Question

A massive oil spill in the gulf unleashes approximately 20,311 barrels of oil into the Gulf each hour. This creates an expanding circular layer of oil on the water’s surface about 1/16 inches thick with the center being the source of the spill. Letting R(t) represent the radius (in miles) t hours after 6:00pm, the growing radius of this oil spill can be modeled by the formula: R(t)=1/2 √(t+1) A.) What time did this spill start? (When was the radius zero?) B.) Fill in the table below: (round your answer to 2 decimal places) Table 1 t hours 0 .5 1 1.5 2 2.5 3 3.5 4 4.5 R(t) miles C.) If left unchecked, how long will it take this oil spill to reach a 2 mile radius? The nearest containment crew is on the Louisiana coast 50 miles away. At 6:00 pm, containment vessels instantly head towards the center of this spill, but the fastest these containment ships can travel is only 15 mph. D.) Write an equation that represents the distance D(t) in miles that the containment vessel is from the center of the spill t hours after 6:00 pm. E.) Fill in the table below: Table 2 t hours 0 .5 1 1.5 2 2.5 3 3.5 4 4.5 D(t) miles F.) When will the containment vessels reach the center of spill? G.) By observing the two tables above, in which 30 minute interval will the containment vessels reach the outer edge of the spill? H.) Algebraically find exactly (to the nearest minute) when the containment vessels will reach the outer edge of the oil spill. You should get two answers…explain them. I.) What is the radius of the oil spill at this time? J.) To manage the spill, one containment vessel is needed every 800 feet around the outer circumference of the spill. How many vessels do they need? (5280 ft. = 1 mile)

Answer 1

R = 1/2 * sqrt(t + 1)

0 = 1/2 * sqrt(t + 1)

t = -1

So, the spill started at t = -1
as in 5 pm

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b)
When t = 0.5, R = 1/2 * sqrt(1.5) = 0.6123724356957945 miles
When t = 1, R = 1/2 * sqrt(2) = 0.7071067811865475 miles
When t = 1.5, R = 1/2 * sqrt(2.5) = 0.7905694150420948 miles
When t = 2, R = 1/2 * sqrt(3) =0.8660254037844386 miles
When t = 2.5, R = 1/2 * sqrt(3.5) = 0.9354143466934853 miles
When t= 3, R = 1/2 * sqrt(4) = 2/2 = 1 mile
When t = 3.5, R = 1/2 * sqrt(4.5) = 1.0606601717798213 miles
When t = 4, R = 1/2 * sqrt(5) = 1.1180339887498948 miles
When t = 4.5, R = 1/2 * sqrt(5.5) = 1.1726039399558574 miles

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c)
2 =1/2 * sqrt(t + 1)
4 = sqrt(t + 1)

Squaring :
16 = t + 1

t = 15 hrs

So, it happens 15 hrs after 6 pm
i.e
9 am the next day

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d)
D = 15*t

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f)
50 = 15*t

t = 50/15

t = 10/3 hrs
to reach the center of the spill

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