Determine the value of x and y when the gradient of the function given by C: x^3 y+xy^3 =27 is undefined.

Solution: The gradient is equal to the derivative of y with respect to x. By implicit differentiation, \left(x^3\cdot\frac{\differentialD y}{\differentialD x}+3x^2y\right)+\left(3xy^2\cdot\frac{\differentialD y}{\differentialD x}+y^3\right)=0 x^3\cdot\frac{\differentialD y}{\differentialD x}+3x^2y+3xy^2\cdot\frac{\differentialD y}{\differentialD x}+y^3=0 x^3\cdot\frac{\differentialD y}{\differentialD x}+3xy^2\cdot\frac{\differentialD y}{\differentialD x}=-\left(3x^2y+y^3\right) \left(x^3+3xy^2\right)\frac{\differentialD y}{\differentialD x}=-\left(3x^2y+y^3\right) \frac{\differentialD y}{\differentialD x}=-\frac{3x^2y+y^3}{x^3+3xy^2} The derivative becomes undefined if the denominator is zero. Therefore, x^3+3xy^2=0 x\left(x^2+3y^2\right)=0 Splitting the factors, x=0 and x^2+3y^2=0 For the second factor, x^2=-3y^2 The equation does not hold for any pairs of x and y on real numbers EXCEPT when x=0 and y=0. Therefore, x=0,y=0 Checking if the values exist in the curve: 0^3\cdot0+0\cdot0^3\stackrel{?}{=}27 0\ne27 Therefore, the ordered pair is not on the curve. Hence, the equation is undefined at the values of x=0 and y=0 due to the fact that there is a discontinuity in the said ordered pair.