What have you tried so far?
Find dy/dx and d^2y/dx^2 as functions of x and y given that xy^2+y=1.
Find the length of the given curve: x=3t^2, y=3t^3-t t=[0,1]
Find the surface area when the following parametric curve is rotated about the x-axis by 360 degrees: x=t-t^2/2, y=4/3t^3/2 t=[0,1]
I am starting into revision for my exams which are looming! And these are a few questions which have stumped me. Any help would be much appreciated
Hello, lyoung!
Find the length of the given curve: .x .= .3t^2, . y .= .3t^3 - t, . t = [0,1]
We have: . dx/dt = 6t, .dy/dt = 9t^2 - 1
Then: .(dx/dt)^2 = 36t^2, . (dy/dt)^2 .= .(9t^2 - 1)^2 .= .81t^4 - 18t^2 + 1
Hence: .(dx/dt)^2 + (dy/dt)^2 .= .81t^4 + 18t^2 + 1 .= .(9t^2 + 1)^2
. . . . . . . . . .__________________
Therefore: . √(dx/dt)^2 + (dy/dt)^2 . = . 9t^2 + 1
Can you finish the problem now?