Question

Use a computer as needed to make plots of the given surfaces and the isothermal or equipotential curves. Try both 3D graphs and contour plots. If the temperature in the \((x, y)\) plane is given by \(T=x y-x,\) sketch a few isothermal curves, say for \(T=0,1,2,-1,-2 .\) Find the direction in which the temperature changes most rapidly with distance from the point \((1,1),\) and the maximum rate of change. Find the directional derivative of \(T\) at (1,1) in the direction of the vector \(3 \mathbf{i}-4 \mathbf{j} .\) Heat flows in the direction \(-\nabla T\) (perpendicular to the isothermals). Sketch a few curves along which heat would flow.

Short answer

The temperature changes most rapidly along the positive y-axis with a maximum rate of 1. The directional derivative in the direction of \(3 \mathbf{i}-4 \mathbf{j}\) is \(-\frac{4}{5}\).

Step-by-step solution

Write the Temperature Function

The temperature function is given by: \[T(x, y) = xy - x\]

Determine the Isothermal Curves

Isothermal curves are given by setting the temperature function equal to a constant value. For example, for \[ T = 0 \], we have \[ xy - x = 0 \] which simplifies to \[ x(y - 1) = 0 \]. Similar steps should be applied for \[ T = 1, 2, -1, -2 \]. Sketch these curves using plots or graphing tools.

Compute the Gradient of T

The gradient of the temperature function \[ abla T \] gives the direction in which the temperature changes most rapidly. Compute the partial derivatives: \[ \frac{\partial T}{\partial x} = y - 1 \] and \[ \frac{\partial T}{\partial y} = x \]. Therefore, \[ abla T = (y-1) \mathbf{i} + x \mathbf{j} \].

Evaluate the Gradient at (1,1)

Substitute \[ x = 1 \] and \[ y = 1 \] into the gradient: \[ abla T (1,1) = (1-1) \mathbf{i} + 1 \mathbf{j} = \mathbf{j} \]. Hence, the temperature changes most rapidly in the direction of the positive y-axis.

Find the Maximum Rate of Change

The maximum rate of change of the temperature at a point is given by the magnitude of the gradient vector at that point. We have: \[ |abla T (1,1)| = | \mathbf{j} | = 1 \]. Therefore, the maximum rate of change is 1.

Compute the Directional Derivative in the Given Direction

The directional derivative of \[ T \] at \[ (1,1) \] in the direction of a vector \[ \mathbf{v} = 3 \mathbf{i} - 4 \mathbf{j} \] is given by: \[D_{\mathbf{v}} T = abla T \cdot \frac{\mathbf{v}}{|\mathbf{v}|} \]. Compute the unit vector by normalizing \[ \mathbf{v} \]: \[ \mathbf{u} = \frac{3 \mathbf{i} - 4 \mathbf{j}}{\sqrt{3^2 + (-4)^2}} = \frac{3 \mathbf{i} - 4 \mathbf{j}}{5} \]. Then, \[ D_{\mathbf{v}} T = abla T (1,1) \cdot \mathbf{u} = 0 \mathbf{i} + 1 \mathbf{j} \cdot \frac{3 \mathbf{i} - 4 \mathbf{j}}{5} = \frac{-4}{5} = -\frac{4}{5} \].

Sketch the Heat Flow Curves

Heat flows in the direction \[ -abla T \]. At any point, this is the direction opposite to the gradient vector. For the point \[ (1,1) \], the direction of heat flow is \[ - \mathbf{j} \] (i.e., along the negative y-axis). Sketch curves combining the isothermal lines and the directions of \[ -abla T \].

Key concepts

temperature gradient

The temperature gradient is a vector that indicates how the temperature changes in space.
It points in the direction of the greatest rate of increase of temperature.
In this context, if the temperature function is given by \(T(x, y) = xy - x\), the gradient, denoted as \(abla T\), gives us crucial information.
The gradient is computed by finding the partial derivatives of \(T\) with respect to \(x\) and \(y\).
Specifically:

• \(\frac{abla T}{abla x} = y - 1\)
• \(\frac{abla T}{abla y} = x\)

Consequently, \(abla T\) is given by \((y-1)\textbf{i} + x\textbf{j}\).
This vector points in the direction of the most rapid temperature increase.
At any given point, such as (1,1), you substitute \(x\) and \(y\) into \(abla T\).
For (1,1), we find \(abla T (1,1) =\textbf{j}\), meaning the temperature increases most rapidly along the positive y-axis.

isothermal curves

Isothermal curves are contour lines connecting points where the temperature is constant.
They are incredibly useful for visualizing temperature distributions across a plane.
For the temperature function \(T(x, y) = xy - x\), isothermal curves are derived by setting \(T\) equal to various constants.
For example:

• For \(T = 0\): \(xy - x = 0 \rightarrow x(y - 1) = 0\)
• For \(T = 1\): \(xy - x = 1 \rightarrow x(y - 1) = 1\)
• For \(T = -1\): \(xy - x = -1 \rightarrow x(y - 1) = -1\)

By solving these equations, we obtain the shapes of the isothermals for each constant value.
For instance, the curve for \(T = 0\) casts \(x=0\) or \(y=1\), which means vertical and horizontal lines respectively.
Plotting these curves provides a clear map of the temperature distribution over a plane.

directional derivatives

The directional derivative of a temperature function describes how the temperature changes in a specific direction.
It is highly valuable when trying to understand variations along a certain path.
For a given point \((1,1)\), to find the directional derivative in the direction of the vector \(3 i - 4 j\), follow these steps:

• Normalize the direction vector \(\textbf{v}\): \(\textbf{u} = \frac{3 i - 4 j}{\r\textbf{v}\r} = \frac{3 i - 4 j}{5}\)
• Compute \(abla T \textbf{(1,1)}\): \((y-1) i + x j\) at \((1,1)\) gives \(\textbf{j}\)
• Dot \(abla T \textbf{(1,1)}\) with \(\textbf{u}\): \(D_{\textbf{v}} T = abla T (1,1) \bullet \textbf{u} = 0 \textbf{ i} + 1 \textbf{ j} \bullet \frac{3i - 4j}{5} = \frac{-4}{5}\)

So, the directional derivative in the direction of \(3 \textbf{i} - 4 \textbf{j}\) is \(-4/5\).
This indicates that the temperature decreases along this vector.

heat flow

Heat flow is a key concept in understanding how heat transfers through a medium.
It always moves in the direction opposite to the gradient of the temperature, denoted as \(-abla T\).
This means heat flows perpendicularly to the isothermal curves, moving towards cooler areas.
In our function \(T(x,y)=xy-x\), heat at the point \((1,1)\) flows in the direction \(- \textbf{j}\), responding to \(abla T(1,1) = (0)\textbf{i} + (1)\textbf{j}\).
This direction is along the negative y-axis.
Sketching curves that display heat flow involves drawing trajectories that cut across isothermal lines perpendicularly.
Graphically depicting these heat flow lines helps us understand how heat distributes across the plane, providing a visual means to analyze temperature changes.