110.202. Calculus III
2004 Summer
Midterm I
7/14/2004 1:00PM
1. Compute the following limits if they exist:
(a)
lim cos xy−1
.
x
(x,y)→(0,0)
r¯
¯
¯ x+y ¯
(b)
lim
¯ x−y ¯ with x 6= y.
(x,y)→(0,0)
[Solution]
(a) Write
µ
¶
cos xy − 1
cos xy − 1
lim
= lim
·y .
(x,y)→(0,0)
(x,y)→(0,0)
x
xy
When (x, y) approaches (0, 0), we have xy approaches 0.
By setting t = xy, we have t → 0 as (x, y) → (0, 0). That
xy−1
means lim cos xy
= lim cos tt−1 = 0. So,
t→0
(x,y)→(0,0)
µ
¶
cos xy − 1
lim
·y
(x,y)→(0,0)
xy
cos xy − 1
· lim y
=
lim
(x,y)→(0,0)
(x,y)→(0,0)
xy
= 0 · 0 = 0.
cos xy − 1
lim
=
(x,y)→(0,0)
x
(b) If we approaches (0, 0) along x = 2y, we have
s¯
¯
p
√
¯ 2y + y ¯
¯
¯=
lim
|3|
=
3.
lim
¯ 2y − y ¯ (2y,y)→(0,0)
(2y,y)→(0,0)
If we approaches (0, 0) along x = 4y, we have,
s¯
s¯ ¯ r
¯
¯ 4y + y ¯
¯5¯
¯
¯ ¯ = 5.
¯=
lim
lim
¯
¯3¯
¯
(4y,y)→(0,0)
(2y,y)→(0,0)
4y − y
3
Therefore, the
¥
lim
(x,y)→(0,0)
1
sin 2x−2x+y
x3 +y
does not exist.
2
2. For the function
f (x, y, z) = exyz
and
¢
¡
g (t) = 6t, 3t2 , t3 ,
find ∇f and g0 and evaluate (f ◦ g)0 (1).
[Solution]
By definition,
¶
µ
∂f ∂f ∂f
,
,
= (yzexyz , xzexyz , xyexyz ) .
∇f =
∂x ∂y ∂z
Since g (1) = (6, 3, 1), we have
∇f |(6,3,1) = (yzexyz , xzexyz , xyexyz )
¢
¡
= 3e18 , 6e18 , 18e18 .
By definition,
¡
¢¯
g0 (1) = 6, 6t, 3t2 ¯t=1 = (6, 6, 3) .
By chain rule, we have
(f ◦ g)0 (1) = ∇f (g (1)) · g0 (1)
¡
¢
= 3e18 , 6e18 , 18e18 · (6, 6, 3)
= 108e18 .
¥
3. Find the equation for the plane tangent to the surface
z = xy
at the point (1, 1, 1).
[Solution]
Let f (x, y) = xy .
The equation of the tangent of the graph of f (x, y) is
∙
¸
∙
¸
∂f
∂f
z = f (x0 , y0 ) +
(x0 , y0 ) (x − x0 ) +
(x0 , y0 ) (y − y0 ) .
∂x
∂y
We have
f (1, 1) = 11 = 1;
¯
∂f
(1, 1) = yxy−1 ¯(1,1) = 1 · 11−1 = 1;
∂x
∂f
(1, 1) = (ln x) xy |(1,1) = (ln 1) 11 = 0.
∂y
3
Therefore, the equation is
z = 1 + 1 (x − 1) + 0 (y − 2)
= x.
So, the equation for the plane tangent to the surface z = xy at
the point (1, 1, 1) is z = x.
¥
4. Determine the second-order Taylor formula for the function
¡ ¢
f (x, y) = cos xy 2
¡
¢
about the point π2 , 1 .
[Solution]
= − sin (xy 2 ) · y 2 = −y 2 sin (xy 2 ) and ∂f
=
We have ∂f
∂x
∂y
2
− sin (xy 2 )·2xy = −2xy sin (xy 2 ). Moreover, ∂∂xf2 = − cos (xy 2 )·
2
y 2 ·y 2 = −y 4 cos (xy 2 ), ∂∂yf2 = −2x [sin (xy 2 ) + y cos (xy 2 ) · 2xy] =
2
2
∂ f
∂ f
−2x sin (xy 2 )+2xy 2 cos (xy 2 ) and ∂x∂y
= ∂y∂x
= − [2y sin (xy 2 ) + y 2 cos (xy 2 ) · 2xy] =
¯
¡
¢
¯
=
−2y sin (xy 2 ) − 2xy 3 cos (xy 2 ). Hence, f π2 , 1 = 0, ∂f
∂x ( π2 ,1)
¯
¯
¯
¯
2 ¯
2 ¯
¯
∂2f ¯
= −π, ∂∂xf2 ¯ π = 0, ∂∂yf2 ¯ π = −π and ∂x∂y
−1, ∂f
¯ π =
∂y ¯ π
( 2 ,1)
( 2 ,1)
( 2 ,1)
¡ π (¢2 ,1)
−2. Therefore, the second-order Taylor formula about 2 , 1 is
´
³π
+ h1 , 1 + h2
f
2
³ π ´ ∂f ³ π ´
∂f ³ π ´
,1 +
, 1 h1 +
, 1 h2
= f
2
∂x 2
∂y 2
1 ∂2f ³ π ´
1 ∂ 2f ³ π ´
∂ 2f ³ π ´
, 1 h1 h1 +
, 1 h2 h2 +
, 1 h1 h2
+
2 ∂x2 2
2 ∂y 2 2
∂x∂y 2
³π
´
+R2
, 1; h1 , h2
2
1
1
= 0 + (−1) h1 + (−π) h2 + (0) h1 h1 + (−π) h2 h2 + (−2) h1 h2
2
2
³π
´
+R2
, 1; h1 , h2
2
³π
´
π
, 1; h1 , h2
= −h1 − πh2 − h22 − 2h1 h2 + R2
2
2
where
¥
R2 ( π2 ,1;h1 ,h2 )
k(h1 ,h2 )k
→ 0 as (h1 , h2 ) → (0, 0).
