Practice Problems for Midterm 2
Here are some problems for you. Only some of these are my own, the rest I gathered from a variety of
sources.
1. Solve the following differential equations. If initial conditions are provided, find the unique solution.
(a)
dy
dx
=
xy+y
x2 +1 ,
2
y(0) = 1. Solve explicitly for y.
x(x +1)
,
y3
(b)
dy
dx
(c)
dy
dx
= xy(y − 2). First try solving without initial conditions, and solving explicitly for y. Then
solve using y(0) = 0.
(d)
dy
dt
=
y(0) = −1. Solve explicitly for y.
= ey−t sec(y)(1 + t2 ). Do not try to solve explicitly for y. This one I mentioned in discussion,
but make sure you are comfortable doing it.
dy
(e) (1 + ex ) dx
− cot2 (y) = 0.
(f)
dy
dx
= y2 x − 1 − y2 + x
(g)
dy
dx
=
(h)
dy
dx
= cos(x + y) − cos(x − y). I will give you that
Z
csc(x)dx = − ln | csc(x) + cot(x)| + C.
ln(xy 2 )
ln(y)
− 2.
2. Consider the differential equation
dP
=P
dt
P
1−
100
.
(a) Find all equilibria.
(b) Let f (P ) = P 1 −
P
100
. By graphing f (P ), determine the stability of the equilibria in (a).
(c) Now analyze the stability using the other approach, i.e. by considering f 0 (P ).
(d) If P (0) = 50, what do you expect
lim P (t)
t→∞
to be? Same question with P (0) = 150, and P (0) = 100.
(e) Solve the differential equation explicitly for P with initial conditions P (0) = 150, P (0) = 50, and
P (0) = 100.
(f) Using your answer in (e), take the limit as t → ∞ in the each of the three cases and check your
answer in (d).
3. Consider the differential equation
dy
= (y − 2) ln(y + 1),
dt
y > −1.
Find all equilibria and analyze stability (using any approach).
4. Suppose


0
 1 

u=
 −1  ,
2
Calculate the following:
(a) u + 2v.
1


−1
 −2 

v=
 0 .
3
(b) A unit vector in the direction of u.
(c) The angle between u and v.
(d) Is there a number a such that the vector


