all points on the line.

1. Recall that for a real valued function from Rn to R1, 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, [ fx(x,y) fy(x,y) ] can be represented by either Df(x,y) or f (x,y), and [ fx(x,y,z) fy(x,y,z) fz(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 all points on the line. If f is a real valued function of two or three variables, then as t varies, f(x + tv) gives the values of f along the line. If v is a unit vector (i.e., ||v||=1), then, [f(x + tv) – f(x)] / t gives the change in f over an interval per unit of measurement of t (for instance, the change in temperature per mile).

2. 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0 We can find this derivative by using the chain rule. By letting c(t) = x + tv , we have c(t) = v . The chain rule then tells us that d — f(c(t)) = f (c(t)) •c(t) = dt f(x + tv) •v , and setting t = 0 to get this derivative at c(0) = x gives 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) = x2 – y2 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) = – 8 which is fy(2,4)

4. Find a formula for the directional derivative of f(x,y,z) = xy3 – z2 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|| = (y3 , 3xy2 , – 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 cos3t , 240(sin t)(cos2t) , – 6 cos t) • ———————————— = 5 320 cos4t– 960(sin2t)(cos2t) + 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.  = 0 (f(x) and v point in the same direction)  =  (f(x) and v point in opposite directions). Find the direction in which f(x,y) =3x2 – 2y2 + xy increases the fastest from the point (–4 , 5). f (x,y) = fx i + fyj = f (–4 , 5) = (6x + y)i + (x– 4y)j – 19i– 24j Find the direction in which T(x,y,z) = e–x + e–2y + e3z decreases the fastest from the point (1 , 1/2 , –1/3). T (x,y,z) = fx i + fyj + fzk = –T (1 , 1/2 , –1/3) = – e–xi – 2e–2yj + 3e3zk (i + 2j – 3k) / e

6. Consider the curve in R2 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)) = Consider z = x2 + y2. One level curve is x2 + y2 = 1. One path on the curve is c(t) = (cos t , sin t) for 0  t  /2. b (cos t)2 + (sin t)2 = 1 d — f(x(t) , y(t)) = dt d — [(cos t)2 + (sin t)2] = 0 dt 0 – 2(cos t)(sin t) + 2(sin t)(cos t) = 0 Df(x,y) Dc(t) = 0 – sin t 2 cos t 2 sin t = 0 cos t f(c(t)) •c (t) = 0 y x

7. Consider the surface in R3 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 Since f(x0 , y0 , z0) is orthogonal to each path which is on the surface w = f(x,y,z) and goes through the point (x0 , y0 , z0), then 0 Df(x,y,z) Dc(t) = 0 f(c(t)) •c (t) = 0 f(x0 , y0 , z0) must be a normal vector to the plane tangent to the surface at the point (x0 , y0 , z0). Look at Theorem 14 on page 167. The fact that f(x0 , y0 , z0) must be orthogonal to each vector with tail (x0 , y0 , z0) and head any point (x,y,z) on the plane tangent to the surface f(x,y,z) = b at the point (x0 , y0 , z0), 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 = x2 + y4 + exy at the point (1,0,2). First, we write the equation describing the graph in the form f(x,y,z) = b: x2 + y4 + exy – z = 0 . f(x,y,z) = (2x + yexy)i + (4y3 + xexy)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