CALCULUS
WORKSHEET 1 ON PARTICLE MOTION
Work these on notebook paper. Use your calculator only on part (f) of problems 1. Do not use your calculator
on the other problems. Write your justifications in a sentence.
1. A particle moves along a horizontal line so that its position at any time is given by
s  t   t 3  12t 2  36t , t  0, where s is measured in meters and t in seconds.
(a) Find the instantaneous velocity at time t and at t = 3 seconds.
(b) When is the particle at rest? Moving to the right? Moving to the left? Justify your answers.
(c) Find the displacement of the particle after the first 8 seconds.
(d) Find the total distance traveled by the particle during the first 8 seconds.
(e) Find the acceleration of the particle at time t and at t = 3 seconds.
(f) Graph the position, velocity, and acceleration functions for 0  t  8 .
(g) When is the particle speeding up? Slowing down? Justify your answers.
__________________________________________________________________________________________
2. The maximum acceleration attained on the interval 0  t  3 by the particle whose velocity is given
by v  t   t 3  3t 2  12t  4 is
(A) 9
(B) 12
(C) 14
(D) 21
(E) 40
_________________________________________________________________________________________
3. The figure on the right shows the position s of a
particle moving along a horizontal line.
(a) When is the particle moving to the left? moving
to the right? standing still? Justify your answer.
(b) For each of v 1.5 , v  2.5 , v  4  , and v  5 , find
the value or explain why it does not exist.
(c) Graph the particle’s velocity.
(d) Graph the particle’s speed.
__________________________________________________________________________________________
4. (2005) A car is traveling on a straight road.
For 0  t  24 seconds, the car’s velocity v  t  ,
in meters per second, is modeled by the
piecewise-linear function defined by the
graph on the right.
(a) For each of v  4  and v  20  , find the value
or explain why it does not exist. Indicate
units of measure.
(b) Let a  t  be the car’s acceleration at time t, in
meters per second per second. For 0 < t < 24,
write a piecewise-defined function for a  t  .
(c) Find the average rate of change of v over the interval 8  t  20 . Does the Mean Value
Theorem guarantee a value of c, for 8 < c < 20, such that v  c  is equal to this average
rate of change? Why or why not?
TURN->>>
5. (Modification of 2009 Form B, Problem 6)
The velocity of a particle moving along the x-axis is modeled by a differentiable function v, where the
position x is measured in meters, and time t is measured in seconds. Selected values of v  t  are
given in the table above.
(a) Use data from the table to estimate the acceleration of the particle at t = 36 seconds. Show the computations
that lead to your answer. Indicate units of measure.
(b) For 0  t  40, must the particle change direction in any of the subintervals indicated by the data in table? If
so, identify the subintervals and explain your reasoning. If not, explain why not.
(c) Based on the values in the table, what is the smallest number of instances at which the velocity v  t  could
equal  9 m / sec on the interval 0  t  40? Justify your answer.
Answers to Worksheet 1 on Particle Motion
1. (a) 3t 2  24t  36,  9m / sec
(b) At rest at t = 2 because v  t  = 0 there. Moving right for [0, 2) and  6,   because v  t  > 0.
Moving left for (2, 6) because v  t  < 0.
(c) 32 meters
(d) 96 meters
(e) 6t  24,  6m / sec2
(f) Graph
(g) Speeding up on (2, 4) because vel. and acc. are both neg. there and on  6,   because vel. and acc. are
both pos. there. Slowing down on [0. 2) because vel. is pos. and acc. is neg. and on (4, 6) because vel. is
neg. and acc. is pos.
2. D
3. (a) Moving left on (2, 3) and (5, 6) because v  t  < 0. Moving right on (0, 1) because v  t  > 0.
Standing still on (1, 2) and (3, 5) because v  t  = 0 there.
(b) 0, - 4, 0, dne because graph of s has a sharp turn there
(c) and (d) Graphs
4. (2005 AB 5)
(a) v  4  does not exist because the graph of v  t  has a sharp turn at t = 4.
v  20   
5
m / sec2 .
2