2
 a 

w=
 1 
a
is perpendicular to u? If so, find it.
5. Consider the plane P which passes through the point (1, 0, 1) and is perpendicular to the vector [1, 2, 2].
(a) Find an equation for P .
(b) Find parametric equations for the line passing through (1, 3, 3) in the direction of [3, 0, −1].
(c) Is there a point of intersection between the line and the plane? If so, find it.
The next three exercises deal with level curves. If you are still uncomfortable with the
subject, please look at the exercises in 10.1.
6. Consider the function f (x, y) = ey−x
2
−1
.
(a) Find the domain and range of f (x, y).
(b) For which values of c does the level curve f (x, y) = c have no points?
(c) What do the level curves of f look like for valid values of c (namely, the ones not in (b)).
(d) Sketch the level curve for c = 1.
7. Let g : R2 → R be defined as
g(x, y) =
x
.
y
(a) Find the domain and range of g.
(b) What does the level curve of g at c = 0 look like?
(c) Sketch the level curve of g with c = 1.
8. (Slightly harder) Let g : R2 → R be defined as
g(x, y) =
sin(x)
.
y
(a) Find the domain and range of g.
(b) What does the level curve of g at c = 0 look like?
(c) Sketch the level curve of g with c = 1.
9. What are the domain and range of the following functions?
p
(a) f (x, y) = 1 − x2 − y 2
p
(b) f (x, y, z) = ln(x + y + z)
(c) f (x, y) = e
√
xy
10. Consider
(
f (x, y) =
x3 y
x6 +y 2
0
(a) Calculate lim(x,y)→(1,1) f (x, y).
2
if (x, y) 6= (0, 0)
.
if (x, y) = (0, 0)
(b) Find lim(x,y)→(0,0) f (x, y) along the curve y = x.
(c) Find lim(x,y)→(0,0) f (x, y) along the curve y = x3 . What can you conclude about lim(x,y)→(0,0) f (x, y)?
(d) Is f (x, y) continuous at (0, 0)? Explain your answer.
11. Calculate the partials fx , fy (and fz where appropriate) for the following functions.
(a) f (x, y) = sin(x2 y).
2
(b) f (x, y) = tan(xy) + exy +
(c) f (x, y, z) =
ln(y)
x
y sin(xz)
.
z
yz
(d) f (x, y, z) = x .
2
12. If f (x, y) = exy , find fxxy . That is, find
∂3f
.
∂y∂x∂x
13. Consider the function
f (x, y) = tan−1 (xy 2 ).
(a) What are the domain and range of f ?
(b) What is
lim
f (x, y)?
(x,y)→(1,1)
(c) What is
lim
f (x, y)?
(x,y)→(0,0)
(d) Show that f (x, y) is differentiable at every point in R2 .
(e) Find the tangent plane to f (x, y) at (0, 0) and (1, 1) (that is, I want two different planes, one for
each point).
(f) Find the linearization of f (x, y) at (1, 1), and use it to approximate f (1.1, .9) (you can leave you
answer in terms of a decimal if you need to).
14. Consider the vector valued function f : R2 → R2 given by
x
e cos(y)
f (x, y) =
.
ex sin(y)
(a) Calculate the Jacobian of f (what the book called the Jacobi matrix).
(b) Calculate the Jacobian at (0, 0) and (1, π/2).
(c) Use these two give linear approximations to f at (0, 0) and (1, π/2).
3
Solutions:
√
−1
(1) (a) y = etan (x) · x2 + 1.
√
√
(b) y = − 4 x4 + 2x2 + 1. Equivalently, y = − x2 + 1.
(c) Without initial conditions: y =
(d)
e
−y
(sin(y)−cos(y))
2
2
1+Cex2
. With the initial conditions: y = 0.
= −e−t (t2 + 2t + 3) + C.
(e) (HINT: At some point you will need to use u = 1 + ex , and use tan2 (y) = sec2 (y) − 1) tan(y) − y =
x − ln(1 + ex ) + C.
2
2
(f) (HINT: Factor the right side) tan−1 (y) = x2 − x + C, or equivalently y = tan x2 − x + C .
(g) (HINT: Logarithm rules) y ln(y) − y = x ln(x) − x + C.
(h) (HINT: cos(x + y) = cos(x) cos(y) − sin(x) sin(y) and cos(x − y) = cos(x) cos(y) + sin(x) sin(y))
Solution: − ln | csc(y) + cot(y)| = 2 cos(x) + C.
(2) (a) The equilibria occur when
dP
dt
= 0, so P = 0 and P = 100.
(b) The equilibrium P = 0 is unstable, and P = 100 is stable.
(c) f 0 (100) < 0 and f 0 (0) > 0, so 0 is unstable and P = 100 is stable, which is what we found in (b).
(d) If we start at 50, we would expect the solutions to move away from 0 and towards 100, so we
expect the limit to be 100. Similarly, we expect the limit to be 100 in each of the other two cases.
(e) We get
P (t) =
Cet
1+
,
1
t
100 Ce
or any simplified version of this. When P (0) = 150, C = −300. If P (0) = 50, then C = 100.
So in each of these cases, we plug these values for C into the equation above to get P (t). In the
case P (0) = 100, we run into trouble, because we can’t solve for C. However, we note that we
divided by 0 in the beginning, and so the unique solution is the constant solution P (t) = 100
when P (0) = 100.
(f) You will have to use L’Hopital’s rule in the first two cases, but you should get 100 in each case.
(3) The equilibria are at y = 0 and y = 2. The easiest approach is to take the derivative of f (y) =
(y − 2) ln(y + 1) to find that 0 is stable and 2 is unstable.
(4) (a) This was a simple calculation:


−2
 −3 

u + 2v = 
 −1  .
8
(b) The length of u is |u| =
(c) The length of v is |v| =
√
√
u·u=
√
6, and so a unit vector in the direction of u is


