MATH 1220—Midterm 1
Thurs., Sept. 20, 2007
• Write your name and ID number at the top of this page.
• Show all your work.
• You may refer to one double-sided sheet of notes during the exam and
nothing else.
• Calculators are not allowed.
• You have until 8:00PM to finish the exam.
• Stay strong, be cool, and don’t forget to breathe.
—Begin Midterm—
1. Find the value of each of the following expressions.
SOLUTION:
For this problem, we use the fact that the functions ln x and ex are
inverses of each other, which means
ln ex = x and eln x = x.
We need to do a little bit of algebra to get expressions (a) and (c) into
the first (left) and second (right) forms above, respectively. For expression (b), the numerator and denominator are already in the second
(right) form above.
1
(a) ln √
3
e
(b)
eln 7
eln 5
= ln e−1/3 = −
=
7
5
− ln 102
(c)
1
e
1
3
= (e−1 )− ln 100 = eln 100 = 100
2. (a) Determine sin−1 (−1).
SOLUTION:
π π
We want the angle whose sine is −1, in the range − , .
2 2
π
π
−1
= −1, so sin (−1) = − .
sin −
2
2
−1
(b) Determine tan
3π
tan
.
4
SOLUTION:
3π
= −1, so we want the angle whose tangent is −1, in the
tan
4
π π
range − ,
.
2 2
π
π
tan −
= −1, so tan−1 (−1) = − .
4
4
(c) Determine sin tan−1
5
2x
in terms of x.
SOLUTION:
5
5
Let θ = tan−1 , which means tan θ =
, and we need to find
2x
2x
sin θ.
Draw a right triangle and label one of the angles θ. Let the side
opposite θ be length 5 and the side adjacent to theta be length
2x, which reflects the fact that tan θ = 5/(2x) (opposite-overadjacent). Then use the √
pythagorean theorem to calculate the
length of the hypotenuse: 25 + 4x2 .
5
Then, sin θ =
(opposite-over-hypotenuse).
25 + 4x2
3. Find the inverse function of
s
f (x) =
x+4
− 2,
3
x ≥ −4.
Be sure to state the domain of your new function.
SOLUTION:
Let y =
q
x+4
3
− 2, and solve for x in terms of y to get
x = 3(y + 2)2 − 4.
Then replace y with x to get
f −1 (x) = 3(x + 2)2 − 4.
For any inverse function f −1 , the domain of f −1 is the same as the range
of f . Looking back at f , we see that the range is f (x) ≥ −2 (because
the square-root portion has to be positive or zero). Therefore, the
domain of f −1 (x) is x ≥ −2, or [−2, ∞)
4. Let
√
f (x) =
x2 + 1
.
(9x − 4)2
Use logarithmic differentiation to find f 0 (x).
SOLUTION:
Let
√
y=
x2 + 1
.
(9x − 4)2
Then take the natural log of both sides, and use log properties to make
things easier for differentiating later.
√
1
x2 + 1
1
2
2
2 −ln(9x−4) =
=
ln(x
+1)
ln(x2 +1)−2 ln(9x−4).
ln y = ln
(9x − 4)2
2
Then differentiate with respect to x, using implicit differentiation (chain
rule) on ln y,
1
1
1
1 dy
= · 2
· 2x − 2 ·
· 9.
y dx
2 x +1
9x − 4
Simplify,
1 dy
x
18
= 2
−
.
y dx
x + 1 9x − 4
Solve for
dy
,
dx
dy
18
x
=y 2
−
.
dx
x + 1 9x − 4
Finally, replace y with f (x) above,
√
x2 + 1
x
18
0
f (x) =
−
.
(9x − 4)2 x2 + 1 9x − 4
5. A population at time t (measured in years) is given by the function
P (t) = 450, 000e.02t .
(a) Find the doubling time for the population. Hint: Use ln 2 ≈ 0.7.
SOLUTION:
Let T be the doubling time. Since P (0) = 450, 000, it must be
true that P (T ) = 900, 000 (doubled from time 0 to time T ). Then,
900, 000 = 450, 000e.02T .
Solve for T:
2 = e.02T
ln 2 = .02T
ln 2
T =
≈ 35 years.
.02
(b) Write down the differential equation and initial condition to which
P (t) above is the solution.
SOLUTION:
P (t) above describes exponential growth, which we derived in class
from a differential equation that assumes that the rate of increase
) is proportional to the size of the population
of the population ( dP
dt
(P ), with proportionality constant k. In this case, k = .02. The
differential equation and initial condition are
dP
= .02P,
dt
P (0) = 450, 000.
6. Calculate the following integrals.
(a)
Z −1
x−1 dx
i−1
= ln |x|
−e
−e
= ln 1 − ln e = 0 − 1 = −1
√
e x
√ dx
(b)
x
SOLUTION:
Use the substitution
Z
u=
so
√
1
x = x2 ,
1 1
1
du = x− 2 = √ dx.
