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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