A curve has parametric equations x= z-cos z , y=sin z , for -pi<z<pi. Find the coordinates of the point at which the gradient of this curve is zero?

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Answer 1 · 2015-08-26T14:23:27

$(\frac{π}{2}, 1)$ $(pi/2, 1)$

For $- π < z < π$ $-pi < z < pi$

$x = z - cos z$ $x= z-cos z$ $" "$ and $" "$ $y = sin z$ $y=sin z$ .

$\frac{d x}{d z} = 1 + sin z$ $dx/dz = 1+sinz$ $" "$ and $" "$ $\frac{d y}{d z} = cos z$ $dy/dz = cos z$ .

So

$\frac{d y}{d x} = \frac{cos z}{1 + sin z}$ $dy/dx = cosz/(1+sinz)$

$cos z = 0$ $cosz = 0$ at $z = - \frac{π}{2} and \frac{π}{2}$ $z = -pi/2 " and " pi/2$

But $\frac{d y}{d x}$ $dy/dx$ does not exist at $z = - \frac{π}{2}$ $z = -pi/2$ .

When $z = \frac{π}{2}$ $z = pi/2$ , we get $(x, y) = (\frac{π}{2}, 1)$ $(x,y) = (pi/2, 1)$