how secant lines become tangent lines. adrienne, samantha, danielle, and eugene, trying to raise...
TRANSCRIPT
How Secant Lines become Tangent Lines
Adrienne, Samantha, Danielle, and Eugene, trying to raise Walkathon money, build a 196 foot high-dive platform in the middle of the plaza. After charging admission for prime bleacher seats in the plaza, they then “persuade” Mr Murphy to be the first high diver.
Abby, Nica, and Hamidou observe Mr Murphy’s “dive” closely enough to form an equation for his height. They find the equation to be given by:
216196)( tts
…and the graph is given by…
216196)( tts
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
What are the coordinates for this point?
(0, 196) because at time = 0, Mr. Murphy is at the top of the 196 foot high platform.
At what time did Mr. Murphy land?
(3.5, 0) which you can find by setting s(t) = 0
What is Mr. Murphy’s average velocity during his 3.5 second plunge?
vavg s(3.5) s(0) feet
3.5 0 sec
vavg 0 196 feet
3.5 0 sec
vavg 196 feet
3.5 sec= −56 feet/sec
Why is the velocity negative?
Because the motion is downward.
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
216196)( tts
Since you are all experts at algebra…
sec05.3
)0()5.3(
feetss
vavg
sec05.3
1960
feet
vavg
sec5.3
196 feetvavg
= 56 feet/sec
How can you represent the average velocity on this graph?
The average velocity can be represented as the slope of the secant line through the initial and final points.
secm from time 0 to time 3.5
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
…can be shown graphically to be…
216196)( tts
Find Mr. Murphy’s average velocity between 1 and 3 seconds.
sec13
)1()3(
feetss
vavg
sec13
18052
feet
vavg
sec2
128 feetvavg
= 64 feet/sec
…the slope of the line through the initial and final points.
secm from time 1 to time 3
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
216196)( tts
Approximate Mr. Murphy’s instantaneous (exact) velocity at 3 seconds.
sec23
)2()3(
feetss
vavg
sec23
13252
feet
vavg
sec1
80 feetvavg
= 80 feet/sec
…which is close to the exact velocity at 3 seconds.
secm from time 2 to time 3
We can draw a secant line close to 3. we’ll start with 2 seconds.
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
216196)( tts
Approximate Mr. Murphy’s instantaneous (exact) velocity at 3 seconds.
sec5.23
)5.2()3(
feetss
vavg
sec5.23
9652
feet
vavg
sec5.0
44 feetvavg
= 88 feet/sec
…which is even closer to the exact velocity at 3 seconds.
secm from time 2.5 to time 3
We can even try a secant line through 2.5 and 3.
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
216196)( tts
I think we’re getting the idea of how to find Mr. Murphy’s instantaneous (exact) velocity at 3 seconds.
sec3
)3()(lim
3
t
feetstsv
t
We would need to get the secant points as close as we can. How would we do that?
By taking the LIMIT as one point approaches the other…
sec3
)3*16196(16196lim
22
3
t
feettv
t
v limt 3
196 16t 2 52 feet
t 3 seclimt 3
144 16t 2
t 3
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
limt 3
16(t 2 9)
t 3
216196)( tts
By taking the LIMIT as one point approaches the other…
3
)9(16lim
2
3
t
tv
t
3
)3)(3(16lim
3
t
ttv
t
)3(16lim
3t
tsec/96 feet
12
12tan
12
limtt
tstsm
tt
tansec mm In other words…
tanm
s(t)
= H
eigh
t off
of
the
grou
nd(i
n fe
et)
t = Time in seconds
We would need to get the secant points as close as we can. How would we do that?
I think we’re getting the idea of how to find Mr. Murphy’s instantaneous (exact) velocity at 3 seconds.
216196)( tts
So from this we find that as the two points on a secant line approach each other, it becomes the tangent line. And the slope of the tangent line is also called…
The Derivative
12
121
12
lim)(xx
yyxf
xx
12 xxh Let hxx 12
12
11 )(lim
12 xx
xfhxfxx
h
xfhxfh
)(lim 11
0
12
12tan
12
limtt
tstsm
tt
…which can also be written as…
h
xfhxfxf
h
)(lim)(
0
And since )(xfy
Hey! This is from Algebra class!
( )sec
f x h f xm
h
f(x)
x
h
xfhxfxf
h
)(lim)(
0
0
( )limtanh
f x h f xm
h
( , ( ))x h f x h
( , ( ))x f x