MATH 107 UMBC WK 6 Real Applications Involving Parabolas Question

Week 6 Discussion Problems
Use the given pair of functions to find the following values if they exist (i.e. are real numbers)
(𝑓 ∘ 𝑔)(2)
(𝑔 ∘ 𝑓)(0)
(𝑓 ∘ 𝑓)(−1)
(𝑔 ∘ 𝑔)(0)
3
01) 𝑓(𝑥) = √3 − 𝑥, 𝑔(𝑥) = 𝑥−1
𝑥
02) 𝑓(𝑥) = 𝑥+4 , 𝑔(𝑥) =
𝑥 2 +2
𝑥+1
Use the given pair of functions to find and simplify expressions for the functions (𝑓 ∘ 𝑔)(𝑥) and
(𝑔 ∘ 𝑓)(𝑥). State the domain of each using interval notation.
03) 𝑓(𝑥) = |𝑥 − 1|, 𝑔(𝑥) = √4 − 𝑥
𝑥
04) 𝑓(𝑥) = 𝑥−3 , 𝑔(𝑥) = 𝑥 2 + 2𝑥
2
05) 𝑓(𝑥) = 𝑥+1 , 𝑔(𝑥) = 𝑥 − 1
1
06) 𝑓(𝑥) = √2𝑥 − 1, 𝑔(𝑥) = 𝑥 2
Demonstrate that the function 𝑓(𝑥) is one-to-one and find its inverse, 𝑓 −1 (𝑥). Check your answers
algebraically and graphically. Verify the that range of f is the domain of f-1 and vice versa.
07) 𝑓(𝑥) =
𝑥+2
5
08) 𝑓(𝑥) =
3𝑥−1
𝑥+3
−1
3
09) 𝑓(𝑥) = √2𝑥 + 1 + 5
10) 𝑓(𝑥) =
√𝑥
2
Use Theorem 6.2 (p. 423) to simplify the following logarithmic expressions
11) log √1000
1
12) log 5 (125)
13) log 9 (3)
14) log 2 (1)
15) log 0.001
16) log √2 (8)
17) log 1 (2)
8
18) log 25 (125)
Find all real solutions.
2
19) 3(𝑥 3 − 1) = 72
20) √𝑥 + 3 + 3 = 2(𝑥 + 1)
21) 𝑥 − 1 = √3 − 𝑥
22) (2𝑥 − 1)4/3 = 16
Real applications involving parabolas
Any problems involving gravity or other types of acceleration can be modeled using a parabolic function.
This function can be written as
𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
When the function is written in this form, the vertex of the parabola will occur at (−
𝑏
𝑏
, 𝑓 (− )).
2𝑎
2𝑎
Alternatively, the function can be converted to standard form for a parabola, 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘,
where the vertex of the parabola occurs at (ℎ, 𝑘).
With this in mind, let’s look at three examples that involve acceleration due to gravity.
I understand functions!
I’m the
KING OF THE WORLD!
With your newfound appreciation for functions, you
climb to the top of a cliff to let the world know about it.
You also bring three balls and a stopwatch to
demonstrate your understanding of how
gravity creates parabolic functions. You know
you can predict how long it will take for the
balls to hit the ground from 400 ft up.
You plan to test three scenarios.
1)
Dropping a ball (initial velocity = 0)
The height of the ball in feet above ground level,
ℎ(𝑡), can be described as a function of time in
seconds after the ball is released, 𝑡. When the
ball hits the ground, its height ℎ(𝑡) will be zero.
1
ℎ(𝑡) = 𝑎𝑡 2 + 𝑏𝑡 + 𝑐
2
𝑎 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦, −32 𝑓𝑡/𝑠 2
𝑏 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 0
𝑓𝑡
𝑠
𝑐 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 ℎ𝑒𝑖𝑔ℎ𝑡, 400 𝑓𝑡
The function becomes ℎ(𝑡) = −16𝑡 2 + 400.
