Calculus 3 Notes Qatar University

TRIGONOMETRY FORMULAS

1)(sin)(cos 22=+ xx )(sec)(tan1 22 xx =+ )(csc1)(cot 22 xx =+

cos( ) cos( )cos( ) sin( )sin( )

sin( ) sin( ) cos( ) cos( )sin( )

x y x y x y

x y x y x y

± =

± = ±

∓

)tan()tan(1

)tan()tan()tan(

yx

yxyx

∓

±=±

)(tan1

)tan(2)2tan(

)(sin21

1)(cos2

)(sin)(cos

)2cos(

)cos()sin(2)2sin(

2

2

2

22

x

xx

x

x

xx

x

xxx

−=

−

−

−

=

=

2 2 2 2 cos( )

sin( ) sin( ) sin( )

c a b ab C

A B C

a b c

= + −

= =

)2cos(1

)2cos(1)(tan

2

)2cos(1)(cos

2

)2cos(1)(sin

2

2

2

x

xx

xx

xx

+

−=

+=

−=

)cos(1

)cos(1

2tan

2

)cos(1

2sin

2

)cos(1

2cos

x

xx

xx

xx

+

−±=

−±=

+±=

( )

( )

( )

( ) )]sin([sin)sin()cos(

)]sin([sin)cos()sin(

)]cos([cos)cos()cos(

)]cos([cos)sin()sin(

21

21

21

21

yxyxyx

yxyxyx

yxyxyx

yxyxyx

−−+=

−++=

++−=

+−−=

−

+−=−

−

+=+

+

−=−

−

+=+

2sin

2sin2)cos()cos(

2cos

2cos2)cos()cos(

2cos

2sin2)sin()sin(

2cos

2sin2)sin()sin(

yxyxyx

yxyxyx

yxyxyx

yxyxyx

For two vectors A and B, A·B = ||A||||B||cos(θ)

F2006 © Department of Mathematics & Statistics – Arizona State University 2

The well known results: soh, cah, toa

soh: s stands for sine, o stands for opposite and h stands for hypotenuse, sino

xh

=

cah: c stands for cosine, a stands for adjacent h stands for hypotenuse, cosa

xh

= h o

toa: t stands for tan, o stands for opposite and a stands for adjacent, tano

xa

= a

Where x is the angle between the hypotenuse and the adjacent.

Other three trigonometric functions have the following relations:

1csc

sin

hx

x o= = ,

1sec

cos

hx

x a= = and

1cot

tan

ax

x o= =

Important values:

0 030

6

π= 045

4

π= 060

3

π= 090

2

π=

sin

0 1

2 2

2

3

2

1

cos

1 3

2

2

2

1

2

0

tan

0 1

3

1

3

undefined

csc

undefined

2

2

2

3

1

sec

1 2

3

2

2

undefined

cot

undefined

3

1 1

3

0

sin( ) [?]sin , cos( ) [?]cos , tan( ) [?] tann x x n x x n x xπ π π± = ± = ± = , the sign ? is for plus or minus

depending on the position of the terminal side. One may remember the four-quadrant rule: (All

Students Take Calculus: A = all, S = sine, T = tan, C = cosine)

sine all

tan cosine

F2006 © Department of Mathematics & Statistics – Arizona State University 3

Example: Find the value of 0sin 300 . We may write 0 0 0 0sin 300 sin(2 180 60 ) [ ]sin 60= ⋅ − = − = -3

2,

in this case the terminal side is in quadrant four where sine is negative.

In the following diagram, each point on the unit circle is labeled first with its coordinates (exact

values), then with the angle in degrees, then with the angle in radians. Points in the lower hemisphere

have both positive and negative angles marked.

Qatar University- CAS- Dept of Math.-Stat- Phy.14/09/14- What I Need to my Calculus I -Dr. A. Hamdi .

