The temperature T
 at a point (x,y)
 is given by the function

T(x,y)={1}/{x^2+2y^2+1}
.

At the point (x,y)=(1,1)
, what is the rate of change of T
 traveling in the direction aligned with the vector ⟨1,2⟩
. If your answer is a decimal, round it to two decimal places. 

Answer = ?

To solve this, we need to find the directional derivative of the temperature function \( T(x,y) = \frac{1}{x^2 + 2y^2 + 1} \) at the point \( (1, 1) \) in the direction of the vector \( \langle 1, 2 \rangle \).

The steps to follow are:

1. **Find the gradient** of \( T(x, y) \), denoted as \( \nabla T(x, y) \).
2. **Normalize the given direction vector** \( \langle 1, 2 \rangle \) to get a unit vector.
3. **Compute the dot product** of the gradient and the unit vector to get the directional derivative.

### Step 1: Gradient of \( T(x, y) \)

The gradient \( \nabla T(x, y) \) is given by:

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

We first find the partial derivatives.

#### Partial derivative with respect to \( x \):

\[
\frac{\partial T}{\partial x} = -\frac{2x}{(x^2 + 2y^2 + 1)^2}
\]

#### Partial derivative with respect to \( y \):

\[
\frac{\partial T}{\partial y} = -\frac{4y}{(x^2 + 2y^2 + 1)^2}
\]

So the gradient is:

\[
\nabla T(x, y) = \left( -\frac{2x}{(x^2 + 2y^2 + 1)^2}, -\frac{4y}{(x^2 + 2y^2 + 1)^2} \right)
\]

At the point \( (1, 1) \):

\[
\nabla T(1, 1) = \left( -\frac{2 \times 1}{(1^2 + 2 \times 1^2 + 1)^2}, -\frac{4 \times 1}{(1^2 + 2 \times 1^2 + 1)^2} \right)
\]

Simplifying the denominator:

\[
1^2 + 2 \times 1^2 + 1 = 4
\]
\[
\nabla T(1, 1) = \left( -\frac{2}{4^2}, -\frac{4}{4^2} \right) = \left( -\frac{2}{16}, -\frac{4}{16} \right) = \left( -\frac{1}{8}, -\frac{1}{4} \right)
\]

So,

\[
\nabla T(1, 1) = \left( -0.125, -0.25 \right)
\]

### Step 2: Normalize the Direction Vector

The given direction vector is \( \langle 1, 2 \rangle \). We normalize it to obtain a unit vector.

The magnitude of \( \langle 1, 2 \rangle \) is:

\[
\| \langle 1, 2 \rangle \| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
\]

So the unit vector in this direction is:

\[
\hat{u} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)
\]

### Step 3: Compute the Directional Derivative

The directional derivative of \( T(x, y) \) at \( (1, 1) \) in the direction of \( \hat{u} \) is the dot product of \( \nabla T(1, 1) \) and \( \hat{u} \):

\[
D_{\hat{u}}T(1, 1) = \nabla T(1, 1) \cdot \hat{u}
\]

\[
D_{\hat{u}}T(1, 1) = \left( -0.125, -0.25 \right) \cdot \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)
\]

\[
D_{\hat{u}}T(1, 1) = \left( -0.125 \times \frac{1}{\sqrt{5}} \right) + \left( -0.25 \times \frac{2}{\sqrt{5}} \right)
\]

\[
D_{\hat{u}}T(1, 1) = -\frac{0.125}{\sqrt{5}} - \frac{0.5}{\sqrt{5}} = -\frac{0.125 + 0.5}{\sqrt{5}} = -\frac{0.625}{\sqrt{5}}
\]

Approximating:

\[
D_{\hat{u}}T(1, 1) \approx -\frac{0.625}{2.236} \approx -0.28
\]

### Final Answer:

The rate of change of \( T \) at \( (1, 1) \) in the direction of the vector \( \langle 1, 2 \rangle \) is approximately \( \boxed{-0.28} \).

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How does the rate of change vary in other directions at \( (1, 1) \)?
2. What is the significance of the directional derivative in this context?
3. How does the temperature function behave near critical points?
4. Can the temperature function have a maximum or minimum at any point?
5. What is the physical meaning of the gradient in this scenario?

**Tip:** When dealing with directional derivatives, always ensure the direction vector is normalized for accurate results.

The temperature T at a point (x,y) is given by the function T(x,y) = 1 / (x^2 + 2y^2 + 1). At the point (x,y) = (1,1), what is the rate of change of T traveling in the direction aligned with the vector ⟨1,2⟩? If your answer is a decimal, round it to two decimal places.

Find the rate of change of the temperature function T(x, y) = 1 / (x^2 + 2y^2 + 1) at the point (1, 1) in the direction of vector ⟨1, 2⟩ using the gradient and directional derivative. This problem involves multivariable calculus concepts, including gradients and unit vectors. Solution rounded to two decimal places.

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