32A WEEK 8 Exercise 1 Suppose that f x y z satisfies f P 2 4 4 at some point P a b c i Is f increasing or decreasing in the direction of the vector v 2 1 3 ii Give two non zero non parallel vectors x y such that f is neither increasing nor de creasing in the direction of x or in the direction of y iii Describe the set of vectors w such that f is neither increasing or decreasing in the direction of w at P Solution i We want the directional derivative in the direction of v To use the formula we first make v a unit vector even though this technically is not necessary since we are only interested in the sign of the derivative and not its exact value So we compute D v v f P f P v v 2 4 4 2 1 3 14 12 14 0 So f is increasing in this direction ii Such vectors x will have to satisfy Dxf P 0 Again we won t worry ourselves about making sure x is a unit vector for this exercise since if we look at the formula we want 0 f P x x which is true if and only if f P x 0 assuming x 0 of course Thus we want to find vectors such that f P x 2 4 4 x1 x2 x3 0 Two such solutions that are indeed not parallel are given by 2 1 0 and 0 1 1 iii Using the same reasoning as in part ii with more suggestive notation we want to find those vectors w x y z such that f P w 2 4 4 x y z 0 2x 4y 4z 0 so this set of vectors forms a plane through the origin with normal vector 2 4 4 f P Exercise 2 Let f x y x2e xy and P 1 2 i Compute f P ii Is f increasing or decreasing in the direction of a unit vector making an angle of 6 radians with f P How about 4 radians 19 37 radians iii Can you give a general description of the behavior of f in the direction of some unit vector u at P in terms of the angle made between u and f P It may help to draw the graph of Duf P as a function of Date November 2022 1 2 Solution i f x y 2xe xy x2 ye xy x2 xe xy 32A WEEK 8 cid 123 cid 122 cid 124 cid 123 cid 122 chain rule cid 125 cid 125 product rule cid 124 Hence f 1 2 2e 2 2 e 2 e 2 0 e 2 ii iii In general if u is a unit vector making an angle with the gradient vector f P the directional derivative in the direction of u at P is f P u f P u cos f P cos since u 1 Consider the function g f P cos e 2 cos As varies from 0 to 2 g gives precisely the value of the directional derivative of f at P in the direction of a unit vector making an angle with the gradient Figure 1 Graph of g with 0 2 scaled by a factor of 10 for visibility 32A WEEK 8 3 Because cos x 0 for 0 x 2 and 3 2 x 2 and cos x 0 for 2 x 3 2 we can figure out whether or not the directional derivative is positive just by plugging in the correct value of into g This tells us that f is increasing if 4 and 6 but that f will be decreasing if 19 37 since 19 37 19 38 1 2 and also 19 37 1 so 2 19 37 Exercise 3 Suppose f x y 9 3x2 4y2 i Find a parameterization for the xy projection r t of the level curve at z 4 dt f r t two ways first using the fact that r t parameterizes a level curve ii Compute d of f and then by using the chain rule iii What does part ii imply about the angle between r t and f r t iv Bonus Suppose that f r t is a continuous vector valued function i e no abrupt jumpts between the vectors f gives as output Using part iii only can you find a f r t f r t at t varies from 0 to 2 Your parameterization for the curve traced out by answer may be off by a factor of 1 Compute f r t directly to check i Fixing z f x y 4 we get the equation 3x2 4y2 5 Thus the xy projection of Solution cid 33 2 the level curve is the ellipse cid 32 x cid 112 5 3 cid 42 cid 114 5 which has a parameterization 1 cid 33 2 cid 32 cid 114 5 4 y cid 112 5 4 cid 43 r t cos t sin t 3 0 t 2 ii iii Because r t parameterizes the xy projection of the level curve z 4 by design f r t 9 5 4 for all 0 t 2 Therefore f r t is a constant function of one variable and so d dt f r t f r t r t so that dt 4 0 On the other hand by the chain rule d dt f r t d 0 f r t r t This means f r t is perpendicular orthogonal to r t at each t This is generally true and is usually surmised by the observation that the gradient is always perpendicular to level curves iv Part iii tells us that f r t is a vector with two components that is always perpen dicular to r t By sketching the ellipse that r t parameterizes say you can see that this means that f r t will always point perpendicularly inward towards the interior of the ellipse or outward Since cid 42 cid 114 5 3 cid 43 cid 114 5 4 r t sin t cos t cid 43 cid 114 5 3 4 32A WEEK 8 a vector that will be perpendicular to r t for each t is given by the vector valued function cid 42 cid 114 5 4 cid 114 5 3 s t cos t sin t f r t since as you can verify r t s t 0 for all t Now for each t f r t has length 1 by design so we see that one possible parameterization of the curve traced out by this vector is cid 43 cid 112 5 4 cos2 t 5 3 sin2 t 1 s t s t cos t sin t 0 t 2 cid 42 cid 114 5 4 The assumption that f r t is continuous i e no sudden jumps is necessary to ensure f r t f r t doesn t abruptly jump from pointing in one direction 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