1. Sign Rules

(+)× (+) → (+) (+)÷ (+) → (+)

(−)× (−) → (+) (−)÷ (−) → (+)

(+)× (−) → (−) (+)÷ (−) → (−)

(−)× (+) → (−) (−)÷ (+) → (−)

2. Removing Brackets

(a) x+ (y) = x+ y, x− (y) = x− y, x+ (−y) = x− y, x− (−y) = x+ y

(b) x× (−y) = (−x)× y = −xy, (−x)× (−y) = +xy

(c)−x

−y=

x

y,

−x

y=

x

−y= −x

y,

(d) (−1)n = +1 if n is even, (−1)n = −1 if n is odd.

3. Fractions

(a)a

b+

c

b=

a+ c

b,

a

b+

c

d=

ad+ bc

bd

(b)a+ c

b=

a

b+

c

b,

a

b× c

d=

ac

bd

(c) a× 1

c=

a

c,

a

bc

d

=a

b÷ c

d=

a

b× d

c=

ad

bc

(d)

a

bc=

a

b÷ c

1=

a

b× 1

c=

a

bc

(e)ac

d

=a

1÷ c

d=

a

1× d

c=

ad

c

1

1 Properties.

Given a, b, c, d four nonzero real numbers.

1.a

b=

c

d⇔ a× d = b× c.

2.a

b=

c

d⇔ b

a=

d

c.

3.a

b=

c

d⇔ a

c=

b

d.

4.a

b=

c

d=

a+ c

b+ d, (b+ d = 0).

5.a

b=

c

d=

a− c

b− d, (b− d = 0).

2 Powers

Let x, y be non null real numbers and n, m be integer numbers.

1. x0 = 1, xm × xn = xn+m,xm

xn= xm−n, x−n =

1

xn

2. (xm)n = xn×m, (xy)n = xnyn,

(x

y

)n

=xn

yn

3. x1n = n

√x

(a) if n is even, x must be greater than or equal to zero.

(b) if n is odd, x ∈ R.

In all this review we suppose that all the nth roots are well defined.

4. xmn = n

√xm = ( n

√x)m.

5. n√xy = n

√x n√y.

6. n

√x

y=

n√x

n√y.

7. n√

m√x = n×n

√x.

8. n√x× m

√x = mn

√xm+n = x

m+nmn .

Remark 1 : Sometimes simple observations can be very helpful

(a− b)2n = (b− a)2n and (a− b)2n+1 = − (b− a)2n+1.

Exercise 1 : Perform the given operations and simplify completely:

2

1.

A =

[−3

4÷ 9

16+ 21

3

]×

√2

√72− 5

√2

.

2. B = x+ (y − x)− (y − z)

3. C = (x+ y − z)− (x− y + z) + (−x+ y + z) + (x+ y − z)

4. D = (x− y)− (y + z − x) + (y + z − t) + (2y − x)

5. E = (a+ 2b− 6a)− 3b− (6a− 6b)

6. G = 5a+ {3b+ [6c− 2a− (a− c)]} − [9a− (7b+ c)]

7. H = 5x2 − 3xt+ t2 − 4x2 + 5xt− (3x2 − 7xt+ 5t2)− 7t2

Exercise 2 (*): If m = ax, n = ay, a2 = (mxny)z, evaluate (1− xyz)2012.

Exercise 3 : Simplify completely and express your result only under positive pow-ers:

1. (632 × 222 × 121× 14)3, (a2bc4)2, (a2bc)3 × (ab3c)2

2.(x−2y)−3

(x2y−1)3,

(x3y4z5

x−3y−5z−6

)−4

3.(2−1x−2y−1)−2(2x−4y3)−2(16x−3y3)0

(2x−4y−6)2

4. 9x4y5z6 ÷ 27x6y9z7

4m3n2÷ 4n2m3

3x2y4z− 1

5.42mp

65np÷ 15a2

26b2÷ 28b2mp

5a.

Exercise 4 (*) : Express in its simplest form

1.

A =

1

ab− 1

ac− 1

bca2 − (b− c)2

a

.

2.

B =1 +

x

1 + x

x+1

1 + x

÷ (x+ 1)2 − x2

x2 + x+ 1.

3.C =

a

b− c

d− e

f

.

3

4.