5, 0  t  4

 b  a  t   0, 4  t  16
 5
 , 16  t  24
 2
5
(c) Ave. rate of change =  m / sec 2 . No, the MVT does not apply for 8 < c < 20 because
6
the graph of v  t  is not differentiable at t = 16.
5. (2009 Form B, Problem 6)
(a)
11 m
8 sec 2
(b) The particle changes direction on (8, 20) because v 8  5 and v  20   10. The particle also
changes direction on (32, 40) because v  32    4 and v  40   7.
m
at least two times on (0, 40). Since v  t  is differentiable, it must be
sec
continuous. v 8  5, v  20   10, and  9 lies between 5 and – 10 so v  t  must equal – 9
(c) v  t  must equal  9
for some t between 8 and 20. Similarly, since v  20   10, v  25   8, and  9 lies between
– 10 and – 8 so v  t  must equal – 9 for some t between 20 and 25 by the IVT>
CALCULUS
WORKSHEET 2 ON PARTICLE MOTION
Work these on notebook paper. Use your calculator on problems 1 - 5, and give decimal answers correct to
three decimal places. Write your justifications in a sentence.
1. A particle moves along a horizontal line so that its position at any time t  0 is given by
s  t   t 3  7t 2  14t  8 , where s is measured in meters and t in seconds.
(a) Find the instantaneous velocity at any time t and when t = 2.
(b) Find the acceleration of the particle at any time t and when t = 2.
(c) When is the particle at rest? When is moving to the right? To the left? Justify your answers.
(d) Find the displacement of the particle during the first two seconds.
(e) Find the total distance traveled by the particle during the first two seconds.
(f) Are the answers to (d) and (e) the same? Explain.
(g) When is the particle speeding up? Slowing down? Justify your answers.
_________________________________________________________________________________________
2. The position of a particle at time t seconds, t  0 , is given by s  t   t 2  sin t , 0  t  3 ,
where t is measured in seconds and s is measured in meters. Find the particle’s acceleration
each time the velocity is zero.
_________________________________________________________________________________________
3. A particle’s velocity at time t seconds, t  0 , is given by v  t   cos t 2  t , 0  t  2 , where
 
t is measured in seconds and v is measured in meters/second. Find the velocity of the particle
each time the acceleration is zero.
__________________________________________________________________________________________
4. (2004) A particle moves along the y-axis so that its velocity at time t  0 is given by
 
v  t   1  tan 1 et .
(a) Find the acceleration of the particle at time t = 2.
(b) Is the speed of the particle increasing or decreasing at time t = 2? Give a reason for your
answer.
(c) Find the time t  0 at which the particle reaches its highest point. Justify your answer.
__________________________________________________________________________________________
5. (Modification of 2005 Form B, Problem 3)
A particle moves along the x-axis so that its velocity at time t, for 0  t  5 , is given by


