1 / 11

f ( x,y ) = x 2 y + y 3 + 2 y

Definition of partial derivative (page 128-129). f ( x,y ) = x 2 y + y 3 + 2 y. ¶ f — = f x = ¶ x. ¶ f — = f y = ¶ y. 2 xy. x 2 + 3 y 2 + 2 y ln2. z = f ( x,y ) = cos( xy ) + x cos 2 y – 3. ¶ z — ( x,y ) = ¶ x. – y sin( xy ) + cos 2 y.

snana
Download Presentation

f ( x,y ) = x 2 y + y 3 + 2 y

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. Definition of partial derivative (page 128-129) f(x,y) = x2y + y3+ 2y ¶f — = fx = ¶x ¶f — = fy = ¶y 2xy x2 + 3y2 + 2yln2 z = f(x,y) = cos(xy) + x cos2y– 3 ¶z —(x,y) = ¶x – y sin(xy) + cos2y – x sin xy – 2x cos y sin y fy(x,y) = ¶z —(x0,y0) = ¶x – y0sin(x0y0) + cos2y0 fy(2,p/2) = – 2sin p – 4 cos (p/2) sin (/2) = 0 y ——— 3(xy)2/3 x ——— 3(xy)2/3 z = f(x,y) = (xy)1/3fx = fy =

  2. Compare the functions y = x2 and y = x2/3 in R2, and decide whether or not each function is differentiable at x = 0. y y Either by looking at graphs or from the definition of derivative, we see that y = x2 is differentiable at x = 0 but y = x2/3 is not. y = x2 y = x2/3 x x Now consider the function z = f(x,y) = (xy)1/3 in R3. What does f(x,y) look like in the xz plane where y = 0? z = 0 z What does f(x,y) look like in the yz plane where x = 0? z = 0 y What does f(x,y) look like in the plane where x = y? z = t2/3 where t represents a variable which gives the value of x = y. x

  3. Now consider the function z = f(x,y) = (xy)1/3 in R3. What does f(x,y) look like in the xz plane where y = 0? z = 0 z What does f(x,y) look like in the yz plane where x = 0? z = 0 y What does f(x,y) look like in the plane where x = y? z = t2/3 where t represents a variable which gives the value of x = y. x We find that for functions in R3, it is possible for a “level curve” in one direction to be differentiable at a point while a “level curve” in another direction is not differentiable at the same point. A function in R2 is differentiable at a point, if and only if there exists a line tangent to the function at the point. A function in R3 is differentiable at a point, if and only if there exists a plane tangent to the function at the point.

  4. In order that a function f(x,y) be differentiable at a point (x0 , y0), the function must be “smooth” in all directions, not just in the x-direction and y-direction. If a function f(x,y) is differentiable at point (x0 , y0), then we can use the value of each partial derivative at the point to find the equation of the plane tangent to the function at the point: The equation of the tangent plane can be written as g(x,y) = ax + by + c. The following must be true: g(x0 , y0) = ax0 + by0 + c = f(x0 , y0) , gx(x0 , y0) = a = fx(x0 , y0) , gy(x0 , y0) = b = fy(x0 , y0) . The equation of the tangent plane is g(x,y) = fx(x0 , y0) x + fy(x0 , y0) y + f(x0 , y0) – fx(x0 , y0) x0–fy(x0 , y0) y0 which can be written z = f(x0 , y0) + fx(x0 , y0) (x–x0) +fy(x0 , y0) (y – y0) .

  5. Definition of differentiable and tangent plane in R3 (page 133) See the derivation on page 132. Example Find the plane tangent to the graph of z = x2 + y3– cos(xy) at the point (8 , –4 , –1) . f(8 , –4) = –1 ¶f — = fx = ¶x ¶f — = fy = ¶y 2x + y sin(xy) 3y2+ x sin(xy) fx(8 , –4) = 16 fy(8 , –4) = 48 The equation of the tangent plane is z = –1 + 16(x– 8) + 48(y + 4) which can be written 16x + 48y – z = –63 .

  6. Look again at the definition of differentiable in R3 on page 133, and observe that we can say f(x,y) is differentiable at (x0 , y0) if This is the difference between the exact function value at point (x , y) and the approximated value from the plane tangent to the function at (x0 , y0). (x–x0) (y – y0) f(x,y) –f(x0 , y0) –fx(x0 , y0) fy(x0 , y0) lim ————————————————————— = 0 (x , y)(xo , y0) ||(x , y) – (xo , y0)|| If x = (x1 , x2 , …, xn) is a vector in Rn, and f(x) is a function from Rn to R, then we define f(x) to differentiable at x0 if f(x) –f(x0) – Df(x0) (x – x0) lim ——————————— = 0 where xxo || x – x0 || Df(x0) is the 1n matrix of partial derivatives at x0 , and (x – x0) is the n1 matrix consisting of the differences between each variable and its specific value at x0 . (NOTE: A 1n matrix times an n1 matrix is the same as the dot product of two vectors.

  7. Suppose x = (x1 , x2 , …, xn) is a vector in Rn, and f(x) is a function from Rn to Rm. Then, we can write f(x) = [f1(x) , f2(x) , … , fm(x)] . We let Df(x) represent the mn matrix with row i consisting of the following partial derivatives: fififi —— —— … —— . x1x2xn We call Df(x) the derivative (matrix) of f , and of course Df(x0) is the derivative (matrix) of f at x0 . Look at the general definition of differentiable on page 134.

  8. Example Find the derivative matrix for w = f(x,y,z) = ( x2+ xy4eyz , ln(xz+y) ) . _ _ | | | | | | | | Df(x,y,z) = | | | | | | | | | | |_ _| 2x + y4eyz 4xy3eyz + xy4zeyz xy5eyz z / (xz+y) 1 / (xz+y) x / (xz+y) Find the linear approximation at the point x0 = (1 , e , 0). x – 1 y – e = z – 0 w = f (x0) + Df (x0) (x – x0) = f (1 , e , 0) + Df (1 , e , 0)

  9. 2 + e4 4e3e5 + 0 1 / e 1 / e x – 1 y – e = z – 0 1 + e4 1 1 + e4 1 (2 + e4)x– 2 – e4 + 4e3y – 4e4 + e5z y/ e– 1 + z / e + = (2 + e4)x+ 4e3y + e5z – 1 – 4e4 (y + z)/ e Observe that since w = f(x,y,z) goes from R3 to R2, this linear approximation is really two linear approximations, one for each of the component functions. That is,

  10. Observe that since w = f(x,y,z) goes from R3 to R2, this linear approximation is really two linear approximations, one for each of the component functions. That is, The linear approximation for f1(x,y,z) = x2+ xy4eyz is w1 = (2 + e4)x+ 4e3y + e5z– (1 +4e4) . The linear approximation for f2(x,y,z) = ln(xz+y) is w2 = (1/e)y + (1/e)z .

  11. Recall that for a function f from Rn to R1, the derivative matrix is 1n. In certain situations, it will be convenient to treat this 1n matrix as a vector. See the definition of a gradient on page 136. Look at Theorem 8 on page 137. Look at Theorem 9 on page 137. Look at the chart at the top of page 138.

More Related