Download Test 3

M-408M
Test 3A - 4A
Old exams
Proposed solutions by the TA
(Disclaimer: this might contain errors. Use at your own risk. Let me know if you find any)
Test 3
Question 4: Show that
x2 − y 2
.
(x,y)→(0,0) x2 + y 2
lim
does not exist.
A: Your job is to find two possible paths taking (x, y) to (0, 0) for which the function
approaches two different values (i.e, two different paths for which the limit above disagree),
yielding that the limit does not exist. Take x = y as x → 0. Then (x, y) → (0, 0), and in
2 −y 2
x2 −x2
0
this path we have that xx2 +y
2 = x2 +x2 = 2x2 = 0 for any (x, y) such that x = y, x 6= 0. Thus
lim(x,x)→(0,0)
x2 −y 2
x2 +y 2
= 0. On the other hand if we pick the y → 0 and x = 0, we have
2
− yy2
2 −y 2
limy→0 (0, y) xx2 +y
2
x2 −y 2
x2 +y 2
=
= −1, for any (x, y) such that x = 0 and y 6= 0. Therefore,
= −1.
(You could also have picked x → 0 and y = 0. In this path, the limit of the function is 1 check that as an exercise).
1
Question 5: True or False0
(t)×r00 (t)||
a) The curvature is κ(t) = ||r ||r
. TRUE.
0 (t)||3
b) The unit tangent vector and the principle normal vector at a point on a vector curve
are perpendicular. TRUE.
c) The gradient at a point on a level curve is perpendicular to the level curve. TRUE.
d) It is always true that fxy = fyx . FALSE, a sufficient conditions is that the functions
are continuous.
e) The normal line to a surface S at a point P is the line passing through P and perpendicular to the tangent plane at P. TRUE.
2
Question 6: Find an equation of the plane tangent to the surface z = 4x2 −y 2 +2y
at the point (−1, 2, 4).
A: Consider an equation F (x, y, z) = 0. Then the plane to the surface defined by the
equation at a point (x0 , y0 , z0 ) is given by
Fx (x0 , y0 , z0 )(x − x0 ) + Fy (x0 , y0 , z0 )(y − y0 ) + Fz (x0 , y0 , z0 )(z − z0 ) = 0.
In our case we were given z = 4x2 −y 2 +2y, thus making a small adjustment our equation
is
z − 4x2 − y 2 − 2y = 0.
Hence F (x, y, z) = z − 4x2 − y 2 − 2y. We are asked the plane tangent to this surface at
the point (−1, 2, 4). Thus, we need to compute Fx (−1, 2, 4), Fy (−1, 2, 4) and Fz (−1, 2, 4).
Note that
Fx = −8x
Fy = −2y − 2
Fz = 1
Hence,
Fx (−1, 2, 4) = 8
Fy (−1, 2, 4) = −6
Fz (−1, 2, 4) = 1
and the equation of the plane is given by
2(x + 1) − 6(y − 2) + 1(z − 4) = 0
which is equivalent to
2x − 6y + z + 10 = 0
3
Question 7: Find the second order partial derivatives of f (x, y) = x3 + x2 y 3 − 2y 2
A: The second-order derivatives are just fxx , fxy , fyx and fyy . Since the function is
continuous we have that fxy = fyx (you can save time by just computing 3 partials instead
of 4). In our case, fx = 3x2 + 2xy 3 and fy = 3x2 y 2 − 4y. Therefore,
fxx = 6x + 2y 3
fyy = 6x2 y − 4
fxy = 6xy 2
(Check that, indeed, fxy = fyx .)
4
Question 8: If z = ex sin y , where x = st2 and y = s2 t, find
A: For a function z = f (x(s, t), y(s, t)) we have
∂z
∂s
∂z
∂t
=
=
∂z ∂x
∂x ∂s
∂z ∂x
∂x ∂t
+
+
∂z
∂y
∂z
∂y
∂y
∂s
∂y
∂t
∂z
∂s
and
(1)
(2)
In our case, f (x(s, t), y(s, t)) = ex sin y, x(t, s) = st2 and y(t, s) = s2 t so
∂z
∂x
∂z
∂y
∂x
∂s
∂y
∂s
∂x
∂t
∂y
∂t
= ex sin y
= ex cos y
= t2
= 2st
= 2st
= s2
(3)
(4)
(5)
(6)
(7)
(8)
Pluggin (3) − (8) into (1) and (2) we find:
∂z
∂s
∂z
∂t
= ex sin(y)t2 + 2ex cos(y)st
= 2ex sin(y)st + ex cos(y)s2
5
∂z
.
∂t
Question 9: Find the local maximum and minimum values and saddle points
of f (x, y) = x4 + y 4 − 4xy + 1.
A: The question asks to find the critical points and say what they are. Setting up the
first derivatives to zero gives us the critical point. The second derivative test tells us what
they are.
