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HARVARD MATH 21A - Directional Derivatives

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Math 21a Directional Derivatives Spring, 20091If f(x, y) = 3x + 7y2and u =35,45, find the directional derivative Duf at (1, 1).2If f(x, y) = sin(xy), find the directional derivative Duf where u =D1√2,1√2E.3You are skiing on a mountain which happens to be the graph of the function f(x, y) =10 − x2− y4. You are at the point (1, 1, 8). If you want to ski down the steepest path, whatdirection should you head?4A fly is flying around a room in which the temperature is given by T (x, y, z) = x2+ y4+ 2z2.The fly is at the point (1, 1, 1) and realizes that he’s cold. In what direction should he fly towarm up most quickly?5Alice the “A” student is debating Chuck the “C” student. Alice says that the direction ofgreatest ascent on the graph z = f (x, y) is in the direction ∇f . Chuck says that instead weshould look at the level surface F (x, y, z) = z − f(x, y) = 0 and go in direction grad F . Whois right? How would you explain this to the student that is wrong?6A racehorse lives in a valley which happens to be the graph of f (x, y) = 3x2+ y2. He isdoomed to wander his racetrack, which is the set of points in the valley where x2+ y2= 1.The racehorse secretly wishes to be a mountain climber, and his fantasy is to escape from hisracetrack and take the steepest path up the mountain. At what point should he make hisescape, and in what direction should he run? (There are actually two possible answers.)Directional Derivatives – Solutions1We use the formula Duf = ∇f ·u. Here ∇f = h3, 14yi, so ∇f (1, 1) ·u = h3, 14i·35,45= 13.2Now ∇f = hy cos(xy), x cos(xy)i, so ∇f · u =x+y√2cos(xy).3Since the problem asks only for a direction, we should give our answer as a unit vector u.We are looking for the unit vector u for which Duf(1, 1) is the most negative. We know thatDuf(1, 1) = ∇f(1, 1) ·u. Let’s first calculate ∇f : it is h−2x, −4y3i, so ∇f (1, 1) = h−2, −4i.Therefore, we want to find the unit vector for which h−2, −4i · u is the most negative.Remember that the dot product of two vectors is given by the formula v · w = |v||w|cos θ.If θ is the angle between h−2, −4i and u, then |h−2, −4i · u| = |h−2, −4i||u|cos θ. Since u isa unit vector, |u| = 1, so we really want to make cos θ as negative as possible. This means weshould take θ = π, so we want to pick the unit vector which goes in the direction opposite ofh−2, −4i. That is, we want a vector which goes in the direction of h2, 4i but which has length1. To get such a vector, we just divide h2, 4i by its length, which is 2√5. So, our answer isD1√5,2√5E.4We want to find the unit vector u which maximizes DuT (1, 1, 1). Using the same idea as inthe last problem, this should be the unit vector in the direction of ∇T (1, 1, 1).Since ∇T = h2x, 4y3, 4zi, we want the unit vector in the direction of h2, 4, 4i, which is13,23,23.5Alice is correct: the direction of greatest ascent on the surface z = f(x, y) is in the direction∇f.What is the problem with Chuck’s answer? He’s looking at the level surface F (x, y, z) = 0,where F (x, y, z) = z −f(x, y). This is the same surface that Alice is looking at. But when hesuggests moving in the ∇F direction, this means moving a direction in space that increasesthe value of F (x, y, z) as quickly as possible. In particular, this means moving off the levelsurface F = 0. This is a different problem altogether.6What the problem is really asking us to do is find the points (x, y) and unit vectors u forwhich the directional derivative Duf(x, y) is biggest, although we’re only allowed to look atpoints where x2+ y2= 1.Let’s first focus on a specific point (x, y) and figure out the biggest the directional derivativeDuf(x, y) could be at that point. We know that the directional derivative is largest in thedirection of the gradient. The unit vector in the direction of the gradient is∇f|∇f|, and thedirectional derivative in this direction is ∇f ·∇f|∇f|. Since ∇f ·∇f is just |∇f|2, the directionalderivative in this direction is really |∇f |. Thus, we could restate the problem as: find thepoints (x, y) for which |∇f(x, y)| is largest.Let’s now calculate the gradient: ∇f (x, y) = h6x, 2yi, so |∇f (x, y)| =p36x2+ 4y2.Thus, we can again restate the problem as: maximize 36x2+ 4y2subject to the constraintthat x2+ y2= 1. Next week, we will learn a method for doing this called the method ofLagrange multipliers. However, we can figure this problem out without Lagrange multipliers.We can write 36x2+ 4y2as 32x2+ 4(x2+ y2); when x2+ y2= 1, this is just equal to 32x2+ 4.So, we’re really trying to maximize 32x2+ 4. We do this by making x2as large as possible.Since (x, y) has to stay on the circle x2+ y2= 1, this means we want x = ±1, which makesy = 0.The unit vectors u are supposed to go in the direction of ∇f = h6x, 2yi. In the case of thepoint (1, 0), this means u = h1, 0i; in the case of the point (−1, 0), this means u = h−1, 0i.So our answer is: the point (1, 0) with the unit vector h1, 0i and the point (−1, 0) with theunit vector h−1,


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HARVARD MATH 21A - Directional Derivatives

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