Asked by Liam
If x=2cos (z) and y= sin (z), then d^2y/dx^2=?
What I got -sin (z) / -8cos (z)*sin (z) simplifies to 1/8cos (z)
I don't think I'm right...
What I got -sin (z) / -8cos (z)*sin (z) simplifies to 1/8cos (z)
I don't think I'm right...
Answers
Answered by
Steve
dy/dt = cosz
dx/dt = -2sinz
dy/dx = -1/2 cotz
d^2y/dx^2 = [d/dt(dy/dx)]/(dx/dt)
= -1/2 csc^2(z) / -2sinz
= -1/4 csc^3(z)
too bad you didn't show any of <u>your</u> work...
dx/dt = -2sinz
dy/dx = -1/2 cotz
d^2y/dx^2 = [d/dt(dy/dx)]/(dx/dt)
= -1/2 csc^2(z) / -2sinz
= -1/4 csc^3(z)
too bad you didn't show any of <u>your</u> work...
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