91Ó°ÊÓ

Partial derivatives are a fundamental concept in calculus. They allow us to understand how a function changes as we vary one of its variables, keeping the others constant. For a function of two variables, like \( f(x, y) \), the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) represent the rate of change of the function with respect to \( x \) and \( y \) respectively. To compute them, we differentiate the function once with respect to each variable, considering the other variables as constants.
In our exercise, the function \( f(x,y) = \sin(x-1)\cos(y) \) requires partial derivatives with respect to \( x \) and \( y \). We find these derivatives as:

• \( \frac{\partial f}{\partial x} = \cos(x-1)\cos(y) \)
• \( \frac{\partial f}{\partial y} = -\sin(x-1)\sin(y) \)

Calculating partial derivatives is essential for constructing Taylor polynomials, as they help determine how the function behaves near a specific point. In our exercise, this point is \((1,0)\). By evaluating partial derivatives at this point, we find the slopes of the function in both the \( x \) and \( y \) directions, aiding in the creation of precise approximations.

Linear approximation is a technique used to approximate a function with a flat plane (or line in one dimension) near a specific point. It is the simplest form of Taylor polynomial and provides a good estimate if the point of interest is very close to where the approximation is based. The linear approximation of a function \( f(x,y) \) at point \((a, b)\) can be described as:

• \( L(x, y) = f(a, b) + \left(\frac{\partial f}{\partial x}(a, b)\right) (x-a) + \left(\frac{\partial f}{\partial y}(a, b)\right) (y-b) \)

For the exercise function, this simplifies to:

Given that \( f(1,0) = 0 \), \( \frac{\partial f}{\partial x} (1,0) = 1 \), and \( \frac{\partial f}{\partial y} (1,0) = 0 \), the linear approximation \( L(x,y) \) becomes:
\[ L(x,y) = 0 + 1\cdot(x-1) + 0\cdot y = x - 1 \]
The result suggests a simple linear function where the value depends solely on \( x \). This is why both \( L(0.9, 0.2) \) and \( Q(0.9, 0.2) \) returned \(-0.1\) except for potential small errors due to not having a complete data set.

Quadratic approximation extends linear approximation by considering the curvature of the function at the point of interest. It involves second-order terms and can provide a more accurate estimate than linear approximation, especially when the point is not too close to the expansion center. For a function \( f(x,y) \), the quadratic approximation at a point \((a, b)\) is:

• \( Q(x, y) = L(x, y) + \frac{1}{2} \cdot \frac{\partial^2 f}{\partial x^2}(a, b)\cdot (x-a)^2 + \frac{1}{2} \cdot \frac{\partial^2 f}{\partial y^2}(a, b)\cdot (y-b)^2 + \frac{\partial^2 f}{\partial x \partial y}(a, b)\cdot (x-a)(y-b) \)

Upon calculation, the second partial derivatives at \((1,0)\) for our function all resulted in zero, simplifying our quadratic approximation. Therefore, the quadratic approximation formula becomes:
\[ Q(x,y) = x - 1 \]
Due to the absence of second-order terms, both quadratic and linear approximations yield the same result in this instance. Hence, when evaluating at \((0.9, 0.2)\), both approximations are similar, producing \(-0.1\). However, by calculating the exact value \(f(0.9, 0.2)\), which is around \(-0.0980\), we can see tiny deviations, showing the improved accuracy of quadratics generally, though in this specific case, no further refinement was needed.