Question

If the temperature is $T=x^2-xy+z^2$, find (a) the direction of heat flow at $(2,1,-1)$; (b) the rate of change of temperature in the direction $j-k$ at $(2,1,-1)$.

Short answer

(a) Direction of heat flow: -3i + 2j + 2k (b) Rate of change of temperature: 0

Step-by-step solution

- Find the Gradient of T

The gradient of a scalar field like temperature is given by the vector of its partial derivatives. Compute the partial derivatives of the function $$T=x^2-xy+z^2$$ with respect to each variable: $ablaT=(\frac{\partial T}{\partial x},\frac{\partial T}{\partial y},\frac{\partial T}{\partial z})$.

- Calculate the Partial Derivatives

First, find the partial derivative with respect to x: $\frac{\partial T}{\partial x}=2x-y$. Next, find the partial derivative with respect to y: $\frac{\partial T}{\partial y}=-x$. Finally, find the partial derivative with respect to z: $\frac{\partial T}{\partial z}=2z$.

- Evaluate the Gradient at the Given Point

Evaluate the partial derivatives at the point $(2,1,-1)$: $\frac{\partial T}{\partial x}{|}_{(2,1,-1)}=2(2)-1=3$, $\frac{\partial T}{\partial y}{|}_{(2,1,-1)}=-2$, $\frac{\partial T}{\partial z}{|}_{(2,1,-1)}=2(-1)=-2$. Thus, the gradient is $ablaT=(3,-2,-2)$.

- Determine the Direction of Heat Flow

The direction of heat flow is in the direction of the negative gradient. Thus, at the point $(2,1,-1)$, the direction of heat flow is $-ablaT=(-3,2,2)$.

- Find the Rate of Change of Temperature in Given Direction

To find the rate of change of temperature in the direction $j-k$, first normalize the direction vector: $\overrightarrow{v}=(0,1,-1)$. The magnitude of $\overrightarrow{v}$ is $\Vert \overrightarrow{v}\Vert =\sqrt{0^2+1^2+(-1{)}^2}=\sqrt{2}$. The unit vector in the direction $j-k$ is then $\hat{v}=(0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$.

- Compute the Directional Derivative

The rate of change of temperature in the direction of the vector $j-k$ is given by the dot product of the gradient and the unit vector: $$ablaT\cdot \hat{v}=(3,-2,-2)\cdot (0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$$ Evaluating the dot product: $$3\times 0+(-2)\times \frac{1}{\sqrt{2}}+(-2)\times \frac{-1}{\sqrt{2}}=0+(-\frac{2}{\sqrt{2}})+\left(\frac{2}{\sqrt{2}}\right)=0$$ Thus, the rate of change of temperature is 0.

Key concepts

Partial Derivatives

In calculus, partial derivatives are used to understand the behavior of multivariable functions. When dealing with functions of several variables, like our temperature function $T=x^2-xy+z^2$, a partial derivative represents the rate at which the function changes as one of the variables changes while keeping the other variables constant. For example, the partial derivative of $T$ with respect to $x$ (denoted $\frac{\partial T}{\partial x}$) can be found by differentiating $T$ with respect to $x$ while treating $y$ and $z$ as constants.
For our function, we compute:
- $\frac{\partial T}{\partial x}=2x-y$
- $\frac{\partial T}{\partial y}=-x$
- $\frac{\partial T}{\partial z}=2z$
Evaluating these at a given point, like (2, 1, -1), can provide crucial insights into how the temperature changes in each direction at that point.

Directional Derivative

The directional derivative extends the concept of partial derivatives to arbitrary directions. It measures the rate at which a function changes in a specific direction. To find the directional derivative, you need a direction vector. For example, in the problem, we are given the direction vector $\mathbf{\text{v}}=j-k=(0,1,-1)$.
First, normalize the direction vector to get the unit vector $\mathbf{\text{hat{v}}}=(0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$.
The directional derivative of our function $T$ in the direction of $\mathbf{\text{v}}$ at point (2, 1, -1) is the dot product of the gradient of $T$ and the unit vector $\mathbf{\text{hat{v}}}$.
The result of this dot product, in our case, is 0, indicating that the temperature does not change in the direction of $\mathbf{\text{v}}$ at the point (2, 1, -1). This is a key concept for understanding how a function behaves not just in alignment with axes but in any chosen direction.

Heat Flow

Heat flow is a physical concept referring to the transfer of thermal energy from one region to another. Mathematically, the direction of heat flow at a specific point in a scalar field like the temperature field $T$ is given by the negative gradient of the temperature at that point. This is because heat naturally flows from regions of higher temperature to regions of lower temperature.
In our problem, we found the gradient of $T$ at (2, 1, -1) to be $ablaT=(3,-2,-2)$. The direction of heat flow is therefore the negative of this gradient, i.e., $-ablaT=(-3,2,2)$.
This tells us that at the point (2, 1, -1), heat flows in the direction $-3,2,2$, illustrating how mathematical concepts can describe and predict physical phenomena such as heat transfer.