Consider f (x, y) = 2x2 − sin(y).
(a) What is the rate of change of f at (2,π) when x is increasing twice as
much as y?
(b) What is the maximum rate of change of f at (2, π)?


We are given the function \( f(x, y) = 2x^2 - \sin(y) \) and asked to address two different questions related to the rate of change of \( f \) at the point \( (2, \pi) \).

### (a) Rate of change of \( f \) when \( x \) is increasing twice as much as \( y \)

We need to compute the total derivative of \( f(x, y) \) using the chain rule and then apply the condition that \( \frac{dx}{dy} = 2 \) (since \( x \) is increasing twice as much as \( y \)).

#### Step 1: Compute the partial derivatives of \( f(x, y) \)

\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2x^2 - \sin(y)) = 4x
\]
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2x^2 - \sin(y)) = -\cos(y)
\]

#### Step 2: Use the chain rule for the total derivative of \( f(x, y) \)

The total derivative of \( f \) with respect to \( y \), given that \( x \) is also changing, is:

\[
\frac{df}{dy} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dy} + \frac{\partial f}{\partial y}
\]

Substitute \( \frac{dx}{dy} = 2 \), and the partial derivatives:

\[
\frac{df}{dy} = 4x \cdot 2 + (-\cos(y)) = 8x - \cos(y)
\]

#### Step 3: Evaluate at the point \( (x, y) = (2, \pi) \)

At \( x = 2 \) and \( y = \pi \):

\[
\frac{df}{dy} = 8(2) - \cos(\pi) = 16 - (-1) = 16 + 1 = 17
\]

Thus, the rate of change of \( f \) at \( (2, \pi) \) when \( x \) is increasing twice as much as \( y \) is **17**.

---

### (b) Maximum rate of change of \( f \) at \( (2, \pi) \)

The maximum rate of change of a function occurs in the direction of the gradient vector. The gradient of \( f(x, y) \) is given by:

\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]

We already computed the partial derivatives:

\[
\frac{\partial f}{\partial x} = 4x, \quad \frac{\partial f}{\partial y} = -\cos(y)
\]

At the point \( (2, \pi) \), we have:

\[
\frac{\partial f}{\partial x} \bigg|_{(2, \pi)} = 4(2) = 8, \quad \frac{\partial f}{\partial y} \bigg|_{(2, \pi)} = -\cos(\pi) = 1
\]

So, the gradient at \( (2, \pi) \) is:

\[
\nabla f(2, \pi) = (8, 1)
\]

The magnitude of the gradient vector gives the maximum rate of change:

\[
|\nabla f(2, \pi)| = \sqrt{8^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65}
\]

Thus, the maximum rate of change of \( f \) at \( (2, \pi) \) is \( \sqrt{65} \).

---

### Summary of Results:
- (a) The rate of change of \( f \) at \( (2, \pi) \) when \( x \) is increasing twice as much as \( y \) is **17**.
- (b) The maximum rate of change of \( f \) at \( (2, \pi) \) is \( \sqrt{65} \).

Would you like a further breakdown of any steps or have questions? 

Here are 5 related questions you might find helpful:
1. How does the total derivative differ from partial derivatives in multivariable functions?
2. What is the interpretation of the gradient in terms of the rate of change of a function?
3. How is the chain rule applied when multiple variables change simultaneously?
4. Can the direction of maximum rate of change ever be negative?
5. What is the geometric meaning of the magnitude of the gradient vector?

**Tip:** The gradient of a function always points in the direction of the steepest ascent at any given point.

Consider f(x, y) = 2x^2 − sin(y). (a) What is the rate of change of f at (2,π) when x is increasing twice as much as y? (b) What is the maximum rate of change of f at (2, π)?

Explore how to compute the rate of change and maximum rate of change for the function f(x, y) = 2x^2 − sin(y) at the point (2, π). This problem covers partial derivatives, the chain rule, and the gradient in multivariable calculus, making it suitable for calculus students at the undergraduate level.

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