Quiz 9 Key
1. Evaluate each expression:
a. ∫7𝑥3+7
𝑥3+ √𝑥3
7 𝑑𝑥
Answer:
∫7𝑥3+7
𝑥3+√𝑥3
7 𝑑𝑥 = 7∫𝑥3 𝑑𝑥 +7∫𝑥−3𝑑𝑥+∫𝑥3
7𝑑𝑥 = 7 ∙𝑥4
4+7 ∙𝑥−2
−2 +𝑥10
7
10
7+𝑐
=7
4𝑥4−7
2𝑥−2 +7
10𝑥10
7+ 𝑐
b. ∫7𝑢+7
√1−𝑢2+cot2𝑢 𝑑𝑢
Answer:
∫7𝑢+7
√1− 𝑢2+cot2𝑢 𝑑𝑢 = ∫ 7𝑢𝑑𝑢+7∫ 1
√1− 𝑢2𝑑𝑢+ ∫cot2𝑢𝑑𝑢
= ∫7𝑢𝑑𝑢+ 7∫ 1
√1−𝑢2𝑑𝑢+∫(−1 +csc2𝑢)𝑑𝑢 =7𝑢
ln7+7sin−1 𝑢 −𝑢 −cot𝑢 +𝑐
2. The gravity on the moon is known to be constant at approximately −1.62 𝑚/𝑠2. Suppose, while on the moon, a
rock is thrown such that, after 1 second, its velocity is 12 𝑚/𝑠 and its height is 17 𝑚 above the moon’s surface.
Using concepts and notation from calculus,
a. Determine the position function of the rock, 𝑠(𝑡)
b. Determine the maximum height of the rock.
Answer:
Determining the position function.
We are told that 𝑎(𝑡)= −1.62. Then 𝑣(𝑡)=∫𝑎(𝑡)𝑑𝑡 =∫−1.62𝑑𝑡 = −1.62𝑡+𝑐. So 𝑣(𝑡)= −1.62𝑡 + 𝑐 (m/s)
Since 𝑣(1)=12 m/s, then −1.62(1)+ 𝑐 = 12. So 𝑐 = 13.62. Then 𝑣(𝑡)= −1.62𝑡 +13.62 (m/s)
Then 𝑠(𝑡)=∫𝑣(𝑡)𝑑𝑡 =∫−1.62𝑡 +13.62𝑑𝑡 = −1.62∙𝑡2
2+13.62∙ 𝑡+ 𝑑. So 𝑠(𝑡)= −0.81𝑡2+13.62𝑡 +𝑑 (m)
Since 𝑠(1)=17 m, then −0.81(1)2+13.62(1)+ 𝑑 = 17. So 𝑑 = 4.19. Then 𝑠(𝑡)= −0.81𝑡2+13.62𝑡 +4.19 (m)
Determining the maximum height of the rock.
Solving 𝑠′(𝑡)= 0 ⇒ 𝑣(𝑡)= 0 ⇒ −1.62𝑡+13.62 = 0 ⇒ 𝑡 ≈ 8.407 s
This time clearly produces a maximum since 𝑠′′(𝑡)= 𝑎(𝑡)< 0 for all 𝑡 [Second Derivative Test]
Then 𝑠(8.407)= −0.81(8.407)2+13.62(8.407)+ 4.19 =61.4m
So the maximum height of the rock is 61.4m.
