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Calculus I Examples For Exam II
| If it is not already on your hard drive, you will need to download
the free DPGraph Viewer to view some of
the pictures linked to on this page. |
QuickTime free download. |
Here is a
Geometer's Sketchpad video with audio (Larger
Version) of
me describing approximating the slope of a tangent line with a secant line.
Here is a
Quicktime Animation showing a secant line
approximating a tangent line.
Here is a
Winplot demonstration of a secant line that can be
used to approach a tangent line. You may need to download the file to your
desktop and then use the freeware
Winplot to open the file (by opening Winplot, clicking on Window,
clicking on 2-dim, clicking on File, clicking on Open, and then opening
SecantSlope from your desktop. You can use the sliders to vary the values
of A and H. The A-value gives you the x-value at a point on the graph of
the function you are looking at. The H-value gives you the x-value of
another point on the graph of the function you are looking at in terms of A + H
(so H is your delta x). The secant line through the points on the graph of
the given function with x-values of A and A + H is graphed and its slope
displayed at the top of the graph. The default function is f(x) = 3sin(x).
This function is named FN and you can edit the definition of FN by clicking on
Equa, choosing
User functions ... ,
highlighting the function and editing its definition. Changing the
definition to x^2 for example will produce a different graph.
Computing the derivative using the limit definition
of derivative
| Derivative Approximation
The following points are the points pictured on
the graph of the function at the right: (-0.5,10.875),
(-0.4,10.816), (-0.3,10.693), (-0.2,10.512), (-0.1,10.279), (0,10),
(0.1,9.681),(0.2,9.328), (0.3,8.947), (0.4,8.544), (0.5,8.125).
Approximate the derivative of the function at the points (-0.3,10.693),
(0,10), and (0.4,8.544).
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Finding the
Equation of a Tangent Line
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Find an
equation of the line tangent to the graph of the function
f(x) =
x3 - 2x2 - 3x + 10 at the point (-1,10).
Click on the picture at the right to see a Quicktime animation of
tangent lines as x
goes
from -2 to 2.4. |
 |
| Differentiation Example--Product
Rule (and Chain Rule)
 |
What would the graph look like for x greater than
4? |
| Differentiation Example--Quotient
Rule 
|
What do you think the graph looks like outside
the viewing window shown above? |
| Differentiation Example--Chain
Rule (Twice) 
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Do you think a graphing calculator could draw an
accurate graph of this function over an x-interval of [0,10]? |
Use implicit differentiation to find the equations
of the tangent and normal lines to the graph of the relation whose equation is
given below at the point (4,2).
 |
In the
graph below the tangent line is drawn in blue and the normal line is drawn
in red.
|
Use implicit differentiation to find an equation of
the line tangent to the graph of the given relation at the indicated point.
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Click on the picture to enlarge. |
Three more implicit differentiation examples
DPGraph
picture of z = xcos(y) + 2y - y2 and z = c.
When c =
0 the intersection of the two graphs will be the graph of this relation. |
|
DPGraph
picture of z = x2cos(y) + 2xy - xy2 and z = c.
When c =
0 the intersection of the two graphs will be the graph of this relation.
Compare
the graph on the right to the graph in example 1 and note the difficulty
at x = 0. This graph was constructed
using Maple. Can you tell what the problem is around the y-axis?
Click on the graph for a larger Maple image.
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DPGraph
picture of z = 2xy - y2 + 3 + 3xy3 - cos(xy2)
and z = c.
When c =
0 the intersection of the two graphs will be the graph of this relation.
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| Quicktime
Animation Average and Instantaneous Speed
Animation--No
Scales |
Click on the picture to see an animation. |
| Just as she is entering a 20 mph
speed zone, Emily notices a police car up ahead parked a little way off
the road. Emily applies the brakes gently to begin slowing down (she
does not want to be too obvious). She first hits the brakes when the
front of her car is at position zero in the picture above and she slows
down for three seconds with a constant negative acceleration. She
travels a total of 111 feet during these three seconds and at the end of
the three seconds the front of her car is level with the radar gun being
used to clock her speed. It turns out that the radar began clocking
her motion at the instant she first started to slow down. The tables
below give distance traveled in feet and time in seconds from the time
when she first applied the brakes.
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Time
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 |
Distance
0.00 3.99 7.96 11.91 15.84 19.75 23.64 27.51 31.36 35.19 39.00 42.79 46.56 50.31 54.04 57.75 61.44 65.11 68.76 72.39 76.00 79.59 83.16 86.71 90.24 93.75 97.24 100.71 104.16 107.59 111.00 |
Time
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
. . . .
