Math 251 : Final Exam 2

1. [12 points]
(a) Let R be the parallelogram with vertices A(0, 0), B(2, 4), C(2, 1) and D(4, 5).
! !
(i) Find the angle between the directed line segments AB and AC.
(ii) Find the area of the parallelogram using vector methods.
Eo .
B (2 9)
od(s)
,

= 12, 3)
⑨
il Ac = (2, 1

Cos
(0 %
>2 I ,

ii) Area = /EXAd

-
>
lYM-o ; Area =I

(b) (i) Show that the distance from the point P(x0 , y0 , z0 ) to the plane ax+by+cz =
d is
|ax0 + by0 + cz0 d|
distance = p .
a2 + b 2 + c 2
(ii)Compute the distance from P(2, 1, 3) to the plane x + y z = 1.

P
ax + by + 1 =
d
↑
G Be point on
/

[b]
= Let a

↓
proje the plane X1 , Y112 ,

Q =
<Xo-X ,, Yo- Y , 20-2 7 , = d

/Projnk/p/axotbyt
/

Th

#
=

ii)
Math 251 : Final Exam 3

2. [8 points]
(a) Find the tangent line to the curve C given by
⌦ ↵
r(t) = t2 4t, et 3 + 2, t 2

at the point Q( 3, 3, 1).

et 3,
v'(t) = 12t -2, 1

i For what value

*
of

3k
to does

3
r(t) =
Q?

4 = =
-
-

pr 3
-

+2
=
3
1
t 2 a
-

r () = < 2
,
1
, 17 ((s) =
[p) s(i)
+

(b) Find an equation for the plane that passes through the point P (2, 1, 3) and
contains the tangent line to C derived in (a).

Ep =
2) , -
9, 27

3) Pl ,i

Y4E
L(s)
sh Px =

3
-

92 =
1 -

6 , 1 , 13)

21/

6x -

y + 132 d
=

6(2) (1) + 1313) d ; d = 28

-
=

-
(x -

y + 12
= 28
 Math 251 : Final Exam 4

3. [10 points]
(a) Show that if u(t) is a vector function of constant magnitude, |u(t)| = c (a
constant), then u0 (t) is orthogonal to u(t) for all t.
(u(t)) = C

both sides
square
property(u(t)k = 2

& (u(t)
·
u(t) =
c)
u St) w(t) + u'(k) u(t)
- = 8
2 (u(t) u'(h))
.
=
0

: u(t) u(t) 8 -
=

: uSt) +
u(t)

(b) Consider the helix C given by

p
x = sin 2t, y= 5t, z = cos 2t, 0  t  ⇡.

Show that the speed v = |v| is constant, and verify that the acceleration is
orthogonal to the velocity, as predicted in part (a). Find the arc length of the
curve C.
r (k) (sinht Est
=
,
(os2r)
,

~ (t) = v (t) <210s2t 55 , - Isin2t]

J v(t)
=
, -
a(t)
↓
"
(t) =
a(t) =
<- 4 Sin2t 0 -cos[i] = -
8 cos2tsi2 + Scos2tsindt =
OV orthogonal
, .

I M

[Irtilde (3dt 3h/ 3

=

Iris S = = =

= =
3
Math 251 : Final Exam 5

4. [14 points]
(a) Let z = f (x2 y2, y2 x2 ), where f is di↵erentiable. Show that z satisfies the
equation
@z @z
y +x = 0.
@x @y
[Hint: set u = x2 y 2 and v = y 2 x2 .]

2
0
) - -+(xy =

(b) Find all critical points of the function f (x, y) = 4xy 2x4 y 2 , and classify
them using the Second Derivative Test. Does the function f have any global
maxima or minima?

fx 8x 0 9(2x) 8x 0 8x3 + 8x 0 ; x = 0,
4y
=

-
- = -

=
= -

0-
& 2x
if x= 0 , y =
0, if x= 1
,
y = 2, if x = +, y = 2
=

fy = (x -
2y = y =

(1 , 21
· we have (0 01 ,
(1 2),

Feature f xx +y y (xyy2
-
= S

fxx = - 29x -
> 18X -
16 = D

D = 16
fyy =
-

2 G (0 0),

32
G (1 ,
2) D =

3)
& ( -2)D
=

,
 Math 251 : Final Exam 6

5. [8 points]
Let C be the curve of intersection of the two surfaces:
(x, Y, z)
g
paraboloid: z = x2 + y 2 and plane: x + y + 2z = 2
h(x y , 2) X + y2 2
sub zexity into place
, = -

(the curve C is an ellipse).

Find the highest and lowest points (above the x-y plane) on the curve C, clearly
explaining your reasoning. Use the method of Lagrange multipliers.