The derivatives are
fx (x, y) = 4x3 − 4y
fy (x, y) = 4y 3 − 4x
fx (x, y) = 0 and fy (x, y) = 0 simultaneously when x3 = y (1) and y 3 = x (2). If we plug
in (2) into (1) we conclude that y 9 = y, which is only true if y = 1, y = 0 or y = −1. Thus
the critical points are (0, 0), (1, 1) and (−1, −1).
Now let’s perform the second derivative test. At a critical point (a, b) compute D as
follows
D(a, b) = fxx (a, b) − fyy (a, b) − (fxy (a, b))2 .
If D(a, b) > 0 and fxx (a, b) > 0, then f (a, b) is a local minimum. If D(a, b) > 0 and
fxx (a, b) < 0 then f (a, b) is a local maximum. If D(a, b) < 0, then f (a, b) is a saddle point.
Remark: If D(a, b) = 0 the test gives no information.
Now if we compute the second-order partial derivatives we find fxx (x, y) = 12x2 , fyy (x, y) =
12y 2 and fxy (x, y) = −4. Then
= fxx (0, 0)fyy (0, 0) − fxy (0, 0)2
= −16
D(1, 1)
= (12) × (12) − 16
= 128
D(−1, −1) = (12) × (12) − 16
= 128
D(0, 0)
Hence, (0, 0) is a saddle point, (1, 1) is a local minimum, and (−1, −1) is another point
of local minimum.
6
Question 10: a) Find the maximum slope on the graph of f (x,√
y) = 2 sin(xy) at
the point P (0, 4). b) Find the directional derivative of f (x, y) = 6x + 3y at the
point (2, 2) in the direction v = i + j.
a) A: Finding the maximum slope of the graph corresponds to maximize the directional
derivative.
By theorem 15 (section 15.6) it takes place when the direction is the same as the gradiant
and is maximum is ||Of ||. Computing the
√ gradient we find Of (x, y) = h2y cos(xy), 2x cos(x, y)i,
so Of (0, 4) = h8, 0i, and ||Of (0, 4)|| = 82 = 8. (alternative (ii)).
b) A: The directional derivative is given by
Of ·
v
||v||
6
3
In our case, Of (x, y) = h 2√6x+3y
, 2√6x+3y
i for all x, y, in particular, Of (2, 2) = h 2√2 2 , 2√1 2 i.
√
√
v
v = i + j or v = h1, 1i and so ||v|| = 12 + 12 = 2. Therefore, ||v||
= h √12 , √12 i, and so
v
Of · ||v||
= h 2√2 2 , 2√1 2 i · h √12 , √12 i = 34 (alternative (iii)).
7
Test 4A
Question 16: Determine the minimum value of f (x, y) = 3x − 4y + 3 subject to
the constraint g(x, y) = x2 + y 2 − 1 = 0.
A: Following the method of Lagrange multiplier we need to find the values of (x, y) and
λ such that
Of (x, y) = λOg(x, y)
with
g(x, y) = 0.
In our case, Of (x, y) = h3, −4i and Og(x, y) = h2x, 2yi. Then we need to find (x, y) and
λ such that 3λ = 2x (1), −4λ = 2y (2) and x2 + y 2 = 1 (3). Let’s solve for λ: From (1),
4
and from (2), y = −2λ. Now using (3), 49 λ2 + 4λ2 = 1, so λ2 = 25
or λ = +/ − 25 . It
x = 3λ
2
3
4
follows that x = +/ − 5 and y = −/ + 5 .
= 28
Thus we need
to evaluate the function at 53 , − 45 and − 53 , 45 , so f 35 , − 54 = 9+16+3
5
5
22
22
and f − 35 , 54 = −9−16+3
=
−
.
Thus
the
miminum
is
−
−
and
it’s
achieved
when
5
5
5
(x, y) = − 35 , 54 .
8
R3R3
Question 17: Evaluate the integral I = 1 1 xy + xy dydx.
R3
A: We first compute I1 = 1 xy + xy dy treating x as a constant. We find
R3
I1 = 1 xy + xy dy
3
y2 = −x ln y|31 + 2x
1 = x[ln 3 − ln 1] + x1 9−1
2
= x ln 3 + x4
R3
Next we compute I = 1 I1 dx,
R3
I = 1 x ln 3 + x4 dx
3
2
= ln 3 x2 + 4 ln x|31
1
= 4 ln 3 + 4 ln 3
= 8 ln 3
9
R1R1
Question 18: Evaluate I = 0 y2 2y cos(x2 )dxdy by chaging the order of integration.
A: The region one wants to integrate is given by D√= {(x, y) : 0 ≤ y ≤ 1; y 2 ≤ x ≤ 1}
which is equivalent to D = {(x, y) : 0 ≤ x ≤ 1; 0 ≤ y ≤ x}.
Thus
RR
R1R1
I = 0 y2 2y cos(x2 )dxdy =
2y cos(x2 )dxdy
R 1 RD√x
= 0 0 2y cos(x2 )dydx
√ R1
x
= 0 y 2 cos(x2 )|0 dx
R1
= 0 x cos(x2 )dx
R1
= 0 21 cos(u)du
(By making u = x2 )
1
1
= 2 sin(u)|0
= 12 sin(1)
10