Going Off on a Tangent Name: ___________________________ Period: ____________

Going Off on a Tangent Name: ___________________________ Period: ____________
Name: ___________________________
Period: ____________
Going Off on a Tangent
One gloomy autumnal afternoon, you’re at a friend’s house, trapped inside because the weather refuses to
1
cooperate with your adventurous intentions. You and your friend are playing a high-stakes game of Yahtzee while
discussing your favorite mathematical curve, which of course is the parabola. During this academic discussion,
your friend suddenly goes off on a tangent and begins talking about slope, of all things. You’re thinking, “Here we
are talking about parabolas, which are nice and curvy, and my friend starts talking about slope, which has to do
with lines—straight, boring lines. Parabolas don’t even have a slope, right? I mean, there we were, on the same
page, and it was riveting, but then my friend just goes off in a totally different direction. Can’t he and/or she stay
focused?”
Regardless of what you might think, I’m glad your friend brought up the subject of slope. You’re right, by the way,
a parabola doesn’t have a slope; it has infinitely many slopes, one for every point on the graph. In other words,
the rate at which the y-values are changing relative to the x-values is not constant like it is for a line.
Consider the linear function ( )
.
-2
-2
8
( )
-1
0
0
2
1
4
2
6
6
1.
From the table of values, by how much is the value of
increasing/decreasing?
2.
From the table of values, by how much is the value of ( )
increasing/decreasing?
3.
What is the rate of change in ( ) relative to ? (
4
2
-5
5
-2
( )
)
-4
-6
Notice from Q3 that the rate of change, or slope, is constant; it is the same everywhere.
Now consider the quadratic function ( )
.
10
( )
8
6
4.
4
2
-5
1
5
Or Scrabble, it doesn’t matter which.
-2
10
-1
4
0
2
Is the rate of change in ( ) relative to
reasoning.
1
4
2
10
constant? Explain your
Your answer to Q4 leads to an important concept in mathematics; that is, nonlinear curves have a variable rate of
change. So instead of trying to find a rate of change for the entire graph, which doesn’t make sense, we try to find
the rate of change at a single point on the curve. This is called the instantaneous rate of change, and to find it, we
need to go off on a tangent.
Consider the illustrations below.
According to the
diagrams, a tangent line
touches the circle, or any
curve, at one and only
one point. This is called
the point of tangency.
Think of the point where
a fully inflated tire would
theoretically touch the
road.
Just like a parabola, a circle does not have a constant rate of change. It does, however, have an instantaneous rate
of change at every point along the curve. Mathematicians discovered that this instantaneous rate of change is
equal to the slope of the tangent line at that point, a discovery that eventually lead to the development of
differential calculus.
Calculating the slope of a line tangent to a circle at a
given point is quite simple. Recall from geometry that a
line tangent to a circle is perpendicular to a radius at
the point of tangency.
6
5.
Find the slope of the tangent line at (4, 1) to a circle with
center at (-1, 2).
4
(-1, 2)
2
(4, 1)
-5
5
-2
A circle has its center at (2, 4). For Q6-Q9, find the slope of the line tangent to the circle at each of the given
points, assuming that each point is the other endpoint of the radius.
6. (7, 14)
7. (-5, 4)
8. (2, 2)
9. ( , )
[2]
Now let’s go off on another tangent. Notice in the diagrams below that a curve has infinitely many tangent lines,
one at each point on the graph. Interestingly, you can create the illusion of a curve by graphing a number of its
tangent lines.
10. We can approximate this curve by graphing a set of
its tangent lines. To do this, use a ruler to draw a
line from
to
. Next draw another line
from
to
. Repeat in a similar manner
to complete the curve.
11. In a manner similar to Q10, use tangent lines to
approximate this graph.
10
10
8
8
6
6
4
4
2
2
10
8
6
4
2
5
5
10
10
That was fun. Now let’s get back to trying to find the slope of a tangent line. With a circle, this was quite simple,
since the tangent line is perpendicular to the radius of the circle at the point of tangency. However, for noncircular
curves, there is no radius for tangent line to be perpendicular to. Moreover, we only have one point with which to
calculate the slope, the point of tangency, and as you are more than likely aware, we need two points to find the
slope of a line. So to find the elusive slope of a tangent line, we must devise a cunning plan that involves going off
on yet another secant, I mean tangent.
[3]
Consider the illustrations below.
According to the diagrams,
a secant line touches a
circle, or any curve at 2
distinct points. Since a
secant line always touches
a curve at two points, it is
an elementary exercise to
find its slope.
