A tiny temperature probe is fully immersed in a flowing fluid and is moving with zero relative velocity with respect to the fluid. The velocity field in the fluid is $\overrightarrow{V}=(2x)\hat{i}+(y+3t)\hat{j},$and the temperature field in the fluid is T = 2x² + xy + 4t, where x and y are the spatial coordinates, and t is the time. The time rate of change of temperature recorded by the probe at (x = 1, y = 1, t = 1) is _______.

The correct answer is 18.

Step-by-Step Explanation:

1. Understanding the Physical Scenario:
The temperature probe is fully immersed in the fluid and moves with the fluid at zero relative velocity. This means the probe behaves as a fluid particle (advecting with the flow), and the rate of change of temperature it records is the material derivative (also called the substantial or total derivative) of the temperature field, $\frac{DT}{Dt}$.

The material derivative of a scalar field like temperature $T(x,y,t)$ in a two-dimensional flow is defined as:
$$\frac{DT}{Dt}=\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}$$
where:
• $u$ and $v$ are the velocity components along the $x$ and $y$ directions, respectively.
• $\frac{\partial T}{\partial t}$ is the local time rate of change of temperature.
• $u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}$ is the convective rate of change of temperature.

2. Identifying Velocity Components:
The given velocity field is:
$$\overrightarrow{V}=\left(2x\right)\hat{i}+(y+3t)\hat{j}$$
Thus, the individual velocity components are:
$u=2x$
$v=y+3t$

3. Computing Partial Derivatives of Temperature:
The temperature field is given by:
$T=2x^2+xy+4t$
Let us find the partial derivatives with respect to time and space:
• Local time rate of change:
$$\frac{\partial T}{\partial t}=\frac{\partial }{\partial t}(2x^2+xy+4t)=4$$
• Temperature gradient in the $x$-direction:
$$\frac{\partial T}{\partial x}=\frac{\partial }{\partial x}(2x^2+xy+4t)=4x+y$$
• Temperature gradient in the $y$-direction:
$$\frac{\partial T}{\partial y}=\frac{\partial }{\partial y}(2x^2+xy+4t)=x$$

4. Evaluating at the Specified Location and Time:
We need to calculate the rate of change recorded by the probe at $(x=1,y=1,t=1)$. Let's compute each component at these values:
• $u=2\left(1\right)=2$
• $v=1+3\left(1\right)=4$
• $\frac{\partial T}{\partial t}=4$
• $\frac{\partial T}{\partial x}=4\left(1\right)+1=5$
• $\frac{\partial T}{\partial y}=1$

5. Calculating the Total Rate of Change:
Substituting these values back into the material derivative formula:
$$\frac{DT}{Dt}=4+\left(2\right)\left(5\right)+\left(4\right)\left(1\right)$$
$$\frac{DT}{Dt}=4+10+4=18$$

Therefore, the time rate of change of temperature recorded by the probe is 18.