MATH 213-A01 - Exam 1 - Solution: June 9, 2011
1. (15 pts) Prove or disprove: the function  is continuous at (0 0) where
( 
||
 ( ) =
0
if  6= 0
if  = 0
Take the limit of  ( ) at (0 0) along the line  =  to show that lim()→(00)  ( ) 6=
 (0 0) :


= lim
= lim 1 = 1 6= 0 =  (0 0)
lim
→0 ||
→0
()→(00) ||
along =
In fact, taking the limit of  ( ) at (0 0) along the line  = − yields yet another limit
lim
()→(00)
along =−

 (−)
−2
= lim
= lim 2 = lim −1 = −1
→0
|| →0 | (−)| →0 
Thus the limit DNE. In any event,  is not continuous at (0 0) 
¢
¡
2. (40 pts) Consider the function  ( ) = ln 2 + 2 
¡√ ¢
3 1 
(a) Sketch the level curve through the point
¡
¡√ ¢
¢
 3 1 = ln(3 + 1) = ln(4)  Then level curve is given by ln 2 +  2 = ln(4)  Simplify
be exponentiating
2
2
ln( + ) = ln(4)
Simplify to get the circle of radius 2
2 +  2 = 4
1
¡√ ¢
3 1 on the same plot as part (a)
+
*√
¿
À¯
¯
¯
¢
→ ¡ 2
→ ³√ ´ −
−
2
3
1
2
¯
¯

=
3 1 = ∇ ln  + 2 ¯

∇
=
√
2 +  2 2 +  2 ¯()=(√31)
2 2
()=( 31)
(b) Calculate and sketch the gradient of  at
√ ®
¡√ ¢
­
(c) Compute the directional derivative of  at
3 1 in the direction 1 − 3 
+
*√
D
³√ ´
√ E
3 1
→

·
1
−
−

3
1
=
3 =0

2 2
¡√ ¢
−
→
(d) What direction →
 maximizes the directional derivative −
3 1 ?
  at
*√
+
3 1

2 2
¢
¡√
¢
¡
3 1 ln 4 
(e) Determine a normal vector of the tangent plane to  = ln 2 +  2 at
+
* √
1
3
→
−
− 1
 = −
2
2
¡
¢
¢
¡√
(f) Determine the tangent plane approximation to  = ln 2 + 2 at
3 1 ln 4 
√ ³
√ ´ 1
3
−
 − 3 − ( − 1) + 1 ( − ln 4) = 0
2
2
Solve for  to get
√
√ ³
√ ´ 1
1
3
3
 − 3 + ( − 1) =
 +  + ln 4 − 2
 = ln 4 +
2
2
2
2
(g) Use part (f) to estmate the value of  (17 11) 
√
3
1
 (17 11) ≈
(17) + (11) + 2 ln 2 − 2 ≈ 141
2
2
√
where I have used 3 ≈ 1732 and ln 2 ≈ 069
(h) Suppose  and  are functions of  such that (0) = 3 0 (0) = 5 (0) = 1 and
0 (0) = −1 Calculate  0 (0).

   
=
+

 
 
From part (b) we have
¯
 ¯¯
=
 ¯((0)(0))=(31)
¯
 ¯¯
=
 ¯
((0)(0))=(31)
2
2(3)
= 06
+ 12
32
2(1)
= 02
32 + 12
Then

(0) =

¯
¯
 ¯¯
 ¯¯


(0) +
(0)
 ¯((0)(0))=(31) 
 ¯((0)(0))=(31) 
= (06) (5) + (02) (−1) = 28

 where  = 
3. (10 pts) Suppose  (  ) =  ln ()  Use the chain rule to calculate

 =  and  = 

     
=
+
+

     
Then

=

µ
¶
µ
¶
1
1

  () +   (0) + [ln ()]  =
+  ln ()



Next substitute for   and  to obtain
¡
¢

() 
=
+  ln () =  +  ln 2 


4. (15 pts) Suppose  ( ) = 2 +  2 − 2 + 1
(a) Determine all points ( ) where  has either a local maximum or a local minimum.
First compute all critical points:

= 2


= 2 − 2

Solve 2 = 0 2 − 2 = 0 to obtain the only critical point ( ) = (0 1) 
Next, use the second derivative test to test for local max or min:
 =
2
= 2
2
 =
2
= 2
 2
 =
 2
=0

2 = 4  0 and 
Since   − 
 = 2  0 then  has a local minimum at (0 1) 
(b) Calculate all of the maximum and minimum values of  from part (a).
 (0 1) = 02 + 11 − (2) (1) + 1 = 0
(c) Calculate the absolute maximum and minimum values of  on the disk 2 +  2 ≤ 4
As the critical point (0 1) is the only critical point interior to the disk 2 + 2 ≤ 4 the
only other possible for a point where  takes on the value less than  (0 1) = 0 would
be on the circle 2 +  2 = 4 So replace 2 in 2 +  2 − 2 + 1 with 2 = 4 −  2 to get
the function just of  :
 () = 4 − 2 + 2 − 2 + 1
Simplify
 () = 5 − 2
On the circle  is constrained to lie between −2 and 2 Thus the minimum of  () when
−2 ≤  ≤ 2 is 1 (when  = 2) It follows that the absolute minimum of  is still achieved
sat the point (0 1) and has value 0
3
5. (15 pts) Evaluate the integral
ZZ
2
R
and (1 1) 
−  where R is the triangle with vertices (0 0)  (0 1) 
Integrate first wrt ; you cannot integrate first wrt  For a fixed  integrate from  = 0 to
 =  (the diagonal line). Then
ZZ
−2

 =
Z 1Z
0
R
=
Z
6. (25 pts) Consider the double integral
0
0
(a) (5 pts) Sketch the region of integration.
2
−  =
√
3
4
2
− 
0
1
0
Z 1Z

¢
1¡
1 − −1
2
2
p

2 +  2
(10 pts) Change the integral to an equivalent one involving polar coordinates.
First replace  and  with  =  cos  and  =  sin  Then we get
Z 1Z
0
0
√
3
2
p
 =
2 +  2
ZZ
2 cos2 


R
For a fixed value of  we integrate  from  = 0 out to the vertical line  = 1 In polar
coordinates the line  = 1 is expressed as 1 =  cos  or by solving for 
=
1
cos 
As  varies from 0 to the angle the hypotenuse makes with the -axis, namely  = 3
we obtain the integral
Z 3 Z sec  2
 cos2 


0
0
(b) (10 pts) Evaluate the integral from part (b). [Hint:
Z
0
3 Z sec  2
 cos2 
0

 =
Z
0
=
Z
3 Z sec 
=
3
3
0
=
=
=
=
=
=
1
3
1
3
sec  = ln (sec  + tan )]
2 cos2 
0
0
Z
R
Z
¯
1 3 ¯¯sec 

cos2 
3 ¯0
1
sec3  cos2 
3
3
sec3  cos2 
0
Z
0
3
sec 
¯3
¯
1
ln (sec  + tan )¯¯
3
0
i

´
1h ³
ln sec + tan
− ln (sec 0 + tan 0)
3
3
3
i
√ ´
1h ³
ln 2 + 3 − ln (1 + tan 0)
3
√ ´
1 ³
ln 2 + 3
3
5