v  t   ln t 2  3t  3 .
(a) Find the acceleration of the particle at time t = 4.
(b) Find all times t in the open interval 0  t  5 at which the particle changes direction. During which
time intervals, for 0  t  5 , does the particle travel to the left? Justify your answer.
(c) Find the average rate of change of v  t  on 1.5  t  3.2.
TURN->>>
6. (Modification of 2008, Problem 4)
A particle moves along the x-axis so that its velocity at time t, for 0  t  6 , is given by a differentiable
function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has
horizontal tangents at t = 1 and t = 4.
(a) On the interval 3  t  4 , is the speed of the particle increasing or decreasing? Give a reason for your
answer.
(b) On the interval 2  t  3 , is the speed of the particle increasing or decreasing? Give a reason for your
answer.
(c) During what intervals, if any, is the acceleration of the particle negative? Justify
Answers to Worksheet 2 on Particle Motion
1. (a) 3t 2  14t  14, 2 m / sec
(b)  6t  14, 2 m / sec2
(c) At rest at t = 1.451 and t = 3.215 because v  t  = 0 there. Moving left for [0, 1.451)
and  3.215,   because v  t  < 0. Moving right for (1.451, 3,215) because v  t  > 0.
(d) – 8 m
(e) 9.262 m
(f) No, the displacement and distance are not the same because the particle changed
direction at t = 1.451.
(g) Slowing down on (0, 1.451) and (2.333, 3.215) because vel. and acc. have opposite signs.
Speeding up on (1.451, 2.333) and  3.215,   because vel. and acc. have the same sign.
2. a  0.45018...  2.435m / sec2
3. 1.600
m
m
, 0.730
sec
sec
4. (a) – 0.133
(b) – 0.436. Speed is increasing at t = 2 because v  t  and a  t  are both negative.
(c) v  t  = 0 when t = 0.443. This is the only critical number. v  t  > 0 for (0, 0.443)
and v  t  < 0 for  0.443,   so the particle reaches its highest point at t = 0.443.
5. (a) 0.714
(b) The particle changes direction at t = 1 and at t = 2 because v  t  changes from positive to negative
or vice versa there. The particle travels to the left on (2, 3) because v  t  < 0 there.
(c) 0.929
6. (2008)
(a) On (3, 4) v  t  > 0 and v  t  is increasing so v  t   a  t   0. Therefore, the speed is increasing
on (3, 4).
(b) On (2, 3), v  t  < 0 and v  t  is increasing so v  t   a  t   0. Therefore, the speed is decreasing
on (2, 3).
(c) The acceleration is negative on (0, 1) and (4, 6) because the velocity is decreasing there.
CALCULUS
WORKSHEET 1 ON OPTIMIZATION
Work the following on notebook paper. Be sure to justify your answers.
1. A particle moves along the x-axis so that its position is given by s  t   2 t 3  7t 2  13t  3 for 0  t  2.5.
(a) Find the minimum velocity of the particle and the time at which it occurs. Justify your answer.
(b) Find the maximum velocity of the particle and the time at which it occurs. Justify your answer.
2. The velocity of a space shuttle from t = 0 sec to t = 125 sec is given by v  t   0.001t 3  0.091t 2  24t  3.3
where v  t  is measured in ft/sec. Find the maximum and minimum values of the acceleration of the shuttle
and the time at which they occur. Justify your answer.
3. A person in a rowboat three miles from the nearest point
on a straight shoreline wishes to reach a house eight miles
further down the shore. If the person can row at a rate
of 2 miles per hour and walk at a rate of 5 miles per hour,
find the least amount of time required to reach the house.
How far from the house should the person land the rowboat?
Justify your answer.
8 mi.
House
3
mi.
Boat
4. A cylindrical container has a volume of 250 cm3 . Find the radius and height of the cylinder so that its
surface area will be a minimum. Justify your answer.
(Volume of a cylinder =  r 2 h . Surface area of a cylinder = 2 rh  2 r 2 .)
5. The sum of the perimeters of an equilateral triangle and a square is 10.
(a) Find the dimensions of the triangle and the square that produce a minimum area. Justify your answer.
(b) Find the dimensions of the triangle and the square that produce a maximum area. Justify your answer.
x2 3
(Area of an equilateral triangle =
, where x = side of triangle)
4
6. A rectangular package to be sent by a postal service can have a maximum
combined length and girth (perimeter of a cross section) of 108 in. Find the
dimensions of the package of maximum volume that can be sent. (Assume
that the cross section is square.) Justify your answer.
Answers to Worksheet 1 on Optimization
Note: Only the answers are shown below. Students must also justify their answer. See the article on AP
Central called “On the Role of Sign Charts in the AP Calculus Exam for Justifying Local or Absolute Extrema”
found at http://apcentral.collegeboard.com/apc/public/repository/signcharts2005apcentr_36691.pdf
to see how justifications for absolute maximums and minimums should be written.
7
.
6
(b) The maximum velocity is 15.5, and it occurs when t  2.5.
ft
2. The maximum acceleration is 48.125
, and it occurs at t = 125 sec.
sec 2
ft
The minimum acceleration is 21.240
, and it occurs at t = 30.333 sec.
sec 2
1. (a) The minimum velocity is 4.833, and it occurs when t 
3. The minimum time occurs when x = 1.309 mi. so the person should land the boat at 6.691 miles from the
house. The time will be 2.975 hours.
4. The surface area will be a minimum when the radius is 3.414 cm and the height is 6.828 cm.
5. (a) The area is a minimum when the sides of the triangle are 1.883 and the sides of the square are 1.087.
(b) The area is a maximum when the sides of the triangle are 0 and the sides of the square are 2.5.
6. The maximum volume occurs when the dimensions are 18 in. x 18 in. x 36 in.
CALCULUS
WORKSHEET 2 ON OPTIMIZATION
Work the following on notebook paper. Be sure to justify your answers.
1. The size of a population of bacteria introduced to a food grows according to the formula P  t  
6000t
,
60  t 2
where t is measured in weeks and 0  t  20. Determine when the bacteria will reach its maximum size.
What is the maximum size of the population? Justify your answer.
2. A particle moves along the x-axis so that its position is given by s  t    t 3  4t 2  10t for 0  t  3.
(a) Find the minimum velocity of the particle and the time at which it occurs. Justify your answer.
(b) Find the maximum velocity of the particle and the time at which it occurs. Justify your answer.
3. A tank with a rectangular base and rectangular sides is open at the top. It is to be constructed so that its width
is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and
$5 per square meter for the sides, what is the cost of the least expensive tank, and what are its dimensions?
Justify your answer.
4. A right triangle has one vertex at the origin and one vertex on the
curve y  cos x for 0.25  x  1. One of the two perpendicular sides of
the triangle lies on the x-axis, and the other side is parallel to the
y-axis. Find the maximum and minimum areas for such a triangle.
Justify your answer.
5. A person in a rowboat two miles from the nearest point
on a straight shoreline wishes to reach a house six miles
further down the shore. If the person can row at a rate
of 3 miles per hour and walk at a rate of 5 miles per hour,
find the least amount of time required to reach the house.
How far from the house should the person land the rowboat?
Justify your answer.
6. An offshore well is located in the ocean at point W that is
six miles from the closest shore point A on a straight
shoreline. The oil is to be piped to a point B on the shore
that is eight miles from A by piping it on a straight line
under water from W to some point P on the shore
between A and B and then on to B via a pipe along the
shoreline. If the cost of laying pipe is $100,000 per mile
under water and $75,000 per mile over land, how far from A
should the point P be located to minimize the cost of lying the
pipe? What will the cost be? Justify your answer.
6 mi.
House
2
mi.
Boat
8 mi.
A
6
mi.
W
7. A piece of wire 40 cm long is to be cut into two pieces. One piece will be bent to form a circle, and
the other will be bent to form a square. Find the radius of the circle and the length of the side of the
square so that the total area is:
(a) a minimum
(b) a maximum
B
Answers to Worksheet 2 on Optimization
Note: Only the answers are shown below. Students must also justify their answer. See the article on AP
Central called “On the Role of Sign Charts in the AP Calculus Exam for Justifying Local or Absolute Extrema”
found at http://apcentral.collegeboard.com/apc/public/repository/signcharts2005apcentr_36691.pdf
to see how justifications for absolute maximums and minimums should be written.
1. The maximum size of the population of bacteria is 387.298 (so 387 bacteria) and it occurs when
t = 7.750 weeks.
2. The minimum velocity is 7, and it occurs when t = 3.
The maximum velocity is 15.333, and it occurs when t 
4
.
3
3. The minimum cost is $330, and it occurs when the dimensions of the tank are 3 m x 3 m x 4 m.
4. The maximum area is 0.281, and it occurs when x = 0.860.
The minimum area is 0.121, and it occurs when x = 0.25.
5. The minimum time is 1.733 hr when x = 1.5 mi. so he should land the boat 4.5 mi. from the house.
6. The minimum cost is $996,862.70 when the point P is 6.803 mi. from B.
7. The minimum area is 56.010 sq. cm. when the radius is 2.800 cm and the side of the square is 5.601 cm.
The maximum area is 127.324 sq. cm. when the radius is 6.366 cm and the side of the square is 0 cm.
CALCULUS
WORKSHEET ON LINEAR APPROXIMATIONS & DIFFERENTIALS
Work the following on notebook paper. Give decimal answers correct to three decimal places.
1. Given f  x   x3 .
(a) Write the equation of the tangent line to f at x = 1
(b) Use your answer to (a) to find an approximation for f 1.1 .
(c) Find the actual value of f 1.1 .
(d) Find the amount of error of your approximation.
_________________________________________________________________________________________
Use a tangent line approximation to approximate the value.
2. 103
3. 3 998
__________________________________________________________________________________
Find the differential dy.
2x
4. y  3x 2
5. y 
6. y  sin 2 x
2
1 x
_________________________________________________________________________________
7. The radius of a sphere increases from 5 cm to 5.2 cm.
(a) Use differentials to estimate the increase in the volume of the sphere. (Volume of a sphere =
4
r )
3
3
(b) Compare the estimate you found in (a) with the true change in the volume, V , and find the error in your
approximation.
8. The radius of a sphere decreases from 7 cm to 6.5 cm.
(a) Use differentials to estimate the decrease in the surface area of the sphere.
(Surface area of a sphere = 4 r 2 )
(b) Compare the estimate you found in (a) with the true change in the surface area, SA , and find the error in
your approximation.
9. The side of an equilateral triangle increases from 20 cm to 20.5 cm.
(a) Use differentials to estimate the increase in the area of the triangle. (Area of an equilateral triangle is
found by A 
x2 3
4
where x = side.)
(b) Compare the estimate you found in (a) with the true change in the area, A , and find the error in your
approximation.
10. In the late 1830s, the French physiologist Jean Poiseuille discovered the formula we use today to predict
how much the radius of a partially clogged artery has to be expanded to restore normal flow. His
formula, V  kr 4 , says that the volume V of fluid flowing through a small pipe or tube in a unit of time
at a fixed pressure is a constant times the fourth power of the tube’s radius r. If r increases from 5 cm
to 5.1 cm, use differentials to estimate the increase in V.
11. A particle moves along the x-axis so that its position is given by s  t    t 3  5t 2  9t  3 for 0  t  4.75.
(a) Find the minimum velocity of the particle and the time at which it occurs. Justify your answer.
(b) Find the maximum velocity of the particle and the time at which it occurs. Justify your answer.
12. A cylindrical container has a volume of 300 cm3 . Find the radius and height of the cylinder so that its
surface area will be a minimum. Justify your answer.
(Volume of a cylinder =  r 2 h . Surface area of a cylinder = 2 rh  2 r 2 .)
Answers to Worksheet on Linear Approximations and DIfferentials
1. (a) y  1  3  x  1
(b) 1.3
(c) 1.331
(d) 0.031
2. 10.15
3. 9.993
4. dy  6 x dx
5. dy 
2  2 x2
1  x 
2 2
dx
6. dy  2sin x cos x dx
7. (a) dV = 62.832 cm3
(b) V = 65.379 cm3 , Error  2.547 cm3
8. (a) dA = – 87.965 cm 2
(b) A  – 84.823 cm 2 , Error  3.142 cm 2
9. (a) dA = 8.660 cm 2
(b) A  8.769 cm 2 , Error  0.108 cm 2
10. dV  50k cm3
11. (a) Minimum velocity =  11.188 when t = 4.75
(b) Maximum velocity = 17.333 when t =
12. r = 3.638 cm and h = 7.256 cm
5
3
CALCULUS
WORKSHEET ON NEWTON’S METHOD & TANGENT LINE APPROXIMATIONS
Work these on notebook paper. Give decimal answers correct to three decimal places.
1. Use Newton’s Method to estimate the zeros. Find x2 and x3 , the second and third approximations to the
zero of the function.
f  x   x 4  10 x 2  11
x1  3.5
__________________________________________________________________________________________
2. Use Newton’s Method to estimate the point(s) of intersection correct to three decimal places.
f  x   x 2 , g  x   cos x
x1 

