From (25), we note that the directional derivative has its maximum positive value when 9 : 0, 2'. 3., when points in the same direction as V f. That means that f increases maximally in the direction of its gradient. In other words, the gradient points from low values toward high values. From (25), we also see that the directional derivative is zero when 6 = 90, since cos 90 = 0. Thus, the gradient is perpendicular to an isoline in two dimensions or an isosurface in three dimensions. An isoline in a plane is a \"line\" (curve) along which a variable (pressure, temperature, etc.) has a constant value. Examples are isobars (lines of constant pressure) and isotherms (lines of constant temperature). In three dimensions, an isosurface is a surface on which a variable is constant. For example, we saw above that the vertical pressure gradient is much larger than the horizontal pressure gradient, so the three-dimensional pressure gradient vector points nearly straight down from lower pressure aloft to greater pressure below, perpendicular to a pressure surface. Question 4. Study equations (13)(23). Then put these notes aside and write the denition of the gradient and the directional derivative. Question 5. Study (25) and the paragraphs after it. Then put these notes aside and write the arguments that justify the following statements: (i) The gradient points from low values toward high values. (ii) The gradient is perpendicular to the isoline (or isosurface) that passes through the local point. (iii) The direction opposite to the gradient is the direction in which the function decreases most rapidly. Question 6. Given (23) and (16), evaluate g for: (i) e : i and (ii) 33 : 7k. are3 Computing the Gradient (We will do this in separate assignments.) If we have values for a variable such as pressure, p, on a (latitude, longitude) grid, we can approx- imate the horizontal pressure gradient using centered finite differences: 3p. 3p.~paipw. \"195' 1+ ~ 27 8:: 8y 215$ 2Ay '1 ( ) where E, W, N, and S stand for the grid points that are east, west, north, and south of the point at which we are approximating the gradient. If we have a weather map on which contour lines are drawn, we can estimate the gradient in a different way. Recall the nite difference formula for the directional derivative 5f ~Af where A f = f (r + Ar) f (r). The easiest way to evaluate A f is to take the diiference between the values on two isolines so that we know exactly what Af is. Most of the time, these will be neighboring isolines so that A f equals the contour increment. If the isolines are quite close, we may chose to measure across twice the contour increment so that we can measure a larger value of Ar more accurately. Knowing that the gradient is perpendicular to the isoline, we measure the distance from the starting to the ending isoline at right angles to the isolines. The magnitude of the gradient is |A f / AH, and the direction of V f points perpendicularly across the local isoline (fl 0')