4
5. Find the critical points of the function
1 1
f (x, y) = xy + +
x y
and then determine whether they are local maxima, local minima, or saddle points.
[Solution]
We have
¶
µ
1
1
∇f = y − 2 , x − 2 .
x
y
By setting ∇f = 0, we have y = x12 and x = y12 . This implies
x = x4 and y = y 4 . Since we know x 6= 0 and y 6= 0 from the
definition of the domain of f , we get x = 1 and y = 1. Hence,
we have only one critical points (1, 1).
Since the discriminant
µ 2 ¶2 ¯¯
∂2f ∂2f
∂ f
¯
∆ (1, 1) =
·
−
¯
∂x2 ∂y 2
∂x∂y ¯
(1,1)
¯
µ ¶ µ ¶
¯
2
2
·
− (1)2 ¯¯
=
3
3
x
y
(1,1)
= 3 > 0,
we know that
¯ (1, 1) is ¯a local extrema.
2
¯
Since ∂∂xf2 ¯
= x23 ¯(1,1) = 2 > 0, we know that (1, 1) is a
(1,1)
local minimum.
¥
6. Find the absolute maximum and minimum for the function
on the ball
f (x, y, z) = x + y − z
ª
©
B = (x, y, z) | x2 + y 2 + z 2 ≤ 1 .
[Solution]
Write B = U ∪ ∂B where U = {(x, y, z) | x2 + y 2 + z 2 < 1}
and ∂B = {(x, y, z) | x2 + y 2 + z 2 = 1}.
Since ∇f = (1, 1, −1) 6= (0, 0, 0), we have no critical points
on U.
To find the critical points on ∂B, let f (x, y, z) = x + y − z
and g (x, y, z) = x2 + y 2 + z 2 − 1. By the method of Lagrange
multiplier, we have
∇f = λ∇g,
5
that is,
(1, 1, −1) = λ (2x, 2y, 2z) .
So, we have the following system of equations:
⎧
1 = 2λx
⎪
⎪
⎨
1 = 2λy
−1
= 2λz
⎪
⎪
⎩ x2 + y 2 + z 2 = 1.
1
= y and
By 2λx = 1, we know λ 6= 0. We have x = 2λ
1
3
z = − 2λ . By the last equation, we have 4λ2 = 1. This implies
√
3
. Moreover, we ³have two critical
that
λ
=
±
2´
´ points (x, y, z) =
³
√1 , √1 , − √1
3
3
3
¥
or (x, y, z) = − √13 , − √13 , √13 .
Since we have only two candidates of absolute maximum
and minimum, one must be the absolute maximum and another
³ must be ´the absolute√minimum. It is easy to see that
f √13 , √13 , − √13 = √33 = 3 is the absolute maximum and
´
³
√
f − √13 , − √13 , √13 = − √33 = − 3 is the absolute minimum.
7. Find the extrema of the function
z = xy
subject to the condition x + y = 1.
[Solution]
Let f (x, y) = xy and g (x, y) = x + y − 1. By the method of
Lagrange multiplier, we have
that is,
∇f = λ∇g,
(y, x) = λ (1, 1) .
So, we have the following system of equations:
⎧
y = λ
⎨
x = λ
⎩ x + y = 1.
¥
Therefore, we have x = y. This implies 2x = 1, that is, x =
1
= y. Since f (1, 0) = 0 where (1, 0) ¡is a point
in the condition
2
¢
1 1
x + y = 1, we know the critical point 2 , 2 has to be maximum
with the maximum value 14 .
6
8. When c 6= 0, the level curve {(x, y) | f (x, y) = c} of the function
2x
f (x, y) = 2
x + y2
is a circle. What it its radius, and where is its center? What
happens when c = 0?
[Solution]
The level curve is f (x, y) = c. So, we have
2x
= c,
2
x + y2
that is,
cx2 + cy 2 = 2x.
By rewriting it into the standard form of a circle, we have
µ
¶2
1
1
+ y2 = 2 .
x−
c
c
¡ ¢
1
Thus, we know the radius is |c| and the center is 1c , 0 .
When c = 0, the level curve becomes
2x
= 0,
2
x + y2
that is,
2x = 0
or
x = 0.
So, the level curve is a straight line x = 0 when c = 0.
¥