MATH 1520 Quiz 1C Solutions
1. Find an equation of the line through the point (3, 2) and perpendicular to the line
3x + 5y = 8. [5]
The slope of the line perpendicular is the negative reciprocal of the line 3x + 5y = 8.
The latter line can be rearranged
3
MATH 1520 A03 Winter 2013 Assignment 2 Solutions
1. At the start of an experiment there are 10000 bacteria. The population is increasing
exponentially. After 2 hours, there are 12000 bacteria. How long will it take for the
population to grow to 17000?
Sin
UNIVERSITY OF MANITOBA
DATE: October 30th, 2013
DEPARTMENT & COURSE NO: MATH 1520
EXAMINATION: Calc. for Mgmt. and Soc. Sci.
MIDTERM
PAGE: 1 of 5
TIME: 1 hour
EXAMINER: N. Harland
x2 3x
[4] 1. Determine the domain of the function f (x) =
.
x2 16
The rest
MATH 1520 Assignment 1 Solutions
1. Find k so that the line through (3, 5) and (1, k) is perpendicular to 5x + 3y = 4.
First we put the line 5x + 3y = 4 into slope-intercept form in order to nd its slope.
5
4
5x + 3y = 4 3y = 5x + 4 y = x +
3
3
hence the
Winter 2010 Midterm
1. Find the values of the following limits. If the limit does not exist, write or or
simply Does Not Exist as appropriate.
1+x2
(a) lim e 1x2
x0
x(x2 9)
x3
x3
(c) lim ln(x + x2 )
(b) lim
x0
x2 + ex
x 1 + x100
(d) lim
2. Find the deriva
MATH 1520 Assignment 4 Solutions
1. Let f (x) =
(x2
x
. Then the derivatives are
1)2
f (x) =
3x2 + 1
12x3 + 12x
and f (x) =
.
(x2 1)3
(x2 1)4
(You do not have to show how to nd the derivatives.) Find the following information
about f.
(a) The function i
April 2007 Final Exam (Slightly Modied)
Part A These questions are worth a small amount. Full marks are given for the correct
answer. Partial marks may be considered if work is shown.
1. A displacement function is x(t) = t3 + 2t2 + 4 metres, with t 0 in s
MATH 1520 A01 Winter 2013 Problem Workshop 2 Solutions
1. (a) The lines have p intercepts 0.2 and 1.2 and slopes 1 and 1 respectively. Hence
the graph is given below.
(b) The equilbrium price can be found by setting the two curves equal to each other
q +
MATH 1520 Problem Workshop 6 Solutions
1. (a)
f (x + h) f (x)
h
(x + h)2 3(x + h) (x2 3x)
= lim
h0
h
2
2
x + 2xh + h 3x 3h x2 + 3x
= lim
h0
h
2xh + h2 3h
= lim
h0
h
h(2x + h 3)
= lim
h0
h
= lim (2x + h 3)
f (x) = lim
h0
h0
= 2x 3.
(b)
f (x + h) f (x)
h
4
MATH 1520 Problem Workshop 5 Solutions
1. For innite limits we factor out the largest power of x in the numerator and denominator. Then we use that lim x1n for any n > 0.
x
5
3
x 2 4 + x x2
4x2 + 5x 3
= lim
2
x
x
3x2 2
x2 3 x2
lim
5
3
4 + x x2
2
x
3 x2
4+
MATH 1520 Quiz 2C Solutions
1. (a) An exponential function has the form y = y0 ekt where y0 is the inital amount which
is this case is 30, 000. Since there is 50, 000 after 7 hours we use t = 7 y = 50, 000
to solve for k.
y = 30000ekt
50000 = 30000e7k
5
=
MATH 1520 Quiz 2A Solutions
1. (a) An exponential function has the form y = y0 ekt where y0 is the inital amount which
is this case is 60, 000. Since there is 20, 000 after 7 hours we use t = 7 y = 20, 000
to solve for k.
y = 60000ekt
20000 = 60000e7k
1
=
MATH 1520 Quiz 1D Solutions
1. Find an equation of the line through the point (4, 3) and perpendicular to the line
2x 5y = 8. [5]
The slope of the line perpendicular is the negative reciprocal of the line 2x 5y = 8.
The latter line can be rearranged
8
2
2
MATH 1520 Quiz 1A Solutions
1. Find an equation of the line through the point (2, 1) and perpendicular to the line
3x 4y = 8. [5]
The slope of the line perpendicular is the negative reciprocal of the line 3x 4y = 8.
The latter line can be rearranged
3
3x
MATH 1520 Quiz 1E Solutions
1. Find an equation of the line through the point (5, 4) and perpendicular to the line
2x 3y = 8. [5]
The slope of the line perpendicular is the negative reciprocal of the line 2x 3y = 8.
The latter line can be rearranged
8
4
2
MATH 1520 Quiz 1B Solutions
1. Find an equation of the line through the point (2, 1) and perpendicular to the line
3x 4y = 8. [5]
The slope of the line perpendicular is the negative reciprocal of the line 3x 4y = 8.
The latter line can be rearranged
3
3x
MATH 1520 Quiz 2B Solutions
1. (a) An exponential function has the form y = y0 ekt where y0 is the inital amount which
is this case is 60, 000. Since there is 40, 000 after 6 hours we use t = 6 y = 40, 000
to solve for k.
y = 60000ekt
40000 = 60000e6k
2
=
MATH 1520 Quiz 2D Solutions
1. (a) An exponential function has the form y = y0 ekt where y0 is the inital amount
which is this case is 60, 000. Since there is 100, 000 after 7 hours we use t = 7
y = 100, 000 to solve for k.
y = 60000ekt
100000 = 60000e7k
MATH 1520 Quiz 3A Solutions
1. (a)
x2 2x 8
x2 (1 2/x 8/x2 )
= lim
x
x
4 2x2
x2 (4/x2 2)
1 2/x 8/x2
= lim
x
4/x2 2
100
=
02
1
=
2
lim
x2 7
is a of non zero/0 form and hence it goes to one of the innites.
x2 x2 4
To determine which, we need to test whether
MATH 1520 Quiz 2A Solutions
1. (a) An exponential function has the form y = y0 ekt where y0 is the inital amount
which is this case is 20, 000. Since the value is 15, 000 after 4 years we use t =
4 y = 15, 000 to solve for k.
y = 20000ekt
15000 = 20000e4k
MATH 1520 Quiz 3B Solutions
1. (a)
2x2 + 2x 8
x2 (2 + 2/x 8/x2 )
= lim
x
x
4x 3x2
x2 (4/x 3)
2 + 2/x 8/x2
= lim
x
4/x 3
2+00
=
03
2
=
3
lim
x2 10
is a of non zero/0 form and hence it goes to one of the innites.
x3
x2 9
To determine which, we need to test
MATH 1520 Problem Workshop 3 Solutions
1. Using the formula A = P ert , we have P = 20, t = 4, r = 0.03, hence after 4 years the
price would be
A = 20e4(0.03) = 20e0.12
Hence after 4 years the price would be $20e0.12 .
mt
r
, we wish to solve for r where
MATH 1520 Problem Workshop 4 Solutions
1. We are looking for what the y values approach as x approaches 3, 0, 2. For part (g)
the limit does not exist since the limit from the left doesnt match the limit on the
right. We get the solutions 1, 1, 1, 1, 2, 2