4
__________________________________________________________________________________________
3. (a) For f  x   2 x3 , find an equation of the tangent line to the graph of f at x = 1.
(b) Use the tangent line equation you found in (a) to approximate f 1.1 .
(c) Find the actual value of f 1.1 by using the function f  x  . What is the error in your linear
approximation?
(d) Fill in the blank with < or >. Tangent line approx. ______ Actual value
What does this tell you about the concavity of f  x  ? What does this tell you about the sign
of f   x  ? Explain.
_________________________________________________________________________________________
4. Let f  x   1  tan x  2 .
(a) Write an equation for the line tangent to the graph of f at the point where x = 0.
(b) Use the tangent line equation you found in (a) to approximate f  0.2  .
3
(c) Find the actual value of f  0.2  by using the function f  x  . What is the error in your linear
approximation?
(d) Fill in the blank with < or >. Tangent line approx. ______ Actual value
What does this tell you about the concavity of f  x  ? What does this tell you about the sign
of f   x  ? Explain.
__________________________________________________________________________________________
5. Consider the curve defined by  8 x 2  5 xy  y3  149.
dy
(a) Find
.
dx
(b) Write an equation for the line tangent to the curve at the point
 4, 1 .
(c) There is a number k such that the point  4.2, k  is on the curve. Using the tangent line found
in part (b), approximate the value of k.
(d) Write an equation that can be solved to find the actual value of k so that the point  4.2, k  is on
the curve, and then use your equation to solve for k.
__________________________________________________________________________________________
6. Suppose that the only information you have about a function f is
that f 1  5 and the graph of f  is given on the right.
(a) Use a linear approximation to estimate f  0.9  and f 1.2  .
(b) Are your estimates in part (a) too large or too small? Explain.
Graph of f 
TURN->>>
7. Use a tangent line approximation to approximate the value of
53.
8. Find the differential given y  x 2  4.
 
