give answer in 2 step with explanation at the end of each step and final answer at the end: The temperature in a chemical reactor is modeled by the function T(x, y) = (x ^ 2)y+4x-6y where x and repres

Question

give answer in 2 step with explanation at the end of each step and final answer at the end: The temperature in a chemical reactor is modeled by the function T(x, y) = (x ^ 2)y+4x-6y where x and represent y spatial coordinates in meters. (a) Compute the gradient vector alpha T(x, y) and explain its physical meaning in context of temperature change in the

reactor. (b) At the point (-1, 1), find the rate of change of temperature in the direction of the vector v - 2, 4 > Express your answer as a directional derivative and explain what this value represents physically. (c) Normalize the vector v and explain why normalization is important when computing directional derivatives. (d) At the point (- 1,1), determine the direction in which the temperature increases most rapidly and compute the maximum rate of increase. Explain why this direction corresponds to the gradient vector. (e) Suppose you want to move from (- 1, 1) to a point where the temperature is exactly 10 units higher than at (- 1, 1). Write an equation that represents all possible points satisfying this condition and describe the geometric shape of this set. (f) Discuss how the gradient and directional derivative concepts cou

VIPSolutions · Accepted Answer

# (a) Compute the Gradient Vector and Its Meaning

**Step 1:** Compute the partial derivatives of $ T(x, y) = x^2 y + 4x - 6y $.

- $ \frac{\partial T}{\partial x} = 2xy + 4 $
- $ \frac{\partial T}{\partial y} x^2 - 6 $

**Explanation:**  
The gradient vector, $ \nabla T(x, y) $, is  
$$
\nabla T(x, y) = \left(2xy + 4, \ x^2 - 6\right)
$$

**Physical meaning:**  
The gradient points in the direction of the greatest rate of increase of temperature. Its magnitude tells how steep the increase is at any point in the reactor.

---

# (b) Directional Derivative at $(-1, 1)$ in Direction of $ \mathbf{v} = \langle 2, 4 \rangle $

**Step 2:**  
First, evaluate the gradient at $(-1, 1)$:

- $ \frac{\partial T}{\partial x}(-1, 1) = 2(-1)(1) + 4 = -2 + 4 = 2 $
- $ \frac{\partial T}{\partial y}(-1, 1) = (-1)^2 - 6 = 1 - 6 = -5 $

So,
$$
\nabla T(-1, 1) = \langle 2, -5 \rangle
$$

The directional derivative in the direction of vector $ \mathbf{v} $ is:
$$
D_{\mathbf{u}}T = \nabla T \cdot \mathbf{u}
$$
where $ \mathbf{u} $ is the **unit vector** in the direction of $ \mathbf{v} $.

First, $ |\mathbf{v}| = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} $.

So $ \mathbf{u} = \left\langle \frac{2}{2\sqrt{5}}, \frac{4}{2\sqrt{5}} \right\rangle = \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle $.

Now,
$$
D_{\mathbf{u}}T = (2)(\frac{1}{\sqrt{5}}) + (-5)(\frac{2}{\sqrt{5}}) = \frac{2}{\sqrt{5}} - \frac{10}{\sqrt{5}} = -\frac{8}{\sqrt{5}}
$$

**Explanation:**  
The directional derivative $ -\frac{8}{\sqrt{5}} $ represents the rate at which temperature decreases (since it's negative) at $(-1, 1)$ in the direction of $ \langle 2, 4 \rangle $.

---

# (c) Normalize the Vector $ \mathbf{v} $ and Explain Importance

**Step 3:**  
To normalize $ \mathbf{v} = \langle 2, 4 \rangle $, divide by its magnitude:
$$
\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle
$$

**Explanation:**  
Normalization ensures the directional derivative measures the rate of change per unit distance, giving a consistent interpretation regardless of the original vector's length.

---

# (d) Direction of Maximum Temperature Increase and Rate

**Step 4:**  
The temperature increases most rapidly in the direction of the gradient vector $ \nabla T(-1, 1) = \langle 2, -5 \rangle $.

The maximum rate of increase is the magnitude:
$$
|\nabla T(-1, 1)| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
$$

**Explanation:**  
The gradient points in the direction of maximum increase because it combines the greatest possible contributions from all directions of change.

---

# (e) Equation for Points With Temperature 10 Units Higher

**Step 5:**  
Let $ (x, y) $ be a point where $ T(x, y) = T(-1, 1) + 10 $.

First, compute $ T(-1, 1) $:
$$
T(-1, 1) = (-1)^2(1) + 4(-1) - 6(1) = 1 - 4 - 6 = -9
$$
So, $ T(x, y) = 1x^2 y + 4x - 6y = -9 + 10 = 1 $.

Equation:
$$
x^2 y + 4x - 6y = 1
$$

**Geometric shape:**  
This equation defines a curve in the $ (x, y) $ plane—a level curve (contour) of the temperature function.

---

# (f) Applications of Gradient and Directional Derivative in Engineering

**Step 6:**  
- **Gradient:** Identifies where and how the temperature increases most rapidly, guiding sensor placement or cooling efforts.
- **Directional Derivative:** Quantifies how temperature changes along any chosen path, useful for predicting heat flow along a specific route.
- **Practical Use:** Optimizing heat distribution, preventing hotspots, and designing efficient cooling/heating systems by following or counteracting the gradient.

---

## Final Answers

1. **Gradient:** $ \nabla T(x, y) = \langle 2xy + 4, \ x^2 - 6 \rangle $
2. **Directional Derivative at $(-1, 1)$ in direction $ \langle 2, 4 \rangle $:** $ -\frac{8}{\sqrt{5}} $
3. **Normalized Direction:** $ \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle $
4. **Maximum Increase Direction and Rate:** Direction $ \langle 2, -5 \rangle $; rate $ \sqrt{29} $
5. **Equation for 10 Unit Increase:** $ x^2 y + 4x - 6y = 1 $, a curve in the plane
6. **Applications:** Used to optimize temperature profiles and design efficient thermal systems in engineering contexts.

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