Directional SlopeThe equation of a linear function f(x, y) can be written asf(x, y) = c + mx + ny,where c, m, and n are constants.Significance of m and n Consider two points, (x, y) and (x+∆x, y +∆y). Let ∆f = f(x+ ∆x, y +∆y) − f(x, y). We may then calculate the change in f as∆f = [c + m(x + ∆x) + n(y + ∆y)] − [c + mx + ny] = m∆x + n∆y.This formula shows us the significance of the parameters m and n. Suppose we consider two pointshaving the same y coordinate. Then ∆y = 0. We can rearrange the last formula to getm =∆f∆x.Thus, m is the slope of the function in the x-direction. Similarly, n is the slope of the functionin they-direction. Another way to state this is that m is the slope of the function in the direction given bythe unit vector~i and n is the slope of the function in the direction given by the unit vector~j.Slope in the direction of a unit vector ~u Consider a specific example. We are interested inknowing the slope at the point (2,1) in the direction of the point (6,4).½½½½½½½½½½½½½½>>(2,1)(6,4)543~usNote that we have two different labels on the ray that begins at (2,1) and passes through the point(6,4). There is a coordinate label s and a unit vector label ~u. This is analogous to the standardcoordinate plane, in which the axes can be labeled with the coordinates x and y and also the unitvectors~i and~j.The slope of f in the horizontal direction can be thought of as the rate of change of f with respectto x. Similarly, the slope of f along the path from (2,1) to (6,4) can be thought of as the rate ofchange of f with respect to s, where s is a coordinate that goes in the given direction. The slope inthe s direction is ∆f /∆s. Here, the change in f is m∆x + n∆y = 4m + 3n. The change in s is thedistance between the points, or 5. So the slope isslope =∆f∆s=4m + 3n5= 0.8m + 0.6n.The slope of f in the horizontal direction can also be thought of as the slop e of f in the~i direction.Similarly, the slope of f along the path from (2,1) to (6,4) can be thought of as the slope of f in the~u direction. The unit vector ~u is obtained by dividing the displacement vector 4~i + 3~j by its length 5.Thus,~u = 0.8~i + 0.6~j.Now we define the gradient vector, which we will call g for now, as~g = m~i + n~j.The components of the gradient vector are the slopes in the corresponding directions. Is there a wayto calculate the slope of f in the ~u direction using the gradient and direction vectors? Clearly theanswer is that the slope is the dot product of the two vectors:slope = ~g · ~u.SummaryThere are two ways to think of the slope of a linear function in a given direction.• We can use a symbol, such as s, to represent distance in the given direction. Then we computethe slope in the given direction as the change in f divided by the change in s.slope =∆f∆s.• We can describe the direction using the unit vector ~u that points in the given direction. Wecan define the gradient vector to be the vector in which each component is the slope in thecorresponding direction; thus, the~i component is the slope in the~i direction and so on. Thenwe compute the slope in the given direction as the dot pro duct of the gradient and directionvectors.slope = ~g · ~u.In 1-variable calculus, we use the concept of the slope of a line to define the derivative, whichrepresents the slope of a curve. Similarly, in multi-variable calculus, we will use the concept of theslope of a plane to define the directional derivative, which represents the slope of a surface. Thematerial here on slope of a linear function thus provides the precalculus foundation for the derivativein multi-dimensional calculus.Exercises1. Let f = 2 + 3x + 2y. Use both definitions of the slope to calculate the slope of f in the directionof the vector 2~i −~j.2. Let f = 5 − x + 2y. Use both definitions of the slope to calculate the slope of f in the directionof the vector~i + 3~j.3. Let f = 3 − 2x − 4y. Use both definitions of the slope to calculate the slope of f in the directionof the vector −2~i − 3~j.4. Let f be defined by the contour diagram in Figure 11.73. Determine the slope of f in thedirection indicated by the ray from the point (-1,-1) through the point (1,2).01/23/2002 Glenn