9. Find the differential given y  cos x3 .
10. A person in a rowboat two miles from the nearest point
on a straight shoreline wishes to reach a house seven miles
further down the shore. If the person can row at a rate
of 3 miles per hour and walk at a rate of 4 miles per hour,
find the least amount of time required to reach the house.
How far from the house should the person land the rowboat?
Justify your answer.
7 mi.
House
2
mi.
Boat
11. The sum of the perimeters of an equilateral triangle and a square is 8.
(a) Find the dimensions of the triangle and the square that produce a minimum area. Justify your answer.
(b) Find the dimensions of the triangle and the square that produce a maximum area. Justify your answer.
Answers to Worksheet on Newton’s Method and Tangent Line Approximations
1. x2  3.337, x3  3.317
2. (.824, .679), (- .824, .679)
3. (a) y  2  6  x  1
(b) 2.6
(c) 2.662, Error  0.062
(d) <. This tells us that the tangent line lies below the graph of f so f is concave up.
4. (a) y  1 
3
 x  0
2
(b) 1.3
(c) 1.319, Error  0.019
(d) <. This tells us that the tangent line lies below the graph of f so f is concave up.
16 x  5 y
5x  3 y 2
(b) y  1  3  x  4 
5. (a)
(c) – 0.4
(d)  8  4.2   5  4.2  k   k 3  149 , k = - 0.373
2
6. (a) f  0.9   4.8, f 1.2   5.4
(b) Since f  is decreasing, f is concave down so the tangent line lies above the graph of f.
Therefore, our estimates are too large.
7. 7
2
or 7.286
7
8. dy  2 x dx
 
