1 / 8

all points on the line.

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,

jaimin
Download Presentation

all points on the line.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related