ℎ(𝑡) = 0 when the ball hits the ground, so
16𝑡 2 = 400
𝑡 2 = 25
𝑡=5
(We only use the positive root because time
cannot be negative!)
ANSWER: It takes 5 seconds for the ball to hit the ground when it is dropped from a height of 400 ft.
Note that the acceleration due to gravity
(often written as 𝑔) will be the same in
all equations and is always negative
because it acts downward. So when
gravity is involved and you are working
with distance in feet and time in seconds,
the first term will be −𝟏𝟔𝒕𝟐 .
The graph to the right shows the
HEIGHT of the ball above the ground as
a function of TIME after it was released.
(The x-axis does NOT show how far
away from the cliff the ball is when it
lands; we assume the ball is falling
STRAIGHT down.
NOTE: The FULL graph of the function ℎ(𝑡) = −16𝑡 2 + 400 passes through all four quadrants.
Because we cannot have negative time or height, we are only interested in Quadrant I.
Much of real-world data involves only non-negative numbers for x and y, so the domain is
often restricted to {𝑥|𝑥 ≥ 0}. Similarly, the range is often limited to {𝑦|𝑦 ≥ 0} when we
consider functions in the real world.
Now let’s consider the next scenario where the ball DOES have an initial velocity.
2) Throwing a ball downward (initial velocity = -40 ft/s)
The acceleration due to gravity will not change,
but now there is a velocity term in the
function. The velocity will be negative
because it is sending the ball downward.
1
ℎ(𝑡) = 𝑎𝑡 2 + 𝑏𝑡 + 𝑐
2
𝑎 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦, −32 𝑓𝑡/𝑠 2
𝑏 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, −𝟒𝟎
𝒇𝒕
𝒔
𝑐 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 ℎ𝑒𝑖𝑔ℎ𝑡, 400 𝑓𝑡
The function becomes
ℎ(𝑡) = −16𝑡 2 − 40𝑡 + 400
ℎ(𝑡) = 0 when the ball hits the ground, so
−16𝑡 2 − 40𝑡 + 400 = 0
𝑡=
40 ± √1600 − 4(−16)(400)
2(−16)
𝑡=
40 ± √27200
−32
𝑡=
40 ± 164.924225
−32
𝑡 = 3.90
Note: When you have a middle term in
the function due to velocity, you will
need to solve using the quadratic
formula. Again, only the positive value
for time is meaningful.
ANSWER: It takes 3.9 seconds for the
ball to hit the ground when it is thrown
straight downward with a velocity of
-40 ft/s from a height of 400 ft.
As expected, this is LESS time than it
takes for a dropped ball to reach the
ground from the same height (which is
shown in red).
3)
Throwing a ball upward (initial velocity = +40 ft/s)
When the ball is thrown upward, the
velocity is positive.
1
ℎ(𝑡) = 𝑎𝑡 2 + 𝑏𝑡 + 𝑐
2
𝑎 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦, −32 𝑓𝑡/𝑠 2
𝑏 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, +𝟒𝟎
𝒇𝒕
𝒔
𝑐 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 ℎ𝑒𝑖𝑔ℎ𝑡, 400 𝑓𝑡
ℎ(𝑡) = −16𝑡 2 + 40𝑡 + 400
ℎ(𝑡) = 0 when the ball hits the ground, so
−16𝑡 2 + 40𝑡 + 400 = 0
𝑡=
−40 ± √1600 − 4(−16)(400)
2(−16)
𝑡=
−40 ± √27200
−32
𝑡=
−40 ± 164.924225
−32
𝑡 = 6.40 (taking only the positive root)
ANSWER: It takes 6.4 seconds for the ball to hit the ground when it is thrown straight upward with a
velocity of 40 ft/s from a height of 400 ft.
As expected, this is MORE time than it
takes for a dropped ball to reach the
ground from the same height (which is
shown in red).
There is also a maximum height
(greater than 400 ft) that the ball
reaches. How long does it take to reach
its maximum height? Can we determine
how high the ball goes?
HINT: Find the coordinates of the
vertex of the parabola.