Chapter 14 Partial myotsana/118BSAC_Lecture.pdfآ  14.3 Partial Derivatives Let z = f(x,y). The...

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Transcript of Chapter 14 Partial myotsana/118BSAC_Lecture.pdfآ  14.3 Partial Derivatives Let z = f(x,y). The...

  • Chapter 14 Partial Derivatives

    14.1 Functions of Several Variables

    A real-valued function of two variables, f , is a rule in terms of x and y, denoted

    by

    z = f(x, y)

    where x and y are real numbers.

    Ex 1. Let f be the function defined by

    f(x, y) = 4xy + 6 sin x− 2y.

    Compute f(0, 1) and f(π, 2).

    Ex 2. Let g be the function defined by

    g(x, y) = ex − 1 x− y .

    Compute g(1, 0), g(1, e) and g(e, 1).

    1

  • The domain of a function f is the set of points (x, y) such that z = f(x, y) is a real

    number. Unless specified, the domain of a function f will be taken to be the largest

    possible set for which the rule defining f is meaningful.

    Ex 3. Find and sketch the domain of each of the following functions.

    1. f(x, y) = 4xy + 6 sin x− 2y

    2. f(x, y) = ex − 1 x− y

    3. f(x, y) = √

    1− x2 − y2

    4. f(x, y) = ln(x2 − y)

    2

  • 14.2 Limits and Continuity

    Ex 4. [Plugging in] Evaluate

    1. lim (x,y)→(1,1)

    3x 2+y2 cos(

    π

    2x+ y )

    2. lim (x,y)→(2,−1)

    ln

    (

    1 + x+ 2y

    3y2 − x

    )

    3. lim (x,y)→(−1,1)

    arctan xy

    2x2 + y2

    3

  • Ex 5. [Factoring] Evaluate

    1. lim (x,y)→(2,1)

    x2 − 4y2 x2 − 2xy

    2. lim (x,y)→(1,−1)

    x+ y

    x− √

    xy + 2y2

    4

  • 3. lim (x,y)→(−1,0)

    arctan y2 − xy y √ 3 + xy

    4. lim (x,y)→(1,2)

    2x3 + 6xy − x2y − 3y2 2x4 + 2xy2 − x3y − y3

    5

  • Limits along curves

    Ex 6. Compute the following limits.

    1. lim (x,y)→(0,0) on y = x

    x2

    x2 + y2

    2. lim (x,y)→(0,0) on x = y2

    2x2y

    x+ y2

    3. lim (x,y)→(0,0) on y = −x

    x2y3

    x6 + y6

    6

  • Ex 7. Let f(x, y) = xy

    x2 + y2 , (x, y) 6= (0, 0).

    Find the limit of f(x, y) when (x, y) → (0, 0) on the following curves.

    1. C1 : X-axis

    2. C2 : y = x

    3. C3 : y = 2x

    7

  • Consider the point (x0, y0). If C1 and C2 are two curves passing through (x0, y0) and

    lim (x,y)→(x0,y0)

    on C1

    f(x, y) 6= lim (x,y)→(x0,y0)

    on C2

    f(x, y)

    then lim (x,y)→(x0,y0)

    f(x, y) does not exist.

    Ex 8. Show that the following limits do not exists.

    1. lim (x,y)→(0,0)

    x

    x2 + y2

    8

  • 2. lim (x,y)→(0,0)

    x4 + y2

    x4 − y2

    3. lim (x,y)→(0,0)

    xy

    x2 + y2

    9

  • 4. lim (x,y)→(0,0)

    x2y

    x4 + y2

    5. lim (x,y)→(0,0)

    sin(x2)

    x2 + y2

    10

  • We say that z = f(x, y) is continuous at the points (x0, y0) if

    lim (x,y)→(x0,y0)

    f(x, y) = f(x0, y0).

    Ex 9. Determine the set of points at which the function is continuous.

    1. f(x, y) = y

    x2 + y2

    2. f(x, y) = ln(x2 − y + 4)

    Do Exercises. 14.1 9–10 (a) and (b), 11, 13–19 (odd)

    Do Exercises. 14.2 5–17, 29–33 (odd)

    11

  • 14.3 Partial Derivatives

    Let z = f(x, y).

    The partial derivative of f with respect to x is

    ∂f

    ∂x = lim

    h→0

    f(x+ h, y)− f(x, y) h

    if limit exists

    and the partial derivative of f with respect to y is

    ∂f

    ∂y = lim

    k→0

    f(x, y + k)− f(x, y) k

    if limit exists

    Note that ∂f ∂x

    is the derivative of f with respect to x when y is a constant. Similarly, ∂f

    ∂y is the derivative of f with respect to y when x is a constant.

    Thus we could apply the derivative formulas for one variable in CAL I for finding the

    partial derivatives of f .

    We write ∂f

    ∂x = fx = f1 = D1f

    and ∂f

    ∂y = fy = f2 = D2f.