9. dy  3x 2 sin x3 dx
10. The last amount of time is 2.191 hr. The person should land 4.732 miles from the house.
11. (a) The area is a minimum when the triangle has sides of 1.507 and the square has sides of 0.870.
(b) The area is a minimum when the triangle has sides of 0 and the square has sides of 2.
CALCULUS BC
REVIEW SHEET - PARTICLE MOTION & 3.7 – 3.9
NAME__________________________________
Work these on notebook paper. Use your TI-89, show all setups, and give decimal answers correct to three
decimal places. Write your justifications in a sentence.
1. A particle moves along a horizontal line so that its position at any time t, 0  t  5, is given by
s  t    t 4  5t 3  7t  3 , where s is measured in meters and t in seconds.
(a) When is the particle moving left? Justify your answers.
(b) Find the displacement of the particle between t  2 seconds and t  4 seconds. Explain the meaning of your
answer.
(c) Find the distance traveled by the particle between t  2 seconds and t  4 seconds. Show the work that
leads to your answer.
(d) When is the particle slowing down? Justify your answer.
2. The position of a particle at time t seconds is given by s  t   sin  4t   t , 0  t  0.8. Find the
particle’s acceleration each time the velocity is zero.
__________________________________________________________________________________________
You must use Calculus on problems 3, and 4. Your work must include a function and a derivative, and your
must write your answer and your justification in a sentence.
3. Egbert is in a rowboat four miles from the nearest point on
Egbert
a straight shoreline. He wishes to reach a house 12 miles
farther down the shore. If Egbert can row at a rate of 3 mi/hr
4
and walk at a rate of 4 mi/hr, find the least amount of time
mi.
required to reach the house. How far from the house should
he land the rowboat? Justify your answer.
12 miles House
4. The rate at which people enter an auditorium for a rock concert is modeled by the function
R  t   1240t 2  625t 3 for 0  t  1.5 hours, where R  t  is measured in people per hour. Find
the time when the rate at which people enter the auditorium is a maximum. Justify your answer.
__________________________________________________________________________________________
5. Use Newton’s method to find x2 , the first approximation, to the zero of the function
f  x   5x3  3x 2  x  4 , using x1  1.6 .
__________________________________________________________________________________________
6. (a) Use a tangent line approximation to estimate the value of
53.
(b) Find the actual value of 53 and the amount of error in your approximation.
__________________________________________________________________________________________
7. Let f  x    8  tan x 
4
3
for 

