Question

a) Find the linear and quadratic approximations of sin(0.96π) tan(0.26π) + (0.96)²(0.26)

b) Compare these approximations with the actual value.

Answer 1

$f(x,y)=sin(\pi x)tan(\pi y)+x^2y$

$x_0=0.96,\ y_0=0.26$

a)

Linear approximation:

$L(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

$f_x(x,y)=\pi cos(\pi x)tan(\pi y)+2xy$

$f_y(x,y)=\frac{\pi sin(\pi x)}{cos^2(\pi y)} +x^2$

$f(x_0,y_0)=sin(0.96\pi )tan(0.26\pi )+0.96^2\cdot 0.26=0.37$

$f_x(x_0,y_0)=\pi cos(0.96\pi )tan(0.26\pi )+2\cdot 0.96\cdot 0.26=-2.82$

$f_y(x_0,y_0)=\frac{\pi sin(0.96\pi )}{cos^2(0.26\pi )} +0.96^2=1.77$

$L(x,y)=0.37-2.82\cdot(x-0.96)+1.77\cdot(y-0.26)$

$L(x,y)=2.62-2.82x+1.77y$

Quadratic approximation:

$Q(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+$

$+\frac{1}{2}f_{xx}(x_0,y_0)(x-x_0)^2+f_{xy}(x_0,y_0)(x-x_0)(y-y_0)+$

$+\frac{1}{2}f_{yy}(x_0,y_0)(y-y_0)^2$

$f_{xx}(x,y)=-\pi^2 sin(\pi x)tan(\pi y)+2y$

$f_{yy}(x,y)=\frac{2\pi^2 sin(\pi x)sin(\pi y)}{cos^3(\pi y)}$

$f_{xy}(x,y)=\frac{\pi^2 cos(\pi x)}{cos^2(\pi y)} +2x$