D =a+ b

a+ b− 1

a− b+1

a+ b

.

Exercise 5 : Which terms should we eliminate from the sum

S =1

2+

1

4+

1

6+

1

8+

1

10+

1

12

so that the remaining sum equals 1?

Exercise 6 : Evaluate the following expression for a =−1

3

1

2− a−1

4−(1

a

)−2 ÷[

1

2−2(2 + a)− 2a−1 − 1

].

Exercise 7 : Evaluate the following expression for a =−1

4a+

[1 +

(3− a

a+ 1

)−1]−1

−1

.

Identities

1.

(x+ y)2 = x2 + 2xy + y2, (x− y)2 = x2 − 2xy + y2, x2 − y2 = (x− y)(x+ y)

2.(x+ y)3 = x3 + 3x2y + 3xy2 + y3 = x3 + y3 + 3xy(x+ y)

3.(x− y)3 = x3 − 3x2y + 3xy2 − y3 = x3 − y3 − 3xy(x− y)

4.x3 + y3 = (x+ y)(x2 − xy + y2), x3 − y3 = (x− y)(x2 + xy + y2)

5. (x+ y + z

)2= x2 + y2 + z2 + 2

(xy + yz + zx

)Exercise 8 : Compute without any calculator:

A = 1234567892 − 123456790× 123456788.

(a)

(b)

4

(c) Exercise 9 : Perform the given operations and simplify completely:

1. a = 3(x− y)2(x+ y)− 3(x+ y)2(x− y)

2. b = (x− y)3 + (x+ y)3 + 3(x+ y)(x− y)2 + 3(x− y)(x+ y)2

Exercise 10 . Rationalize and simplify

α1 =2a

−b−√b2 − 4ac

, α2 =2a

−b+√b2 − 4ac

.

Exercise 11 : Factor each expression

1. 13x− 169y + 39z

2. ba− 2ab+ 3ba+ 10ab

3. y −my + by

4. xy2 + x2y − x4y4

5. bx+ cx

6. ax2 + bx− cx

7. ax2 + abx− a

8. 3x(x− 13)− (3x− 1)

9. xα+βyα + xβyα+β

10. x3y + x2y3 + x2y

11. 4x2(y2 − 2) + 2yx(−2 + y2)

12. x(y − 1) + z(1− y)− y + 1

13. 2(x− y)2 − 5x(x− y)

14. α(3 + x)− β√3 + x+ 2γ

√3 + x

15. xz + xt+ yz + yt

16. xz − xt+ yz − yt

17. x3 − 2x2 + 2x− 4

18. x2y2 − (x+ y)xy + x+ y − 1

19. x2 − 16

20. x2 − 15

21. (x2 − x)2 − 9

22. (x2 + 2x)2 − 9

23. (x+ y)2 − (z + t)2

5

24. (x+ y)2 − (x− y)2

25. 16x2 − (3x+ 5y)2

26. 81(x2 + 2x)2 − 256

27. (x+ 3y)2 − 25(y − x)2

28. (x2 + y2 − z2)2 − 4x2y2

29. (8x2 − 7y2)2 − (7x2 + 2y2)2

30. (x2 + y2 − 8)2 − (2xy − 8)2

31. 4(xy + zt)2 − (x2 + y2 − z2 − t2)2

32. 4x12 + 25y10 − 20x6y5

33. 8x6 − 2y4x2

34. x3y − y4

35. 1 + (x− y)3

36. 2x4 + 2xy3

37. 180x4y3 − 20x2y5

38. −x3 − 1 + 2x2

39. 25x2 − y4 − 25y2 + x2y2

40. x2 + z2 − y2 + 2xz

41. 4x2y2 − (x2 + y2 − z2)2

Exercise 12 ∗. Factor

E = 6− (x2 − x− 3)(2x2 − 10− 2x), F = x(x+ 1)(x+ 2)(x+ 3) + 1.

3 First Order Equations

In this section we are concerned by solving a first order equation

bx+ c = 0, b, c ∈ R, b = 0.