x

.
2
2
(a) Write the equation for the line tangent to the graph of f at the point where x = 0.
(b) Use the equation found in part (a) to approximate f  0.324  .
(c) Find the exact value of f  0.324  by using the function f  x  .
(d) What is the amount of error in the approximation you found in (b)?
(e) Is the approximation found in (b) greater or less than the actual value? What does this tell you about the sign
of f   x  on the interval around x = 0? Explain your answer.
Answers
1. (a) v  t    4t 3  15t 2  7  0 when t  0.7657... and 3.616...
The particle is moving left for 0  t  0.766 and for 3.616  t  5 because v  t   0 there.
(b) Displacement = s  4   s  2   39  13  26 m
When t = 4 sec, the particle is 26 m to the right of its position at t = 2 sec.
(c)
t
s t 
2
13
Distance
30.12469…
3.616… 43.1246…
+ 4.12469…
4
39
34.24939…
Distance = 34.249 m
(d) a  t   12t 2  30t  0 when t  2.5.
The particle is slowing down for 0  t  0.766 and 2.5  t  3.616 because v  t  and a  t  have
opposite signs there.
2. v  t   4 cos  4t   1  0 when t  0.4558... sec
a  0.4558...  15.492
3. time 
m
sec2
distance
rate
x
T
x  16 12  x
where 0  x  12

3
4
T
2
4.333…
0
4.5355… 3.8819…
T 
x
3 x  16
2

1
 0 when x  4.5355...
4
4.2163…
12
T must have its minimum at x = 0, 4.5355…, or 12.
T is a minimum when x = 4.536 mi. The rowboat should land 12  x  7.464 mi. from the house,
and the time is 3.882 hr. (Candidates Test)
4. R  t   2480t  1875t 2  0 when t  1.3226...
R must have its maximum at t = 0, 1.3226…, or 1.5.
The rate at which people enter the auditorium is a maximum
at t = 1.323 hr.
t
0
R t 
0
1.3226… 723.1048…
1.5
680.625…
5. f   x   15x 2  6 x  1
x2  1.251
6. Let f  x  
f  x 
x and x  49.
1
2 x
, f   49  
1
, and f  49   7
14
1
 x  49
14
1
2
53  7   4   7 or 7.286
14
7
Actual value is 53  7.280...
Error  0.006
Tangent line: y  7 

1
4
8  tan x  3 sec2 x
3
1
4
8
f   0    8  0  3 1 
3
3
7. (a) f   x  
f  0  8
4
3
 16
Tangent line equation: y  16 
(b) f  0.324   16 
(c)

8
 x  0
3
8
 0.324  16.864
3
f  0.324   8  tan 0.324 
4
3
 16.902
(d) Error  0.038
(e) less
The tangent line lies below the curve so f is concave up. Therefore f   x  is positive for
the interval around x  0.