1. Recall that for a real valued function from R n to R 1, such as f(x,y) or f(x,y,z), the derivative matrix can be treated as a vector called the gradient instead of as a 1  n matrix. That is, [ f x (x,y) f y (x,y) ] can be represented by either Df(x,y) or  f (x,y), and [ f x (x,y,z) f y (x,y,z) f z (x,y,z) ] can be represented by either Df(x,y,z) or  f (x,y,z). Page 163 in the text repeats the definition of a gradient. Recall that if x is a point on a line, and v is in the direction of the line, then x + tv defines If f is a real valued function of two or three variables, then as t varies, f(x + tv) gives If v is a unit vector (i.e., ||v||=1), then, [f(x + tv) – f(x)] / t gives the all points on the line. the values of f along the line. change in f over an interval per unit of measurement of t (for instance, the change in temperature per mile).

2. We can find this derivative by using the chain rule. By letting c(t) = x + tv, we have c  (t) = The directional derivative of f at x in the direction of unit vector v (definition, page 164) is lim f(x + tv) – f(x) ——————. t t0t0 v. The chain rule then tells us that d — f(c(t)) =  f (c(t)) c  (t) = dt and setting t = 0 to get this derivative at c(0) = x gives  f(x + tv) v,  f(x) v. Look at Theorem 12 on page 165 (and note that v should be identified as a unit vector in the theorem).

3. Find the directional derivative of f(x,y) = x 2 – y 2 at the point x = (2, 4) in the direction of v = (5, –12)  f (x,y) = ||v|| = v = (5, 3)||v|| = v = (0, 2) ||v|| = (2x, –2y) 13  f(x) u = (4, –8) (5/13, –12/13) =116 / 13  34  f(x) u =(4, –8) (5/  34, 3/  34) =– 4 /  34 2  f(x) u = (4, –8) (0, 1) =– 8which is f y (2,4)

4. Find a formula for the directional derivative of f(x,y,z) = xy 3 – z 2 at each point on the path c(t) = (5 sin t, 4 cos t, 3 cos t) for 0  t, in the direction of the tangent vector to the path.  f (x,y,z) = x = v = ||v|| = (y 3, 3xy 2, – 2z) c(t) = (5 sin t, 4 cos t, 3 cos t) c  (t) = (5 cos t, – 4 sin t, – 3 sin t) 5  f(x) u = (5 cos t, – 4 sin t, – 3 sin t) (64 cos 3 t, 240(sin t)(cos 2 t), – 6 cos t) ———————————— = 5 320 cos 4 t – 960(sin 2 t)(cos 2 t) + 18(cos t)(sin t) —————————————————— 5

5. Note that if  is the angle between the vectors  f(x) and a unit vector v, then  f(x) v = ||  f(x)|| ||v|| cos  = ||  f(x)|| cos . This is maximized when and is minimized when On page 166, look at Theorem 13 and the comment which immediately follows. Find the direction in which f(x,y) = 3x 2 – 2y 2 + xy increases the fastest from the point (–4, 5).  f (x,y) = f x i + f y j =  f (–4, 5) = (6x + y)i + (x – 4y)j – 19i – 24j Find the direction in which T(x,y,z) = e –x + e –2y + e 3z decreases the fastest from the point (1, 1/2, –1/3).  T (x,y,z) = f x i + f y j + f z k = –  T (1, 1/2, –1/3) = – e –x i – 2e –2y j + 3e 3z k (i + 2j – 3k) / e  = 0 (  f(x) and v point in the same direction)  =  (  f(x) and v point in opposite directions).

6. Consider the curve in R 2 defined by f(x,y) = b (which is a level curve of the function z = f(x,y)). If c(t) = (x(t), y(t)) describes a path on the curve f(x,y) = b, then f(x(t), y(t)) = f(c(t)) = b d — f(x(t), y(t)) = dt  f(c(t)) c  (t) = 0 0 Df(x,y) Dc(t) = 0 Consider z = x 2 + y 2. One level curve is x 2 + y 2 = 1. One path on the curve is c(t) = (cos t, sin t) for 0  t   /2. y x (cos t) 2 + (sin t) 2 = 1 d — [(cos t) 2 + (sin t) 2 ] = 0 dt – 2(cos t)(sin t) + 2(sin t)(cos t) = 0 – sin t 2 cos t 2 sin t = 0 cos t

7. Consider the surface in R 3 defined by f(x,y,z) = b (which is a level surface of the function w = f(x,y,z)). If c(t) = (x(t), y(t), z(t)) describes a path on the surface f(x,y,z) = b, then f(x(t), y(t), z(t)) = f(c(t)) = b d — f(x(t), y(t), z(t)) = dt  f(c(t)) c  (t) = 0 0 Df(x,y,z) Dc(t) = 0 Since  f(x 0, y 0, z 0 ) is orthogonal to each path which is on the surface w = f(x,y,z) and goes through the point (x 0, y 0, z 0 ), then  f(x 0, y 0, z 0 ) must be a normal vector to the plane tangent to the surface at the point (x 0, y 0, z 0 ). Look at Theorem 14 on page 167. The fact that  f(x 0, y 0, z 0 ) must be orthogonal to each vector with tail (x 0, y 0, z 0 ) and head any point (x,y,z) on the plane tangent to the surface f(x,y,z) = b at the point (x 0, y 0, z 0 ), leads to the method for obtaining this tangent plane stated at the bottom of page 167.

8. Find the equation of the plane tangent to the the graph of z = x 2 + y 4 + e xy at the point (1,0,2). First, we write the equation describing the graph in the form f(x,y,z) = b: x 2 + y 4 + e xy – z = 0.  f(x,y,z) = (2x + ye xy )i + (4y 3 + xe xy )j – k  f (1,0,2) = 2i + j – k The tangent plane is (2, 1, –1) (x – 1, y – 0, z – 2) = 0 2x + y – z = 0