Solving an equation of this type is equivalent to make the variable-the unknown-”free”, in other terms, we need to separate the unknown from the other constants orparameters (if any) in the given equation. The process of making free the variableis based on the following properties:

1. x+ α = β ⇔ x = β−α.

2. x− α = β ⇔ x = β+α.

6

3. α x = β ⇔ x =β

α, α = 0.

4.x

α= β ⇔ x = α β, α = 0.

5.γ x

α= β ⇔ x = β

α

γ, αγ = 0.

For example, let us start by these equations:

Example 1 . Solve for x : −x+ 3 = 5.

Sol.−x+ 3 = 5 ⇔ −x = 5− 3 ⇔ −x = 2 ⇔ x = −2.

In the last step, we multiplied the both sides by −1 to cancel the ”-” beside x.

Example 2 . Solve for x : 2x− 3 = 6.

Sol.

2x− 3 = 6 ⇔ 2x = 6 + 3 ⇔ 2x = 9 ⇔ x =9

2.

Example 3 . Solve for x :2x

3− 1

5= 2.

Sol.2x

3− 1

5= 2 ⇔ 2x

3= 2 +

1

5⇔ x =

11

5÷ 2

3⇔ x =

11× 3

5× 2=

33

10.

Exercise 13 : If the numerator of the fraction is increased by 4, the fraction in-creases by 2/3, then find the denominator.

3.1 Equations Reducible to linear Equations

In many situations, solving a given equation, may be transformed and or reducedto solve a certain class of linear equations. We will try to make a brief tour aboutsome common situations.The Zero product Rule

A×B = 0 ⇔ A = 0 or B = 0

Also, it may be useful to know also

A×B = 0 ⇔ A = 0 and B = 0

The following examples illustrate some situations.

Example 4 . Solve (x− 1)(x+ 3) = 0.

Sol. This equation is equivalent to two linear equations:

(x− 1)(x+ 3) = 0 ⇔ x− 1 = 0 or x+ 3 = 0 ⇔ x = 1, x = −3 ⇒ S = {−3, 1}.

Example 5 . Solve (2x2 − 3x)(x− 1)(x+ 3) = 0.

7

Sol. Here, one observe that the term 2x2−3x = x(2x−3) and the equation becomes:

x(2x− 3)(x− 1)(x+ 3) = 0 ⇔

x = 0 or2x− 3 = 0 orx− 1 = 0 orx+ 3 = 0

thus S = {0, 3/2, 1,−3}. In the next example, we will state a general rule a class ofequations that are reducible to linear equations.

Example 6 . Solve for X ∈ RX2 = Y.

Sol. If Y is negative, we have a negative quantity equals a positive one, which isimpossible, thus S = ∅.Now, let us consider Y ≥ 0. Our equation can be rewritten as follows:

X2 − Y = 0,

and using the fact that 0 ≤ Y = (√Y )2, we have

X2 − (√Y )2 = 0 ⇔

(X −

√Y)(

X +√Y)= 0

and thus,X = ±

√Y .

This can formulated in general

X2 = Y ⇔ X = ±√Y , Y ≥ 0.

3.2 Quadratic Equations

By a quadratic equation we mean a problem of finding the roots (or zeros) of apolynomial of degree 2

Q(x) = ax2 + bx+ c, a = 0.

∆ = b2 − 4ac Factoring form Roots

∆ < 0 No factoring No real roots

∆ = 0 ax2 + bx+ c = a

(x+

b

2a

)2

x1 = x2 =−b

2a

∆ > 0 ax2 + bx+ c = a(x− x1)(x− x2) roots x1,2 =−b±

√∆

2a

Example 7 . Solve for x

x2 − 3 = 0, 2 + x2 = 0, (x− 1)2 = 5, (2x− 1)2 = (4− x)2.

Example 8 . Solve for x:

x2 − 7x+ 12 = 0, x2 + 7x+ 12 = 0, x2 − 7x− 12 = 0,

Example 9 . Solve for t

(3t− 1)2 − 5t(1− 3t) = 9t2 − 1, (1 + t)2 + 5 = 0, 4t2 − t− 3 = 0.