0√
 1/ 6 
u
√ .
=
|u|  −1/√ 6 
2/ 6
14, and u · v = 4, and since u · v = |u||v| cos(θ), we get
4
θ = cos−1 √ √
.
6 14
(d) By setting u · w = 0, we see the answer is yes, and a = 1/3.
4
(5) (a) An equation for P is
P : (x − 1) + 2y + 2(z − 1) = 0.
(b) The parametric equations are given by
x = 3t + 1,
y = 3,
z = 3 − t.
(c) Plug in the equations for x, y, z in (b) into the equation for the plane in (a) and see if there are
any solutions for t. Indeed, there is one, and it comes when t = −10, so the point of intersection
is (−29, 3, 13).
(6) (a) The domain is all of R2 and the range is {z ∈ R : z > 0} (i.e. the positive real numbers).
2
(b) Setting f (x, y) = c gives ey−x −1 = c. Since the range of the exponential function consists of
positive real numbers, if c ≤ 0, then the level curve will have no points on it.
(c) If c > 0, then the level curve is the parabola y = x2 + [1 + ln(c)].
(d) When c = 1, we get ln(c) = 0, so your graph should be that of y = x2 + 1.
(7) (a) The domain is {(x, y) ∈ R2 : y 6= 0} (i.e. any point with y 6= 0). The range is all of R.
(b) The level curve at c = 0 looks like the vertical line x = 0 with the point (0, 0) removed (since
y cannot equal 0).
(c) It should be the graph of y = x with a hole at the origin (since y cannot equal 0).
(8) (a) The domain is {(x, y) ∈ R2 : y 6= 0} (i.e. any point with y 6= 0). The range is all of R.
(b) The level curve at c = 0 looks like a bunch of vertical lines at x = 0, ±π, ±2π, ..., except the
point y = 0 is removed from each one.
(c) It should be the graph of y = sin(x) minus the points where y = 0 (i.e. the points where
graph of y = sin(x) crosses the x-axis should be removed, and there should just be a hole in
the graph).
(9) (a) The domain is {(x, y) ∈ R2 : x2 + y 2 ≤ 1}, and the range is {z ∈ R : 0 ≤ z ≤ 1}.
(b) The domain is {(x, y, z) ∈ R3 : x + y + z > 1} and the range is the nonnegative reals:
{z ∈ R : z ≥ 0}.
(c) The domain is {(x, y) ∈ R2 : xy ≥ 0} and the range is the positive real numbers greater or
equal to 1 {z ∈ R : z ≥ 1}.
(10) (a) The function is nice and continuous at (1, 1), and so the limit equals the value of the function,
which is 1/2.
(b) The limit should be 0.
(c) The limit should be 1/2. Since 0 6= 1/2, we get different limits along different curves, so the
limit does not exist.
(d) No, because in order for the function to be continuous, the limit needed to equal the value of
the function. We have f (0, 0) = 0 but the limit does not exist.
(11) (a) fx = 2xy cos(x2 y), fy = x2 cos(x2 y)
2
(b) fx = y sec2 (xy) + y 2 exy −
ln(y)
x2 ,
2
fy = x sec2 (xy) + 2xyexy +
1
xy .
, fz = y[zx cos(xz)−sin(xz)]
.
(c) fx = y cos(xz), fy = sin(xz)
z
z2
(d) fx = yzxyz−1 , fy = zxyz ln(x), fz = yxyz ln(x).
2
2
(12) fxxy = 4y 3 exy + 2xy 5 exy .
(13) (a) The domain of f (x, y) is all of R2 , and the range is the interval (−π/2, π/2 in R.
(b) f (x, y) is continuous everywhere as g(x) = tan−1 (x) is a continuous function in R, so to
evaluate the limit we can just plug in the point, and tan−1 (1) = π/4.
(c) By the same logic as in (b), the limit is tan−1 (0) = 0.
5
(d) Observe that the partials are
fx =
y2
,
1 + x2 y 4
fy =
2xy
,
1 + x2 y 4
which are both continuous everywhere as the denominator can never be 0 (the monomial
x2 y 4 ≥ 0 always). Therefore f is differentiable everywhere.
(e) We calculate the partials already. We get fx (0, 0) = fy (0, 0) = 0, and since f (x, y) = 0,
the plane at (0, 0) is simply z = 0. At (1, 1), fx (1, 1) = 21 , and fy (1, 1) = 1, and since
f (1, 1) = π/4, the plane is
z−
1
π
= (x − 1) + (y − 1).
4
2
(f) The linearization consists solely of isolating z in the previous part, so the linearization is
z = L(x, y) =
So
L(1.1, .9) =
π 1
+ (x − 1) + (y − 1).
4
2
π
1
π 1
+ (1.1 − 1) + (.9 − 1) = − .
4
2
4
20
(14) (a) We get the Jacobian as
(Df )(x, y) =
ex cos(y) −ex sin(y)
ex sin(y) ex cos(y)
(b) At (0, 0) and (1, π/2), we get the matrices
1 0
,
0 1
0
e
−e
0
.
,
respectively.
(c) Notice f (0, 0) =
1
0
and f (1, π/2) =
L(x, y) =
and
L(x, y) =
0
e
respectively.
6
1
0
+
0
e
, and so the linear approximations are
+
1
0
0
e
−e
0
0
1
x
y
x−1
y − π/2
,