2.90
2.91
2.92
2.93
2.94
2.95
2.96
2.97
2.98
2.99
3.00
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Distance
0.0000
0.3999
0.7996
1.1991
1.5984
1.9975
2.3964
2.7951
3.1936
3.5919
3.9900
. . . . .
.
107.5900
107.9319
108.2736
108.6151
108.9564
109.2975
109.6384
109.9791
110.3196
110.6599
111.0000
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| What was Emily's average speed in
feet per second and in miles per hour from time t = 0 seconds to time t =
3 seconds? What would you estimate to have been Emily's speed in
feet per second and in miles per hour at time t = 0 seconds? What
would you estimate to have been Emily's speed in feet per second and in
miles per hour at time t = 3 seconds? Estimate Emily's speed in feet
per second and in miles per hour at time t = 2 seconds. I will call
the position function s(t).
Answers
Average speed from t = 0 to t = 3 seconds:
Speed at t = 0 seconds:
That
guess turns out to be pretty good since the radar, measuring to an
accuracy of 7 digits to the right of the decimal, indicated that
s(0.0000001)
= 0.000004.
Speed at t = 3 seconds:
This
guess also seems to be pretty good since the radar, measuring to an
accuracy of 7 digits to the right of the decimal, indicated that
s(2.9999999)
= 110.9999966. Can you see why this makes the 34 ft/sec estimate of
the speed at t = 3 seconds seem pretty accurate?
Speed at t = 2 seconds:
Since we have less precise data around t = 2
seconds it would probably be more accurate to look at s(2.1) and s(1.9).
Do you think Emily got pulled over and if so do
you think she got a ticket or a warning? Do we have enough
information to make an educated guess at the answer to that
question? The idea for this example was inspired by a terrific presentation
on speed from Calculus Quest.
Notice that in this example the average speed
over the three second time interval is equal to the average of the
estimated instantaneous speeds at the beginning and end of the time
interval. Would this always be the case? Would it be possible
for someone to be traveling at 40 ft/sec at time t = 0 seconds, 34 ft/sec
at time t = 3 seconds, and yet the average speed over this 3 second
interval be 38 ft/sec or even be 42 ft/sec?
EC Find an equation representation
of the position function, s(t), that would nicely fit the data given
above. |
| Average and Instantaneous Speed
The position function shown below gives the
position of an object in feet as a function of time in seconds.
Determine when the object is traveling to the right (positive direction)
and when the object is traveling to the left (negative direction).
Determine the average speed from t = 1 to t = 5. Determine the
instantaneous speed at t = 1 and at t = 5.
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The
graph below shows position (not distance traveled) as a function of
time. Click here or on
the picture to see a linear motion animation. In the linear motion
animation the animated point on the left vertical axis represents the
elapsed time. Quicktime
animation
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A Vertically
Launched Projectile
Quicktime Animation (In the animation the
point traveling down the vertical axis indicates the speed of the projectile.)
The function above (called a position function)
gives the height above the ground (in feet) of a projectile launched vertically
(straight up) with an initial velocity of 192 ft/sec and an initial height of
100 feet above the ground (air resistance is being ignored here). The
variable "t" represents time in seconds with t = 0 corresponding to the moment
the projectile was launched. Here are the types of questions you need to
be able to answer.
(A) What is the average speed of the projectile
during the first 3 seconds after launch?
(B) What is the average speed of the projectile from
t = 1 second to t = 4 seconds?
(C) What is the instantaneous speed of the
projectile at t = 3 seconds?
(D) What will be the maximum height attained by the
projectile?
(E) How long until the projectile comes back down
and hits the ground?
Solutions
(A) We need to find the total distance traveled
during the first three seconds and divide this total distance traveled by the
time interval (3 seconds). This means we need to compute
(B)
(C) The derivative of the position function
will give us the instantaneous speed. Thus we need to evaluate the
derivative at t = 3 seconds (i.e., substitute 3 for t).
(D) At some point in time the projectile is going to
stop going up and start coming back down. At the instant it stops going up
its speed will be zero ft/sec. When it starts coming back down its speed
(velocity in the sense that direction is now being considered) will be negative.
At the instant when the speed is zero the projectile will be at its highest
point (maximum height). We need to find the value of time "t" when the
derivative (which gives velocity) is zero. We can then evaluate the
position function (which gives height above the ground) at that value of t
to find the maximum height.
(E) The projectile will hit the ground when the
value of the position function is zero.
The t-value of -0.5 would correspond to before the
projectile was launched and is not in the domain of our position function
describing this motion. The answer must come from the positive solution to
the equation above. Thus the projectile comes back down and hits the
ground in 12.5 seconds.
| Section
2.6 Number 2a
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| In the
picture above the blue point is the point where x = 3. Click on the
picture (Quicktime version) to see an animation of a point moving along the graph of the
parabola as a function of time (t) where
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return
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