8

3.3 Rational Equations

By ”Rational equation”, we mean an equation involving at least a ratio of twopolynomials. The general technique to solve these equations is the use the ”crossproduct”

A

B=

C

D⇔ AD = BC OR

E

F= 0 ⇔ E = 0.

and to pay attention to those values that cancel the denominators.

Example 10 : Let the following equation

x− 1

x+ 2=

x

x− 6.

Sol. First of all, the values −2 and 6 are excluded from the solution set becausethey make null the denominators. The equation becomes:

0 =x− 1

x+ 2− x

x− 6=

(x− 1)(x− 6)− x(x+ 2)

(x+ 2)(x− 6).

which is equivalent to

(x− 1)(x− 6)− x(x+ 2) = 0 ⇔ x2 − 7x+ 6− x2 − 2x = 0 ⇔ −9x+ 6 = 0,

and then

x =6

9=

2

3⇒ S = {2/3}.

One can also use the main property of fractions to solve the equation. Indeed,

(x− 1)(x− 6) = x(x+ 2) ⇔ x2 − 7x+ 6 = x2 + 2x ⇔ 6 = 9x ⇔ x =6

9=

2

3,

thus, S = {2/3}.

Example 11 . Solve8− 5x

4(x− 2)+

x

x+ 2+ 1 = 0.

Sol. We start first by rejecting x = ±2 from the solution set and to get rid of thedenominators, we multiply all the terms by the common denominator 4(x−2)(x+2)to get then,

(8− 5x)(x+ 2) + 4x(x− 2) + 4(x− 2)(x+ 2) = 0 ⇔ 3x2 − 10x = 0,

i.e.,⇔ x(3x− 10) = 0 ⇔ x = 0, x = 10/3 ⇒ S = {0, 10/3}.

Example 12 . Solvex− 1

3− x=

2

3− x.

Sol. One can solve this problem as follows: Since the denominators are the same,

x− 1

3− x=

2

3− x⇔ x− 1 = 2 ⇒ x = 3,

but the value x = 3 is not allowed because it involves division by zero, then there isno solution.

9

4 Exercises

Exercise 14 . Solve for x

1.

2(x− 1)− 5(5− x) = 3(2x− 3),2x

3− 1

5=

x

15+

1

3.

2.(1− 2x)2 − 2x(x− 3) = 2(x2 − 1), 2− (1− x)3 = (x2 − 1)(x− 3).

3.2 + x√

2=

3 + x√3

, x =x

2+

x

3+

x

6+

1

7.

Exercise 15 . Solve for u

(a2 + 2)(u− 1) = x− 2, a(u− 2) = u+ a− 1, a2u+ 4 = 16u+ a, a ∈ R.

Exercise 16 . Solve for x

x+ 1

2x− 2+

x2 + 1

x2 − 1− x− 1

2x+ 2=

4x

x2 − 1.

Exercise 17 . Solve for λ

1

2

{1

2

[1

2

(1

2λ− 1

1

2

)− 1

1

2

]− 1

1

2

}= 0.

Exercise 18 . Solve for τ

1

2

{1

2

[1

2

(1

2τ − 2

1

2

)− 2

1

2

]− 2

1

2

}= 0.

Exercise 19 . Solve for ω

1

3

{1

3

[1

3

(1

3ω + 2

)+ 2

]+ 2

}+ 2 = 1.

Exercise 20 . Solve for η

1

9

{1

7

[1

5

(1

3(η + 2) + 4

)+ 6

]+ 8

}− 1 = 0.

Exercise 21 . If a, b, c are positive constants, solve for ξ

ξ − (a+ b)

c+

ξ − (b+ c)

a+

ξ − (c+ a)

b= 3, abc = 0.

Exercise 22 . Solve for z

−3z2+5z+2 = 0,z2

3+4z

5− 1

12= 0, 4.3z2−5.1z+1.428 = 0, (z−1)(z+2) = 70.