( )
Let’s say we are trying to find the slope of the line tangent to ( )
at the point (1, 4). Again this is made difficult by the fact that we only have
one point, but we need two points to calculate slope. So let’s forget about
the tangent line for a minute and instead try to calculate the slope of the
secant line through two points on the graph.
12. Find the slope of the secant line through the points (0, 2) and (1, 4).
with secant line
10
8
6
4
13. How do you think the slope of this secant line would compare to the
slope of the tangent line at (1, 4)?
2
-5
5
Your answer to Q13 leads to another important concept. In this case, we can
approach the slope of the tangent line by using a series of secant lines that get
closer and closer to the tangent line.
14. For the table below, evaluate the function ( )
at each -value.
Then compute the slope of the secant line,
, through (1, 4) and each
point from the table.
0
0.25
0.5
0.75
0.8
0.9
0.95
( )
2
4.6
4.4
4.2
4
3.8
3.6
3.4
3.2
3
2.8
2.6
15. What value does the slope of the secant lines seem to be approaching?
2.4
2.2
2
1.8
1.6
Let’s confirm your answer to Q15 by approaching the tangent line using secant lines from above the point (1, 4)
rather than from below.
16. For the table below, evaluate the function ( )
at each -value. Then compute the slope of the
secant line,
, through (1, 4) and each point from the table.
2
1.75
1.5
1.25
1.2
1.125
1.1
1.025
( )
10
[4]
17. What value does the slope of the secant lines seem to be approaching?
18. What seems to be the slope of the tangent line at (1, 4)?
Notice that with -values in the table from Q16, you are just getting closer and closer to the -value of the point of
tangency. The first -value added 1 to the -coordinate of the point of tangency, whereas the last -value added
0.025 to the -coordinate of the point of tangency. It’s this value that we add to the point of tangency that’s
actually getting smaller and smaller. Let’s define a variable for that amount and call it . Thus, the -coordinate of
our new point would be
. The variable can be whatever we want it to be. For example, if
, then we
get
, the first value in your table. If
, then we get
, the last value
in your table.
Notice in the diagram, our secant line is basically
( )) and our new
formed by our point of tangency (
(
)).
point (
f (x + h)
Now we just have to compute the slope between these
two points and then make the value of really, really
small.
f (x)
x
x+h
h
19. Let be some nonzero real number. Evaluate (
your answer.
). Note that you should get an algebraic expression for
20. Find the slope of the secant line through (1, 4) and (
expression that you will need to simplify.
21. Evaluate the slope formula from Q20 at
,
compare to the slopes from the tables in Q14 and Q16?
22. If
, what is the slope of the secant line?
23. How does this compare with your answer to Q18?
[5]
(
,
)). Here, you should also get an algebraic
, and
. How do these values
This idea of approaching a value without actually reaching it is called a limit, and it is a central idea in calculus.
24. Use the method from Q14-Q18 or Q19-Q22 to find the slope of the line tangent to ( )
point (2, 10).
at the
Application:
In physics, the position (distance from the ground, let’s say) of an object (a football, a rocket, a BASE-jumping
lunatic) over time can be modeled by a function. If this function is linear, which is rarely the case, the slope of the
line would indicate the speed of the object. For a nonlinear function, the speed of the object is calculated by
finding the instantaneous rate of change at a particular point. This means that the speed of an object at any point
is equal to the slope of the line tangent to the function at that point.
25. If a stone is dropped from a height of 275 feet, its distance from the
ground ( ) can be modeled by the function ( )
,
where is time, measured in seconds. Find the speed of the rock at
seconds. In other words, find the slope of the tangent line at the
point (3, ( )).
In conclusion, we have discovered that nonlinear functions do not have a constant rate of change. Instead we
consider an instantaneous rate of change at a particular point on the function which is equal to the slope of the
tangent line at that point. To find this slope, since we only have one point, we use a series of secant lines that
approach the tangent line from above and below the point of tangency. When you take calculus in a few years,
you will find a much faster way of computing the slope of a tangent line, which involves finding the derivative of
2
the function. Until then, let’s return to your friendly game of Yahtzee and that discussion you were having with
your friend about parabolas. As it turns out, your friend’s comment about slope was not as tangential as it seemed
at the time. Perhaps your friend knows a lot more than he and/or she lets on. Who is this friend of yours? Maybe
he/she could help you with this project.
2
Or Scrabble.
[6]
[7]
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