Exercise 1.1.5-8.
In Exercises 5-8, f(x) is the number of meters a bus has gone at a time x (in seconds). Find the
instantaneous velocity at the given time x0.
Exercise 5. f(x)=x2+3x;x0=2
Solution:
ΔyΔx=f(2+Δx)−f(2)Δx=(10+7Δx+(Δx)2)−10Δx=7Δx+(Δx)2Δx=7+Δx
Let Δx become very small. The required instantaneous velocity at x0=2 is 7 meters
per second.
Exercise 2. y=3x2+x
Solution:
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Δy=f(2+0.5)−f(2)=7.25. Avg. Velocity = ΔyΔx=7.250.5=14.5
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Δy=f(2+0.01)−f(2)=0.1303. Avg. Velocity = ΔyΔx=0.13030.01=13.01
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Δy=f(4+0.1)−f(4)=2.53. Avg. Velocity = ΔyΔx=2.530.1=25.3
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Δy=f(4+0.01)−f(4)=0.2503. Avg. Velocity = ΔyΔx=0.25030.01=25.03
Exercise 3. y=x2+10x
Solution:
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Δy=f(2+0.5)−f(2)=7.25. Avg. Velocity = ΔyΔx=7.250.5=14.5
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Δy=f(2+0.01)−f(2)=0.1401. Avg. Velocity = ΔyΔx=0.14010.01=14.01
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Δy=f(4+0.1)−f(4)=1.81. Avg. Velocity = ΔyΔx=1.810.1=18.1
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Δy=f(4+0.01)−f(4)=0.1801. Avg. Velocity = ΔyΔx=0.25030.01=18.01
Exercise 4. y=2x
Solution:
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Δy=f(2+0.5)−f(2)=1. Avg. Velocity = ΔyΔx=10.5=0.5
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Δy=f(2+0.01)−f(2)=0.02. Avg. Velocity = ΔyΔx=0.020.01=2
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Δy=f(4+0.1)−f(4)=.2. Avg. Velocity = ΔyΔx=.20.1=2/p>
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Δy=f(4+0.01)−f(4)=0.02. Avg. Velocity = ΔyΔx=0.020.01=2