Find determinant of the local curvature function:f(x, y) = 5x^3 -2xy - y^5 when (x, y) = (2, 1)*

To find the determinant of the local curvature function, we first need to find the second partial derivatives of the function f(x, y) = 5x^3 -2xy - y^5.

The second partial derivatives are:

f_xx = ∂²f/∂x² = 30x f_yy = ∂²f/∂y² = -20y^3 f_xy = ∂²f/∂x∂y = -2

Then, we evaluate these at the point (x, y) = (2, 1):

f_xx(2,1) = 302 = 60 f_yy(2,1) = -201^3 = -20 f_xy(2,1) = -2

Finally, we compute the determinant of the Hessian matrix, which is defined as:

D = f_xx*f_yy - (f_xy)^2

Substituting the values we found:

D = 60*(-20) - (-2)^2 = -1200 - 4 = -1204

So, the determinant of the local curvature function at the point (2,1) is -1204.