*sor retti- The

is

OR
height
given
this
by
is strie the square of the distance function

f(x 4 2) ,
,
=
2 -
This is the paraboloid height restricted by the plane and the top of the parabeloid

Main

Method 1 ifxiY 2) ,
= 2

constraint
Constraint

g
(x , y , 2) =
X +2 0 =
h(X , y, 2) = X + y + 2z =
2

-f =
xg + uh

fx =
Xgx + uhx fy xgy mhy
+ fr xgztehz

I
= =

0 = 2xx + M
o =
2xy +
m 1 = -
x + 2m
M
T
x = 2rs-1
un

useless if 2H) 2 i =

Joxy +u
Y=1
2exte X -1 2= we have (12) & fil
=
= , ,

&xx &Xy
if x = = y= 1 z =
2/= i
f(x y z)
↑
, ,
= , ,
= 2

X =
Y - -

1 .
-
1
, 2) =
2 High
Hi , =
E Low

Plug into both constraints

*+ *2 = 0 x+ x + 2z = 2
2x2 - 2 = 0 2x + 2z = 2

2= 2x2 - 2x + 2x ) =

4 + 2x 2= 0 ;
-
x = 2 1 -

Method 2

DO SAME
THING BUT

LET f(xiy 2)
,
+ X + y+ 22
Math 251 : Final Exam 7

6. [10 points] p p p
p
Consider the surface S described by x + y + z = c.
(a) Find the tangent plane to S at the point (x0 , y0 , z0 ).
fz(x0 20)(2 20)
↑ (X fa(x0 Yo 20)(X - Xol fy(X0 40 20)(y
40) 40,

2)
-
- +
+ , ,
,

= ,
y
,
, ,

* X
-

xo) + 2(y -

yo) +
2 tzz -

2) =
0

(b) Find the intercepts of the tangent plane in the x-, y- and z-axes.
Let Y = z= 0

x) - E = 0
(x
-

X = Mo +20 + Xa

Let X =
z =
0
Y=Yo +2
+ Yo

Let x =
y= O

. No
2 to + 20

(c) Show that the sum of the x-, y- and z-intercepts is a constant, independent of
the point (x0 , y0 , z0 ) .

come back
Math 251 : Final Exam 8

7. [12 points]
(a) Consider the double integral
Z 1 Z 2
2y x3
e dxdy.
0
p
2 y x2

Sketch the region over which the integration is being performed, and evaluate
the integral. x =
2
y =
f

----- edy!
zi
2
i think there is a mistake

in the integral
Math 251 : Final Exam 3 2x 9
3y
-
-

2 =

(b) Find the surface area of the part of the plane 2x + 3y + z = 7 that lies above the
region R in the x-y plane bounded by the parabola y = x2 , the line x + y = 2
and the x-axis. Ye 2 - X

Z 2

((y)" axdy
2 X = X
N
-

S = x+ x -
2= 0 ;
X= -

2, 1
&
2 -

Is i
=

en

-
Freyax
=

No2-x-d
x
+ ( +
Y

ma(2x Ex -x(2)
-
-

=
Mq(2 -

z -

y+ 4 + 2 -)
==
Math 251 : Final Exam 10

8. [10 points]
Consider the hemisphere of radius a above the x-y plane,

x 2 + y 2 + z 2 = a2 , z 0.

If 0 < h < a, find the volume of the spherical segment (“cap”) of the hemisphere
lying above the plane z = h.
For which value of h is the volume of the top “cap” portion equal to half the total
volume of the hemisphere?

i don't feel like if

Math 251 : Final Exam 11

9. [8 points]
Consider the triple integral described in rectangular coordinates by
Z Z p 2 Z p 4 x2 8 x2 y 2

p p xz dzdydx.
0 4 x2 x2 +y 2

Without attempting to evaluate the integral, write down the form of the triple
integral, explicitly indicating the limits of integration, in
(a) Cylindrical coordinates r, ✓, z dV =
rozdrdo

Fr R
S)) resozidaddo

-
- or

Xy

XX

(b) Spherical coordinates ⇢, ✓, .

() Smpsindpaddo tanp =
& given from bound

since r= thym I =

tand =
E = 1 : 0 7
=

+=
05
z =
22 = 8 x2 y2
-

x+ y+ 22 = 8 = p
P 5 =

so
op
and -

E0-E

: psind (psidd a
Math 251 : Final Exam 12

10. [8 points]
Consider again the parallelogram R with vertices A(0, 0), B(2, 4), C(2, 1) and D(4, 5),
as studied in question 1(a).
(a) Sketch the region R in the x-y plane.
-
&

2 !

(b) Sketch the region S, the image of R in the u-v plane under the transformation
x = 2v 2u, y = 4v u. Find the Jacobian of the transformation.
X = 2V 2y-

y = &V a
-

> u = qu y
- - El

X = 2v 8v +2y
-

E2V
= Plug into El

<(4)
m =
Y
mullion sys
-

=
Ex
v
1901
(v) (0 0),
↑

12 1) (Mr 7-1 , 0)
,

12 ,
4144 10 , 1)
Hi. &
caus
Y4

19 , 9)27-1 11 ,

J =

- -

8 +2
J
=

=
-

1 6)
-
6

=
6

(c) If the Jacobian is constant, then the area of R in the x-y plane is simply the
absolute value of the Jacobian times the area in the u-v plane. Use this fact to
compute the area of R.

Area in uV plane = (1) (D = /

(b)(1) = 6