Last update: 17.11.2014
Mathematics 2:
Answers to (selected) exercises of Review Homework 61
Exercise 1. [Domain
of Function, Partial Derivatives, Tangent Plane] Given the function
x
f (x, y) = ln 2y−1 + ln(1 + y 2 ) + xy.
a) Determine the domain and plot it.
b) Calculate first order partial derivatives of f at (1, 2).
c) Calculate the equation of the plane tangent to the graph of f at (1, 2).
Sol.: a) We have two logarithmic functions, whose arguments need to be strictly positive. Hence
x
2
2
>0 & 1+y >0
D = (x, y) ∈ R :
2y − 1
The second inequality 1 + y 2 > 0 is satisfied for all (x, y) ∈ R2 , and may be neglected. The
first inequality is satisfied when numerator and denominator are either both positive or both
negative
x
x>0
x<0
>0⇔
∨
2y
−
1
>
0
2y
−1<0
2y − 1
Summing up, the domain of the function is
1
1
D = (x, y) : x ∈ (0, +∞)&y ∈ ( , +∞) ∪ (x, y) : x ∈ (−∞, 0)&y ∈ (−∞, )
2
2
This is shown in Figure 1 by the shaded area.
b) First order partial derivatives of f are
1
2y − 1
′
+y
=1+2=3
fx (1, 2) =
x
2y − 1
(x,y)=(1,2)
2y − 1
−2x
2y
17
′
fy (1, 2) =
+
+x
=
2
2
x
(2y − 1)
1+y
15
(x,y)=(1,2)
1
The solutions have been kindly provided by Prof. Faggian
1
3.2
2.4
1.6
0.8
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
-0.8
-1.6
-2.4
-3.2
Figure 1: Domain for Example 1.
c) Using the equation
z − z0 = fx′ (x0 , y0 )(x − x0 ) + fy′ (x0 , y0 )(y − y0 )
and recalling that in our case (x0 , y0 ) = (1, 2) and z0 = f (1, 2) = ln(5/3) + 2, we obtain the
equation of the tangent plane
z − ln(5/3) − 2 = 3(x − 1) +
17
(y − 2)
15
Exercise 2. (Exam 27.04.2012) Given the function f (x, y) = x +
ln (4−x2 −y 2 )
√
.
1−x2
a) Determine the domain and represent it graphically.
b) Calculate first order partial derivatives at (0, 0).
c) Calculate the equation of the tangent plane to the graph of f at (0, 0, f (0, 0)).
Sol.: a) We write down the conditions for the domain, as we did in the past exercises, for the
numerator and denominator
D = (x, y) ∈ R2 : 1 − x2 > 0 & 4 − x2 − y 2 > 0
With respect to the first condition, the acceptable couples are (x, y) in R2 such that −1 < x <
1, y ∈ R, shaded in blue. The second condition identifies points (x, y) that are inside the circle
centered at the origin (0, 0) and with radius 2. The intersection of the two sets gives the domain
of the function. Hence
D = (x, y) ∈ R2 : −1 < x < 1& x2 + y 2 < 4
2
3.2
2.4
1.6
0.8
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
-0.8
-1.6
-2.4
-3.2
Figure 2: Domain for Example 2.
b) The first order partial derivatives are:
√
−2x
2 −1/2 ln(4 − x2 − y 2 )
1 − x2 4−x
2 −y 2 + x(1 − x )
′
fx (x, y) = 1 +
1 − x2
−2y
fy′ (x, y) = √
2
1 − x (4 − x2 − y 2 )
If we plug in the coordinates (0,0) into the above formula, we get fx′ (0, 0) = 1 and fy′ (0, 0) = 0.
c) Since f (0, 0) = ln 4, we obtain the equation of the tangent plane by making use of the formula
of the previous exercise
z = x + ln 4
Exercise 3. (Exam 27.04.2012) Given f (x, y) = x +
√
4−y 2
.
ln (16−x2 −y 2 )
a) Determine the domain D and represent it graphically in the plane.
b) Calculate the first order partial derivatives of f .
c) Say what are the interior points and boundary points of D. Is D open, closed, bounded,
compact? Justify each answer.
3
Sol.: a) Writing down the conditions, we have
D = (x, y) ∈ R2 : 4 − y 2 ≥ 0 & ln(16 − x2 − y 2 ) 6= 0 & 16 − x2 − y 2 > 0
The first condition implies that y 2 ≤ 4 therefore −2 ≤ y ≤ 2 as shaded by red. The second
condition implies that the argument of the logarithm function must √
be different from 1 so none
of the pairs (x, y) on the circle centered at the origin and radius of 15 are in the domain set.
