# 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

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