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MATH1013 Quizzes

Quiz 7: Exponential growth and decay
Question 1 Questions
Consider y = ekx. Find dy dx.
a)
dy dx = ekx
 b)
dy dx = kekx + C
c)
dy dx = 1 kekx + C
 d)
dy dx = kekx

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Consider y = lnkx. Find dy dx.
a)
dy dx = 1 x
 b)
dy dx = k x
c)
dy dx = 1 kx
 d)
dy dx = 1 x + 1 k

Choice (a) is correct!
Note that y = lnkx = lnx + lnk. Hence dy dx = 1 x + 0.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find y = k xdx.
a)
y = lnx + kx
 b)
y = lnkx + C
c)
y = klnx + C
 d)
y = kx2 + C

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find y =e4xdx.
a)
y = 4e4x + C
 b)
y = 1 4e4x + C
c)
y = Ae4x
 d)
y = A 4 e4x

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Consider the expression klnx + C where k and C are positive constants. Which of the following statements are true ?
a)
klnx + C = lnAxk where A is a constant.
b)
klnx + C = lnCxk
c)
klnx + C = lnxk + C
d)
klnx + C = klnCx

There is at least one mistake.
For example, choice (a) should be True.
klnx + C = lnxk + C = lnxk + lnA, where lnA = C. So klnx + C = lnAxk
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
Correct!
  1. True klnx + C = lnxk + C = lnxk + lnA, where lnA = C. So klnx + C = lnAxk
  2. False
  3. True
  4. False
Solve the equation e2t + 5 = 20.
a)
t = ln4 2
 b)
t = ln25 2
c)
t = ln7.5
 d)
t = ln15 2

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
e2t = 15 2t = ln15 t = ln15 2 .
Solve the equation ln2t + ln3 = 8.
a)
t = e8 6
 b)
t = e8 2 3 2
c)
t = e4 3 2
 d)
t = e2 23

Choice (a) is correct!
ln2t + ln3 = ln6t = 8 6t = e8 t = e8 6 .
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Consider a population P where P = P0e0.02t and t is measured in years. How long will it take for the population to double ?
a)
1.96 years.
 b)
34.66 years.
c)
2.3 years.
 d)
There is not enough information given.

Choice (a) is incorrect
Choice (b) is correct!
P = 2P0 = P0e0.02t e0.02t = 2 0.02t = ln2
t = ln2 0.02 = 34.66.
Choice (c) is incorrect
Choice (d) is incorrect
The amount of a given radioactive isotope at time t (in years) is given by P = P0ekt where k = 5 × 106. What is the half-life of the isotope ?
a)
139000 years.
 b)
6.1 years
c)
12.9 years
 d)
There is not enough information given.

Choice (a) is correct!
P = 1 2P0 = P0ekt ekt = 1 2 kt = ln 1 2
kt = ln2 t = ln2 5 × 106 = 138629.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
A population P grows exponentially. If it takes 15 years for the population to double, how long does it take for the population to treble ?
a)
22.50 years.
 b)
23.77 years
c)
16.48
 d)
There is not enough information given.

Choice (a) is incorrect
Choice (b) is correct!
Since P grows exponentially P = P0ekt for some k a positive constant.
P = 2P0 = P0e15k e15k = 2 k = ln2 15
P = 3P0 = P0ekt ekt = 3 ln2 15 t = ln3 t = 15ln3 ln2 = 23.77.
Choice (c) is incorrect
Choice (d) is correct!