The last condition identifies an open circle with radius 4, that is, only points inside the circle
are accepted (in blue). Summing up
D = (x, y) ∈ R2 : −2 ≤ y ≤ 2, x2 + y 2 6= 15, x2 + y 2 < 16
5
2.5
-10
-7.5
-5
-2.5
0
2.5
5
7.5
10
-2.5
-5
Figure 3: Domain for Example 3.
b) The first order partial derivatives are:
p
2x 4 − y 2
=1+
(16 − x2 − y 2 ) ln2 (16 − x2 − y 2 )
p
−2y
−y(4 − y 2 )−1/2 ln(16 − x2 − y 2 ) − 4 − y 2 (16−x
2 −y 2 )
′
fy (x, y) =
2
ln (16 − x2 − y 2 )
fx′ (x, y)
c) The set D has interior points consisting of all points that are not boundary points. It has the
border points which are of the four arcs between the two horizontal lines y = 2 and y = −2 and
those of segments detached from the circle of radius 4 on the horizontal lines (a total of four
arcs and two segments ). Of these, the two segments are contained in D with the exception of
the 4 points where each intersects the two circumferences. Therefore D contains some but not
4
all of its boundary points, so it is not closed. Since it is not consisting of only interior points, it
is not open. Moreover, since it is not closed, it cannot be compact. In the end, it is bounded
because it is entirely contained in a closed circle of radius 4 and the center of the origin.
Exercise 4. (Exam 10.01.2011) Determine and plot the domain of the function f (x, y) =
and calculate first order partial derivatives at (2,2).
q
x2 −1
y+2
Sol.: a)
x2 − 1
(x, y) ∈ R2 :
≥ 0 & y + 2 6= 0
y+2
It is important that the denominator be different from zero, therefore y 6= −2. Under the square
root must be positive so (x2 − 1)(y + 2) ≥ 0
D=
x2 − 1 ≥ 0
∨
y+2>0
x2 − 1 ≤ 0
y+2<0
⇔
⇔
x ≥ 1 or x ≤ −1
∨
y > −2
−1 ≤ x ≤ 1
y < −2
3.2
2.4
1.6
0.8
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
-0.8
-1.6
-2.4
-3.2
Figure 4: Domain for Example 4.
First order partial derivatives
r
x
1
y+2
=
⇒ fx′ (2, 2) = √
y + 2 x2 − 1
3
r
√
2
1 y+2 x −1
3
⇒ fy′ (2, 2) = −
fy′ (x, y) = −
2 x2 − 1 (y + 2)2
16
fx′ (x, y)
5
4
4.8
Exercise 5. (Exam 10.01.2012) Given the function
r
f (x, y) = e
x(y 2 +1)
y+2
+1
a) determine and plot the domain f .
b) write the equation of the 2-level curve and plot it.
c) say if the function admits global minimum.
Sol.: a)
2
2 x(y + 1)
≥ 0 & y + 2 6= 0
D = (x, y) ∈ R :
y+2
It is important that the denominator be different from zero, therefore y 6= −2. The argument
of the square root must be positive so x(y 2 + 1)(y + 2) ≥ 0. Since y 2 + 1 is always positive, we
just need to write conditions on x and y + 2 such that the expressions are both positive or both
negative, yielding a positive product. Summing up
D = (x, y) ∈ R2 : x ≥ 0, y > −2 ∪ (x, y) ∈ R2 : x ≤ 0, y < −2
5
2.5
-10
-7.5
-5
-2.5
0
2.5
5
-2.5
-5
Figure 5: Domain for Example 5.
b) The equation of the required level curve is f (x, y) = 2, more explicitly
6
7.5
10
Sol.:
r
e
x(y 2 +1)
y+2
+1
=
r
2⇔e
x(y 2 +1)
y+2
=1⇔
s
x(y 2 + 1)
=0
y+2
⇔ x = 0, y 6= 2
q 2
x(y +1)
= 0, as a square root is always
Sol.: c) Note that the function is minimal when
y+2
positive or null; then minimum points are exactly all points on the level curve at height 2.
Exercise 6. (Exam 9.10.2012) Given z =
1
2
ln(x2 + y 2 ). Prove that ∂ 2 z/∂x2 + ∂ 2 z/∂y 2 = 0.
Sol.:
∂z
∂x
∂z
∂y
∂2z
∂x2
∂2z
∂y 2
x
x2 + y 2
y
= 2
x + y2
(x2 + y 2 ) − 2x2
=
(x2 + y 2 )2
(x2 + y 2 ) − 2y 2
=
(x2 + y 2 )2
=
∂ 2 z/∂x2 + ∂ 2 z/∂y 2 =
4(x2 + y 2 ) − 4x2 − 4y 2
=0
(x2 + y 2 )2
Exercise 7. Determine and plot in R2 the domain of f (x, y) =
q
3
log10
6x−2
8
−y .
Sol.: Notice that odd root can accept positive as well as negative values. So their domain is R.
