/dx produced the following - x / ((1-x^2)^0.5). Alternatively if the same equation is presented in parametric form it would be x= cos(t) y = sin(t). If the latter is differentiated it produces -cos(t)/sin(t) as the slope which is -Cot(t). I am seeking to get a better appreciation of what happens should one wish to integrate the equation of the slope in either form ie. Integral of (-x / (1-x^2)^0.5 ) or Integral of -Cot(t) and whether this is consistent with the conventional outcome as regards derivation of Area. Hope this detail is feasible to follow. Regards Ed

Answer 3

one second

Answer 4

it would be nice to integrate a parametrically defined curve

Answer 5

to integrate -x / sqrt (1-x^2) you use trig substitution I believe

Answer 6

yeah thats easier to integrate the parametric integral

Answer 7

cot x is easy to integrate let u = sin t, so we get -ln (sin(t)) , then plug in the limits

Answer 8

so if you want to integrate from x= 0 to x = 1, we have
0 = cos(t) ---> t = pi/2, 1= cos (t) --> t = 0, so this will flip the integral

Answer 9

ok i did it wrong