Exercise 23 . Solve for u

(5−√2) u2−10u+5+

√2 = 0, u2+(3a−4b)u−12ab = 0, 2 u3−3(2a−1) u2−9a u = 0.

10

Exercise 24 . Solve for z

3z2 =1

2, (2z − 1)2 + (1− 2z)− 6 = 0, z2 + 2z + 1− α4 = 0.

Exercise 25 . Find the domain of each expression

x2 −√3

x2 − 3x+ 1,

x2 − 3x2

x2 + x− 1,

x2 −√3

2x2 −√3x+ 1

.

Exercise 26 .

1. Show that U = V , where

U = (5x2 + 2x− 3)2 − (2x2 + x− 1)2, V = (x+ 1)2[(5x− 3)2 − (2x− 1)2

]2. Factor U and V .

3. Solve U = 0 or V = 0.

4. Evaluate U or V for x = 0, x = −1, x =√2− 1.

5. Simplify

W =(5x2 + 2x− 3)2 − (2x2 + x− 1)2

(7x− 4)2(3x− 2)2.

6. Solve W = 1/21.

Exercise 27 . Evaluate

f(x+ h)− f(x)

h, for each function f(x) = x2−3, f(x) = x3+2x, f(x) =

1

x− 1, f(x) =

√x+ 2.

11

Dr. A. Hamdi——Oct. 23, 2014- Homework 1 and 2- Math 211Due Time November 6th for HW1 and November 30th for HW2Choose 10 questions for HW1 from Q1 to Q18 but HW2 is from Q19 to Q23

1. Find the norm of 2−→a − 3−→b if −→a = ⟨2,−1, 3⟩,

−→b = ⟨−2, 1, 0⟩.

2. Given ∥−→a ∥ = 5, ∥−→b ∥ = 8,

(−→a ,

−→b ) = 150o. Evaluate ∥−→a +

−→b ∥.

3. Given three unit vectors −→a ,−→b , −→c satisfying −→a +

−→b +−→c =

−→0 . Evaluate

−→a •−→b +

−→b • −→c +−→c • −→a .

4. Given A(5, 1, 2) and B(−3, 5, 3), find an equation of the sphere with centerthe midpoint of the segment line [AB] and is tangent to the xy-plane.

5. Determine the projection of−→b on −→a if −→a = ⟨1,−2, 2⟩,

−→b = ⟨3, 4,−5⟩.

6. Find the distance from the point P (2, 3, 4) to the plane 3x− y − 4z + 6 = 0.

7. Under what conditions are −→u +−→v and −→u −−→v orthogonal?

8. Determine whether −→u = ⟨−1, 0, 2⟩, −→v = ⟨0, 1, 1⟩, −→w = ⟨−2, 1,−1⟩ lie in thesame plane.

9. Find the point where the line L through P0 = (2, 1, 3) and P1 = (4,−2, 5)meets the plane 2x+ y − 4z + 5 = 0.

10. Find an equation of the plane that contains (3, 2,−1), (−1, 1− 2) and that is

parallel to−→i −−→

j −−→k

11. Determine whether the following lines intersect, and if so, find their intersec-tion point.

L1 :x− 1

1=

y − 1

1=

z − 4

7, L2 :

x+ 4

5=

2− y

2=

1− z

4,

12. Verify the identity

−→a x(−→b x−→c ) = (−→a • −→c )

−→b − (−→a •

−→b )−→c

for −→a =−→i +

−→j , b = 2

−→i −

−→k , −→c =

−→j −

−→k .

13. If ∥−→a +−→b ∥ = ∥−→a −

−→b ∥ show that −→a ⊥

−→b .

14. Find the value of p for which the vectors −→a = 3−→i + 2

−→j + 9

−→k and−→

b =−→i + p

−→j + 3

−→k are:

(a) perpendicular (b) parallel.

15. For any vectors −→a and−→b , prove that:

∥−→a +−→b ∥ = ∥−→a −

−→b ∥ ⇔ −→a ⊥

−→b

Interpret the result geometrically.

