Final answer:
The quadratic Taylor polynomial for (a) (x+1)^3(y+2) around (0,0) is 2 + 3xy + x^3y, and for (b) cos(x)cos(3y), it is 1 - 1/2x^2 - 9/2y^2.
Step-by-step explanation:
To find the quadratic Taylor polynomial for a function near a given point, we need to use the Taylor series expansion up to the second order for functions of two variables. We are finding the Taylor polynomial near (0,0) for the given functions.
Part (a)
For f(x,y)=(x+1)^3(y+2), we expand it around the point (0,0). The quadratic Taylor polynomial consists of the function evaluated at the point, the first order derivatives evaluated at the point times their respective variables, and the second order derivatives evaluated at the point times their respective variables squared or multiplied by each other (in the case of mixed partial derivatives). The respective Taylor polynomial up to the second degree for f(x,y) about the point (0,0) is:
f(0,0) + fx(0,0)x + fy(0,0)y + 1/2fxx(0,0)x^2 + fxy(0,0)xy + 1/2fyy(0,0)y^2
Calculating the derivatives and evaluating at the point (0,0) gives us:
2 + 3(x+1)^2(y+2)x + (x+1)^3y + ...
The polynomial simplifies to:
2 + 3xy + x^3y
Part (b)
For f(x,y)=cos(x)cos(3y), we again expand around (0,0) similarly. The expansion up to second order terms is:
cos(0)cos(0) + fx(0,0)x + fy(0,0)y + 1/2fxx(0,0)x^2 + fxy(0,0)xy +1/2fyy(0,0)y^2
Upon evaluating the derivatives at the point (0,0), we find the Taylor polynomial:
1 - 1/2x^2 - 9/2y^2