Name __Solutions____________________________________________________________ Chapter 6a
1) A man drops a penny off of the Empire State building in New York City. Ignoring air resistance, the
acceleration of the penny is 32 ft/s2 downward. The building is 1250 feet tall.
a) Find the equation for the height of the penny off the ground.
b) Find the distance the penny travels in the first 3 seconds.
If we call 𝑓(𝑡) the height of the penny, we know:
𝑓 ′′ (𝑡) = −32
𝑓 ′ (0) = 0
𝑓(0) = 1250
𝑓 ′ (𝑡) = ∫ −32𝑑𝑡 = −32𝑡 + 𝐶
−32 ⋅ 0 + 𝐶 = 0
𝐶=0
𝑓(𝑡) = ∫ −32𝑡 𝑑𝑡 = −16𝑡 2 + 𝐶
−16 ⋅ 02 + 𝐶 = 1250
𝐶 = 1250
𝑓(𝑡) = −16𝑡 2 + 1250
For the distance the penny travels, we have:
3
∫ −32𝑡 𝑑𝑡 = −16𝑡
0
The penny travels 144 feet downward.
3
2|
0
= −144
𝑡
2) A town of 40 people grows at a rate of 𝑟(𝑡) = 20 − 5 people/year for 0 ≤ 𝑡 ≤ 150.
a) Find an equation for the population of the town after 𝑡 years.
b) How many people does the town have in 20 years?
c) Around 𝑡 = 100 years there was a very contentious political topic that all the townspeople
debated at length. What do you suspect they were talking about?
𝑡
𝑡2
𝑃(𝑡) = ∫ 𝑟(𝑡)𝑑𝑡 = ∫ 20 − 𝑑𝑡 = 20𝑡 −
+𝐶
5
10
02
20 ⋅ 0 −
+ 𝐶 = 40
10
𝐶 = 40
𝑡2
𝑃(𝑡) = 20𝑡 −
+ 40
10
After 20 years we have 𝑃(20) = 20 ⋅ 20 −
At 𝑡 = 100, note that 𝑟(100) = 20 −
100
5
202
10
+ 40 = 400 people
= 0. Using the first or second derivative test, we see that this
is in fact a maximum for 𝑃(𝑡). The contentious issue is probably that the population of the town reached
a maximum and started to decrease!
3) Find the area of the shaded region enclosed by the three curves below. (Answer check: 193)
𝑦 = 4√𝑥
𝑥2
𝑦=
2
𝑦 = −2𝑥 + 6
2
4
∫ 4√𝑥 − (−2𝑥 + 6)𝑑𝑥 + ∫ 4√𝑥 − (
1
2
Note that this integral can also be computed using 𝑑𝑦.
𝑥2
19
) 𝑑𝑥 =
2
3
4) Find the area of the shaded region enclosed by the two curves below. (Answer check: 0.669)
𝑦 = 𝑥2
𝑥 = 2 sin2(𝑦)
The 𝑥-coordinates of the two marked points are:
𝑥 ≈ 0.8
𝑥 ≈ 1.5
2.25
∫ 2 sin2 (𝑦) − √𝑦 𝑑𝑦 = 0.669
0.64
5) Use calculus to find the volume of a pyramid whose base is a right isosceles triangle with base length
5. The height of the pyramid is 20. (Answer check: 83. 3̅)
If we put the right angle at the origin, and lay the triangle down on its side as illustrated below, we have
triangular cross sections.
1
1
We sum up triangular cross sections with volume 2 [𝑏𝑎𝑠𝑒][ℎ𝑒𝑖𝑔ℎ𝑡][𝑑𝑒𝑝𝑡ℎ] = 2 𝑏 ⋅ ℎ ⋅ 𝑑𝑥
1
The base and height are both given by 𝑦 = 5 − 4 𝑥. Hence we have:
20
∫
0
1
𝑥 2
(5 − ) 𝑑𝑥 = 83. 3̅
2
4
1
6) Suppose a football-shaped object is created by rotating the curve 𝑦 = 9 (𝑥 − 3)2 + 1 around the 𝑥axis. Find the volume of this object. (Answer check: 56𝜋
)
5
Here we’re adding up circular cross sections of volume:
𝜋[𝑟𝑎𝑑𝑖𝑢𝑠]2 [𝑤𝑖𝑑𝑡ℎ]
1
The radius is given by 𝑦 = 9 (𝑥 − 3)2 + 1, while the width is 𝑑𝑥.
6
1
2
∫0 𝜋 (9 (𝑥 − 3)2 + 1) 𝑑𝑥 =
56𝜋
5
However…. I wrote the equation down incorrectly so the above volume is actually wrong.
The keen mind will notice that the curve I gave and the graph shown don’t match up. The curve is
missing a negative sign, it should have been:
1
𝑦 = − (𝑥 − 3)2 + 1
9
This gives the following volume:
6
2
1
16𝜋
2
∫ 𝜋 (− (𝑥 − 3) + 1) 𝑑𝑥 =
9
5
0
If you noticed this, you received an automatic A+ on this assignment.