    E.g. f(x, y) = y2 sin x f(x, y) = √ x ln y

    12

  • Ex 10. Find the partial derivatives of

    1. f(x, y) = x3 √ y + ey + arctan x

    2. f(x, y) = e(x sin y)

    13

  • 3. f(x, y) = ( 3 √ x+ y3)4

    4. f(x, y) = x2 tan(x+ 2y)

    14

  • 5. f(x, y) = x+ ln y √ y

    6. f(x, y) = ysinx

    15

  • 7. f(x, y, z) = z(sec y2)(ln x) + (cos z)y

    16

  • Higher-Order Partial Derivatives

    fxx = ∂2z

    ∂x2

    fx = ∂z

    ∂x

    ✉✉✉✉✉✉✉✉✉

    ■■ ■■

    ■■ ■■

    fxy = ∂2z

    ∂y∂x

    z = f(x, y)

    ✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟

    ✻✻ ✻✻

    ✻✻ ✻✻

    ✻✻ ✻✻

    ✻✻ ✻✻

    ✻✻ ✻✻

    fyx = ∂2z

    ∂x∂y

    fy = ∂z

    ∂y

    ✈✈✈✈✈✈✈✈✈

    ❍❍ ❍❍

    ❍❍ ❍❍

    fyy = ∂2z

    ∂y2

    The last four vertices of the above diagrams are called the second order partial deriva-

    tives of f .

    Ex 11. Find the second order partial derivatives of f .

    1. f(x, y) = x2y3 + √ x ln y

    17

  • 2. f(x, y) = x arctan y − y4e2x

    3. f(x, y) = ln(3x+ y2)

    18

  • Ex 12. Let f(x, y) = (sin y)(tan x) + (x2 + 3)y. Find fxy and fyx.

    Do Exercises. 14.3 15–27 (odd), 31–37 (odd), 41, 43, 51–57 (odd), 63–69 (odd)

    19

  • Interpretations of Partial Derivatives

    Partial derivatives can also be interpreted as rates of change. If z = f(x, y), then ∂f

    ∂x = the rate of change of z with respect to x when y is fixed, and

    ∂f

    ∂y = the rate of change of z with respect to y when x is fixed.

    Ex 13. Let z = f(x, y) = 4− x2 − 2y2. At the point (1, 1, 1), we have

    ∂z

    ∂x = −2x and ∂z

    ∂x

    (1,1)

    = −2

    is the slope of the tangent line to the curve C1 which is the intersection of the plane

    y = 1 and z = f(x, y) as shown.

    Similarly, ∂z

    ∂y = −4y ∂z

    ∂x

    (1,1)

    = −4

    is the slope of the tangent line to the curve C2 which is the intersection of the plane

    x = 1 and z = f(x, y) as shown.

    20

  • In general, if we have a surface z = f(x, y), at the point (a, b) we have

    fx(a, b) = the slope of the tangent line to the curve which is the intersection of the

    plane y = b and z = f(x, y) and

    fy(a, b) = the slope of the tangent line to the curve which is the intersection of the

    plane x = a and z = f(x, y)

    Ex 14. Find the slope of the line tangent to the curve in which

    1. the Y Z plane intersects the surface z = arctan(ye2x) at the point (0, 1, π/4)

    2. the plane y = −1 intersects the surface 4x2− 3y2+ z = 5 at the point (1,−1, 4)

    21

  • 14.4 Tangent Planes and Linear Approximations

    Tangent Planes

    Suppose f has continuous partial derivatives. An equation of the tangent plane to

    the surface z = f(x, y) at the point P (x0, y0, z0) is

    z − z0 = fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0).

    22

  • Ex 15. Find an equation of a plane tangent to

    1. z = x2 + y2 at the point (2,−1, 5)

    2. z = x− y + arctan(xy) at the point (1, 1, π/4)

    Do Exercises. 14.4 1–6 (odd)

    23

  • Linear Approximations

    From tangent plane to z = f(x, y) at (a, b, c)

    z − c = fx(a, b)(x− a) + fy(a, b)(y − b),

    we may write

    dz = fx dx+ fy dy,

    called the total differential of f . For ∆x and ∆y are small, we have

    ∆z = f(a+∆x, b+∆y)− f(a, b)

    ≈ fx(a, b)∆x+ fy(a, b)∆y.

    That is,

    f(a+∆x, b+∆y) ≈ f(a, b) + fx(a, b)∆x+ fy(a, b)∆y.

    24

  • Ex 16. 1. If z = f(x, y) = x2 + 3xy − y2, find the differential dz.

    2. If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values

    dz and ∆z.

    25

  • Ex 17. Approximate √

    (3.05)2 − (1.97)3.

    26

  • Ex 18. Find the linear approximation of the function

    f(x, y) = y2

    (1 + sin x)y

    at (0, 2) and use it to approximate f(0.03, 1.98).

    Do Exercises. 14.4 11–15 (odd), 25–31 (odd)

    27

  • 14.5 The Chain Rule

    In CAL I, for y = f(x) and x = g(t), we have

    dy

    dt =

    dy

    dx

    dx

    dt .

    y dy dx

    x dx dt

    t

    z ∂z ∂x

    ⑧⑧ ⑧⑧ ⑧⑧ ⑧⑧

    ∂z ∂y

    ❄❄ ❄❄

    ❄❄ ❄❄

    x dx dt

    y dy dt

    t t

    The Chain Rule (Case 1) Suppose that z = f(x, y) is a differentiable function of x

    and y, where x = g(t) and y = h(t) are both differentiable functions of t. Then

    dz

    dt =

    ∂z

    ∂x

    dx

    dt +

    ∂z

    ∂y

    dy

    dt .

    Ex 19. 1. If z = x2 − 2xy + xy3, x = t3 + 1 and y = 1/t, compute dz dt

    at t = 1.

    2. If z = x2 + wey + sin(xw), x = 2t, y = t2 and w = sec t, compute dz

    dt .

    28