But the logarithm inside the root only accepts strictly positive values 6x−2
8 − y > 0. Hence we
1
obtain an open half plane whose boundary is the straight line of equation y < 3x
4 − 4 , that is
3x 1
2
−
D = (x, y) ∈ R : y <
4
4
Exercise 8. Determine and plot in
R2
q
the domain of f (x, y) = 6 log10
4x2 −8 +
x3 −8 y.
Sol.: The domain of absolute value function is (−∞, +∞). However, its argument is an even
root of a logarithm function, which only accepts positive values; the logarithm itself need a
7
5
2.5
-10
-7.5
-5
-2.5
0
2.5
5
7.5
10
-2.5
-5
Figure 6: Domain for Example 7.
strictly positive argument. In addition, the denominator of the quotient must be different from
zero.
4x2 − 8
4x2 − 8
2
3
D =
(x, y) ∈ R : log10 3
≥0& 3
> 0 & x − 8 6= 0
x −8
x −8
4x2 − 8
4x2 − 8
3
2
≥1& 3
> 0 & x − 8 6= 0
=
(x, y) ∈ R :
x3 − 8
x −8
4x2 − 8
2
=
(x, y) ∈ R :
≥ 1 & x 6= 2
x3 − 8
where the last line was obtained by noting that the first condition overpowers the second, being
more restrictive, as for the denominator x3 − 8 6= 0 ⇔ x 6= 2. We further expand the condition
4x2 − 8
x3 − 8
≥
⇔
Summing up
4x2 − x3
x2 (4 − x)
≥
0
⇔
≥0
x3 − 8
x3 − 8
x−4
≤0⇔2<x≤4
x3 − 8
1⇔
D = (x, y) ∈ R2 : 2 < x ≤ 4
Exercise 9. Determine and plot in R2 the domain of f (x, y) =
Sol.: (Sketch) We have
D =
=
=
=
q
log10
(6x−x2 )
.
8
2
(6x − x2 )
2 (6x − x )
>0&
≥1
(x, y) ∈ R :
8
8
(6x − x2 )
(x, y) ∈ R2 :
≥1
8
2
2
(x, y) ∈ R : x − 6x + 8 ≤ 0
(x, y) ∈ R2 : 2 ≤ x ≤ 4
8
1.5
1
0.5
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
-0.5
-1
-1.5
Figure 7: Domain for Example 8.
Exercise 10. (Difficult) Given the function f (x, y) =
√
4−x2 −y 2
:
ln(y−2x2 +8)
a) determine and plot the domain;
b) calculate the partial derivatives of the f ;
c) determine the interior and boundary points and whether the domain is closed, open, bounded,
compact?
Sol.: a) We decompose the quotient function into two parts. In the numerator, under the square
root must be always positive or zero. The denominator i) must be always different from zero ii)
the argument of logarithm must be strictly positive. Hence we write all conditions as below
D = (x, y) ∈ R2 : 4 − x2 − y 2 ≥ 0 & ln(y − 2x2 + 8) 6= 0 & y − 2x2 + 8 > 0
As for the first part, x2 + y 2 ≤ 4 forms a circle with radius 2 and origin of (0,0), where all the
points inside the circle satisfy the inequality condition as shown by red shaded area in Figure 9.
For the second inequality, we find the locus of all points where the natural logarithm in the
dominator will be different from zero. This is shown by the black line in the figure.
ln(y − 2x2 + 8) 6= 0 ⇒ y − 2x2 + 8 6= 1 ⇒ y 6= 2x2 − 7
Finally, if you rearrange the last equation, we will have y > 2(x2 − 4). This is a parabola
that starts from y = −8 and crosses the x-axis at points -2 and 2. The points that satisfy the
inequality are shown by the blue area in the Figure 9.
9
3.2
2.4
1.6
0.8
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
5.6
6.4
-0.8
-1.6
-2.4
Figure 8: Domain of function in Example 10.
2.5
-7.5
-5
-2.5
0
2.5
5
7.5
10
-2.5
-5
-7.5
Figure 9: Domain for Example 11.
b) We calculate the first order partial derivatives with respect to two variables x and y.
p
1
−4x
−1/2(2x)(4 − x2 − y 2 )− 2 ln(y − 2x2 + 8) − y−2x
4 − x2 − y 2
2 +8
∂f (x, y)
=
,
∂x
(ln(y − 2x2 + 8))2
√
4x 4−x2 −y 2
−x ln(y−2x2 +8)
√
+ y−2x2 +8
2
2
4−x −y
.
=
(ln(y − 2x2 + 8))2
10
and with respect to y
1
−1/2(2y)(4 − x2 − y 2 )− 2 ln(y − 2x2 + 8) −
∂f (x, y)
=
∂y
(ln(y − 2x2 + 8))2
√
4−x2 −y 2
−y ln(y−2x2 +8)
√
− y−2x2 +8
2
2
4−x −y
.
=
(ln(y − 2x2 + 8))2
11
1
y−2x2 +8
p
4 − x2 − y 2
,