2
2 x
√
Z
Z √
Z
1
e x
√ dx = 2 e x √ dx = 2 eu du
x
2 x
= 2eu + C = 2e
(c)
Z
√
x
+C
sinh x cosh x dx
SOLUTION:
Use the substitution u = sinh x, so du = cosh x dx.
Then,
Z
sinh x cosh x dx =
Note that
1
2
Z
1
1
u du = u2 + C = sinh2 x + C
2
2
cosh2 x + C is another way to express the answer.
7. Find
dy
for each of the following:
dx
(a) y = log3 (sin x)
SOLUTION:
By the definition of log,
3y = sin x
Take the natural log of both sides:
ln 3y = ln(sin x)
y ln 3 = ln(sin x)
y=
ln(sin x)
ln 3
Now,
1
dy
=
· cos x
dx
(ln 3)(sin x)
(b) y = 3sin x
SOLUTION:
Take the natural log of both sides,
ln y = ln 3sin x
ln y = (sin x)(ln 3)
Now differentiate,
1 dy
= (cos x)(ln 3)
y dx
dy
= 3sin x (cos x)(ln 3)
dx
(c) y = (ln x)ln x
SOLUTION:
Take the natural log of both sides,
h
ln y = ln (ln x)ln x
i
ln y = ln x ln(ln x)
Now differentiate, using the product rule on the right-hand side
1 dy
1
1 1
= ln(ln x) + ln x ·
·
y dx
x
ln x x
dy
ln(ln x) + 1
= (ln x)ln x
dx
x
!
8. Find the particular solution to the differential equation
x
dy
− 3y = x3 + 2x,
dx
given that y = 0 when x = 1.
SOLUTION:
First we need to put the equation into the form
dy
dx
+ P (x)y = Q(x):
dy
3
− y = x2 + 2
dx x
So P (x) = − x3 , and the integrating factor is
R
e
−3
x
dx
= e−3 ln x = eln x
−3
= x−3
Multiply both sides by the integrating factor and proceed:
x−3
dy
− 3x−2 y = x−1 + 2x−3
dx
d −3
(x y) = x−1 + 2x−3
dx
−3
x y=
Z
[x−1 + 2x−3 ] dx
x−3 y = ln x − x−2 + C
y = x3 ln x − x + Cx3
Now use the initial condition to solve for C,
0 = 0 − 1 + C,
so C = 1
Finally,
y = x3 ln x − x + x3
9. Re-write the differential equation in the previous problem in the form
suitable for using Euler’s method. Then perform one step of Euler’s
method with stepsize h = 1 to approximate y(2).
SOLUTION:
dy
= f (x, y), so rearrange the
The form suitable for Euler’s method is dx
differential equation in the previous problem to get,
dy
x3 + 2x + 3y
=
dx
x
Use the initial condition in the previous problem to start off Euler’s
method: x0 = 1, y0 = 0
With h = 1, x1 = x0 + h = 2, and
y1 = y0 + hf (x0 , y0 ) = 0 + 1 · f (1, 0) =
So
y(2) ≈ 3
1+2+0
=3
1
10. Calculate the following integrals.
(a)
e4x
dx
9 − e4x
SOLUTION:
Use the substitution u = 9 − e4x , making du = −4e4x dx
Then,
Z
√
1
e4x
1Z
1Z 1
4x
√
√
√ du
dx = −
(−4e ) dx = −
4
4
u
9 − e4x
9 − e4x
Z
1
1
1 Z −1
=−
u 2 du = − · 2u 2 + C
4
4
√
1
=−
9 − e4x + C
2
(b)
e2x
dx
9 − e4x
SOLUTION:
This time, use the substitution u = e2x , making du = 2e2x dx
Then,
Z
√
Z
√
e2x
1
1
1Z
1Z
2x
q
√
dx
=
(2e
)
dx
=
du
2
2
9 − e4x
9 − u2
9 − (e2x )2
e2x
1
= sin−1
2
3
!
+C
(c)
1
dx
9 + 16x − 4x2
SOLUTION:
We can rearrange the quadratic expression under the square root
to put it in the form a2 − u2 :
Z
√
9
25
9 + 16 − 4x = −4 x − 4x −
= −4 x2 − 4x + 4 −
4
4
2
2
25
=4
− (x − 2)2
4
The integral becomes,
1
Z
q
4[ 25
− (x − 2)2 ]
4
dx
=
1Z
1
q
dx
25
2
− (x − 2)2
4
Use the substitution u = x − 2, so that du = dx, and proceed,
!
1
5/2
1Z
1
q
+C
du = sin−1
25
2
2
u
− u2
4
!
1
5
= sin−1
+C
2
2(x − 2)