Temperature and Terrain Direction Problems

Reading a Gradient as Local Field Information

A temperature map or elevation map gives values at different locations, but a gradient gives more specific local information: it describes how the field changes as you move from a particular point. With that information, you can compare different headings, choose a direction of fastest increase or decrease, or find a heading along which the field has no instantaneous change.

The same calculation has different physical interpretations depending on the field. For temperature, a rate might be measured in degrees Celsius per kilometre. For terrain elevation, it might be metres of elevation change per kilometre travelled. In each case, the gradient points in the direction of greatest instantaneous increase, and its magnitude is the greatest rate per unit distance.

Headings and Directional Rates

Use coordinates in which the first component points east and the second points north. A heading is represented by a unit vector \(\mathbf{u}=(u_1,u_2)\), where \(u_1^2+u_2^2=1\). The unit length matters: the rate is measured per unit distance travelled, not per unit of an arbitrarily scaled vector.

For a compass bearing measured clockwise from north, the east–north components of the unit heading at bearing \(\theta\) are \((\sin\theta,\cos\theta)\). Thus a bearing of \(0^\circ\) points north, \(90^\circ\) points east, and \(180^\circ\) points south. If a problem gives a heading directly as a vector, first check that it has unit length; if not, divide it by its magnitude.

The dot product makes the rate calculation direct. If \(\nabla f(P)=(a,b)\), then the rate along \(\mathbf{u}=(u_1,u_2)\) is \(au_1+bu_2\). A positive result means the field increases in that direction, a negative result means it decreases, and zero means there is no instantaneous change along that heading.

Why the Dot-Product Formula Works

The formula comes from following the field along a straight path through \(P\). For a unit heading \(\mathbf{u}\), write that path as \(\mathbf{r}(t)=P+t\mathbf{u}\). Because \(\mathbf{u}\) has length one, changing \(t\) by a small amount corresponds to travelling that same amount of distance along the path. The chain rule gives the rate of the field along this path:

At \(t=0\), the path is at \(P\), and \(\mathbf{r}'(t)=\mathbf{u}\). Therefore the instantaneous rate there is \(\nabla f(P)\cdot\mathbf{u}\), as claimed. This also explains why the heading must have unit length when the result is to be interpreted as a rate per unit distance.

Worked Example: Comparing Temperature Changes Along Headings

Fastest Increase and Decrease

For a nonzero gradient, the direction of fastest increase is the unit vector pointing along the gradient. The direction of fastest decrease is the opposite unit vector. The corresponding rates are the magnitude of the gradient and its negative. This conclusion follows from the dot-product inequality: for unit \(\mathbf{u}\), \(\nabla f(P)\cdot\mathbf{u}\leq \lVert\nabla f(P)\rVert\), with equality when \(\mathbf{u}\) points along the gradient.

Worked Example: Choosing a Direction on a Slope

Headings with Zero Instantaneous Change

A heading has zero instantaneous change exactly when it is perpendicular to the gradient, because the dot product of perpendicular vectors is zero. In two dimensions, if the gradient is nonzero, there are two opposite unit headings perpendicular to it. These directions are tangent to the local level curve of the field. Zero instantaneous change is a local statement: it does not say that the field remains constant over an entire route.

Worked Example: Finding Temperature-Neutral Headings

A Common Pitfall: Forgetting to Normalize

A direction vector can point the right way but still fail to represent a unit heading. For example, using \((3,4)\) directly with the gradient \((3,4)\) gives the dot product \(3(3)+4(4)=25\). That is not the rate per unit distance: the vector \((3,4)\) has length \(5\), so this calculation scales the rate by \(5\).

Normalize the vector first: \((3,4)/5=(3/5,4/5)\). The correct rate is then

This agrees with the gradient magnitude, \(\sqrt{3^2+4^2}=5\), because the normalized heading points in the direction of steepest increase. Always check the length of a heading vector before interpreting its dot product as a rate per unit distance.

Using Rates Responsibly

Gradient data describe change at a particular point, not necessarily over a long journey. A rate of \(2\) °C/km means that temperature initially changes by about \(2\) °C for each kilometre travelled in that direction near the point. If the heading or the gradient changes farther along the route, the same rate need not continue to apply.

In practical comparisons, retain the units and the sign. A positive elevation rate indicates uphill travel; a negative one indicates downhill travel. A positive temperature rate indicates warming along the heading, and a negative one indicates cooling. When comparing headings for fastest increase or decrease, compare the rates themselves, not just the east or north components separately.