METHODS OF CALCULUS SUMMER 2012
EXAM-3
NAME: ......................................................................Z#:...........................................
Maximum Points: 100
Points Obtained.................
Time: 1hr 15 minutes
You MAY use a scientic calculator but NOT a graphing or programmable calculator.
ALL WORK MUST BE CLEARLY SHOWN FOR EACH QUESTION. DO NOT DETACH PAGES ! ! ! 1) Solve for x:
(4×2.5 = 10 Points)
a) e3x = 1
Ans: Taking ln both sides we get 3x = ln1
or
3x = 0
(because ln1=0)
x=0
b) lnx − lnx2 + ln2 = 0
2
Ans:ln x×2
x2 = 0 =⇒ ln x = 0 .
=⇒ x = 2
c) ln(ln(2x))= 0
Ans:ln(2x) = 1 =⇒ 2x = e
x=
e
2
d) (1 + x) × 2−x − 5 × 2−x = 0
2−x (1 + x − 5) = 0
x − 4 = 0 because 2−x 6= 0.
Ans: x = 4
2) Dierentiate the following functions.
a) y = 3ex − 7x
1
(4×2.5=10 Points)
y 0 = 3ex − 7
b) y = ln(x + 3) + ln5
Ans: y 0 =
1
x+3
c) y = ex +2x−5
6
y 0 = (6x5 + 2)ex
6
+2x−5
d) y = ln(e2x (x3 + 1))
Ans: y = ln(e2x ) + ln(x3 + 1)
y = 2x+
y0 = 2 +
ln(x3
+ 1)
3x2
x3 +1
3) Find the slope of the tangent line to the curve y = xex at (1, e).
...................................................................................................(10 points)
y 0 = ex + xex
0
So at (1, e), y = e + 1 × e = 2e is the required slope of the tangent line
to the curve at (1, e).
Ans: Here
4) Find the relative extrema of the function f (x) = xlnx − x.
..........................................................................................(15 points)
Ans: Here f 0 (x) = lnx + x x1 − 1 = lnx + 1 − 1 = lnx
For the critical number we put f 0 (x) = 0 and solve for x .
Thus lnx = 0 =⇒ x = 1.
Again f 00 (x) = x1 .
Testing the second derivative at the critical number x = 1, we get
f 00 (1) = 1 > 0.
2
Thus the function attains its minimum value at x = 1 and the minimum value is f (1) = 1 × ln1 − 1 = −1.
Thus the relative minimum point is (1, −1).
2
2
(x+2)
] at x = 0.(10 points)
5) Find the derivative of f (x) = ln[ (x+1)
(x+3)3
Ans: Here f (x) = ln(x + 1)2 + ln(x + 2)2 − ln(x + 3)3 = 2ln(x + 1) +
2ln(x + 2) − 3ln(x + 3)
f 0 (x) =
2
x+1
+
2
x+2
−
3
x+3
f 0 (0) = 2 + 1 − 1 = 2.
6) Let P (t) be the population (in millions) of a certain city t years
after 1990, and suppose that P (t) satises the dierential equation
P 0 (t) = 0.02P (t), P (0) = 3, 000, 000.
(5 × 3 = 15 points)
a) Find the formula for P (t).
Ans: P (t) = 3, 000, 000e0.02t
b) What was the initial population, that is, the population in 1990?
Ans: 3,000,000
c) What is the growth constant?
Ans: 0.02
d) Use the dierential equation to determine how fast the population
is growing when it reaches 4 million people.
Ans: P 0 (t) = 0.02 × 4, 000, 000 = 80, 000
e) How large is the population when it is growing at the rate of 70,000
people per year?
Ans:
70, 000 = 0.02 × P (t)
3
Thus P (t) = 3, 500, 000
7) The decay constant for the radioactive element cesium 137 is 0.023
when time is measured in years. Find its half-life.
(10 Points)
Ans: λ = 0.023
P (t) = P0 e−λt
P0
2
= P0 e−0.023t
−0.023t = ln( 12 )
t = 30.1 yrs
8) One thousand dollars is invested at 5% interest compounded continuously.
........................................................................... (3+3+4+5=15 points)
a) Give the formula for A(t), the compound amount after t years.
Ans: A(t) = 1, 000e0.05t
b) How much will be in the account after 6 years?
Ans: A(6) = 1, 000e0.05×6 ≈ $1349.86
c) After 6 years at what rate will A(t) be growing?
Ans: A0 (t) = 0.05A(t)
A0 (6) = 0.05 × A(6) = 0.05 × 1349.86 ≈ $67.48 per year
d) How long is required for the initial investment to double?
Ans: 2000 = 1000e0.05t
e0.05t = 2
ln2 ≈ 13.86
t = 0.05
9) Fill in the blanks below:
(5×2=10 points)
a) ln(ln(e))=0
4
b) e2ln(3) = ...
c) dtd [lnt] =
d)
d
x
dx [2 ]at
1
t
x = 0 is ln2
√
e) ln( e) = 12
GOOD LUCK!
5