16. If −→a and−→b are such that ∥−→a +

−→b ∥ = ∥−→a ∥, prove that: 2−→a +

−→b is perpen-

dicular to−→b .

17. If −→a ,−→b and −→c are non-null vectors in space such that −→a ·

(−→b ×−→c

)= 0, and

−→b and −→c are not parallel vectors, prove that the three vectors are coplanar.

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18. For any vector −→a , evaluate:

∥−→a ×−→i ∥2 + ∥−→a ×−→

j ∥2 + ∥−→a ×−→k ∥2 − 2∥−→a ∥2.

19. Find and sketch the domain of just one of the given functions

f(x, y) =

√xy

ln(x2 + y2 − 4

) , g(x, y) = ln(x+ y + 1) + cos−1(x2 + y2 − 1).

20. Find each limit (if it exists)

lim(x,y)→(0,0)

x2 + y2

x2 + |xy|+ y2, lim

(x,y)→(0,0)

3x4y2

7x8 + y4, lim(x,y)→(0,0)

x4 + 2x2y2

x2 + y2, lim(x,y)→(0,0)

xy

sin(x2 + 2y2).

21. Determine α for which f is continuous at the point (1, 2), where

f(x, y) =

x2 − 1 + y − xy

x2 − 3x+ 2 + xy − yif (x, y) = (1, 2)

α if (x, y) = (1, 2).

22. Find the directional derivative of f defined by: f(x, y, z) = (x + y)(y + z) at

the point P (5, 7, 1) in the direction of the vector −→v = 3−→i − 4

−→j + 12

−→k .

23. (a) Find zxy at x = 1, y = 0, if z = u3 + v3, u = x2 + tan−1(y) andv = x+ e−y.

(b) If f(x, y, z) = sin(cz)eax+by where c2 = a2 + b2, then show that for anyx, y, z

fxx + fyy + fzz = 0.

24. Challenging Questions

(a) Solve for −→r : −→r ×−→b = −→a ×

−→b , −→r •−→c = 0 assuming that the non null

vectors−→b and −→c are not perpendicular.

(b) Show that the altitude of the triangle ABC through the vertex A is equalto:

∥−→a ×−→b +

−→b ×−→c +−→c ×−→a ∥

∥−→b −−→c ∥

where the position vectors of A, B and C are respectively −→a ,−→b and −→c .

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Qatar University- CAS- Dept of Math.-stat- Phy.22/05/14- MidTerm-2-Calculus II(M211) -Dr. A. Hamdi -90 min.

1. Find an equation of the normal line to the graph of the equation:

9x2 − 4y2 − 25z2 = 40, at P (4, 1,−2).

2. Find the point(s) on the surface x2 + 4y2 − z2 = 4 at which the tangent planeis parallel to the plane 4x+ 4y + 2z = 10.

3. Find the local extrema and saddle point(s) if any of the functionf(x, y) = x4 + 2y2 − 4xy.

4. Find the extrema of f(x, y) = x2 + 2x+ y2 subject to x2 + 4y2 − 24 = 0.

5. Let f(x, y, z) = x(xz + y2)− 3 cos(πyz).

(i) Determine the maximum rate of change of f at P (1,−1, 1).

(ii) Show that for any vector unit vector −→u we haveD−→u f(1,−1, 1) ∈ [−

√14,

√14].

(iii) Use linear approximation to estimate f(0.97,−1.02, 1.01).

6. Let z = f(x, y) be a function with continuous second order partial derivatives,where x = s+ t and y = s− t. Show that

zss − ztt − 4zxy = 0.

7. Evaluate∫ ∫

(R)

√x2 + y2dA; (R) is bounded by y =

√2x− x2, y = 0 and the

line y = x.

8. Evaluate∫ 2

0

∫ 4

y2y cos(x2)dxdy.

9. Use double integration to find the area bounded by y =√x, y+x = 0, x = 1

and x = 4.

10. Find the volume of the solid in the first octant that is bounded by the threecoordinate planes and the plane 2x+ y + z = 5.

11. Bonus Use polar coordinates to evaluate∫ 4

0

∫ √25−x2

3dydx.

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