Using the first order Taylor approximation at the point (﻿x equals﻿0.5, ﻿y equals﻿0.6) for ﻿f left parenthesis x comma y right parenthesis equals sin left parenthesis straight pi y x squared right parenthesis﻿, estimate ﻿f left parenthesis﻿0.6﻿comma﻿0.7﻿right parenthesis﻿.

To estimate \( f(0.6, 0.7) \) using the first-order Taylor approximation at the point \((x_0, y_0) = (0.5, 0.6)\) for the function \( f(x, y) = \sin(\pi y x^2) \), we follow these steps:

### Step 1: Compute \( f(x_0, y_0) \)
Evaluate the function at the point \((x_0, y_0) = (0.5, 0.6)\):
\[
f(0.5, 0.6) = \sin(\pi \cdot 0.6 \cdot (0.5)^2) = \sin\left(\pi \cdot 0.6 \cdot 0.25\right) = \sin\left(\frac{3\pi}{20}\right)
\]

### Step 2: Compute the partial derivatives at \((x_0, y_0)\)

#### Partial derivative with respect to \(x\):
\[
f_x(x, y) = \frac{\partial}{\partial x} \sin(\pi y x^2) = \cos(\pi y x^2) \cdot \frac{\partial}{\partial x}(\pi y x^2) = 2\pi y x \cos(\pi y x^2)
\]
At \((x_0, y_0) = (0.5, 0.6)\), this becomes:
\[
f_x(0.5, 0.6) = 2\pi \cdot 0.6 \cdot 0.5 \cos\left(\frac{3\pi}{20}\right) = 0.6\pi \cos\left(\frac{3\pi}{20}\right)
\]

#### Partial derivative with respect to \(y\):
\[
f_y(x, y) = \frac{\partial}{\partial y} \sin(\pi y x^2) = \cos(\pi y x^2) \cdot \frac{\partial}{\partial y}(\pi y x^2) = \pi x^2 \cos(\pi y x^2)
\]
At \((x_0, y_0) = (0.5, 0.6)\), this becomes:
\[
f_y(0.5, 0.6) = \pi \cdot (0.5)^2 \cos\left(\frac{3\pi}{20}\right) = \frac{\pi}{4} \cos\left(\frac{3\pi}{20}\right)
\]

### Step 3: Use the Taylor approximation formula
The first-order Taylor approximation of \(f(x, y)\) at \((x_0, y_0)\) is given by:
\[
f(x, y) \approx f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
\]
Substituting the values:
\[
f(0.6, 0.7) \approx \sin\left(\frac{3\pi}{20}\right) + 0.6\pi \cos\left(\frac{3\pi}{20}\right) \cdot (0.6 - 0.5) + \frac{\pi}{4} \cos\left(\frac{3\pi}{20}\right) \cdot (0.7 - 0.6)
\]

Simplifying further:
\[
f(0.6, 0.7) \approx \sin\left(\frac{3\pi}{20}\right) + 0.6\pi \cos\left(\frac{3\pi}{20}\right) \cdot 0.1 + \frac{\pi}{4} \cos\left(\frac{3\pi}{20}\right) \cdot 0.1
\]
\[
f(0.6, 0.7) \approx \sin\left(\frac{3\pi}{20}\right) + 0.1\pi \left(0.6 + \frac{1}{4}\right)\cos\left(\frac{3\pi}{20}\right)
\]
\[
f(0.6, 0.7) \approx \sin\left(\frac{3\pi}{20}\right) + \frac{11\pi}{40} \cos\left(\frac{3\pi}{20}\right)
\]

### Final Expression
Thus, the estimate for \( f(0.6, 0.7) \) is:
\[
f(0.6, 0.7) \approx \sin\left(\frac{3\pi}{20}\right) + \frac{11\pi}{40} \cos\left(\frac{3\pi}{20}\right)
\]

Would you like to see the numerical value of this expression or need more details?

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**Related Questions:**

1. How does the Taylor series approximation change if we consider higher-order terms?
2. What are the implications of using a first-order Taylor approximation instead of a higher-order one?
3. How does the approximation error relate to the choice of the expansion point \((x_0, y_0)\)?
4. How does the behavior of the function \( f(x, y) = \sin(\pi y x^2) \) influence the accuracy of the Taylor approximation?
5. Can we use the second-order Taylor series to get a more accurate estimate for \( f(0.6, 0.7) \)?

**Tip:** When estimating a function value using a Taylor approximation, the closer the point of interest is to the expansion point, the more accurate the approximation.

Learn how to estimate f(0.6, 0.7) using the first-order Taylor approximation for the function f(x, y) = sin(pi * y * x^2). This problem involves computing partial derivatives, evaluating trigonometric functions, and applying the Taylor series formula. Suitable for advanced undergraduate students and those interested in mathematical analysis.

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