I completed most but really need help with the rest An oil t

I completed most but really need help with the rest

An oil tanker strikes a sand bar that rips a hole in the hull of the ship. Oil begins leaking out of the tanker with the spilled oil forming a circle around the tanker. The radius of the circle is increasing at the rate of 2.2 feet per hour. Write the area of the circle as a function of the radius r. Write the radius of the circle as a function of time, t. What if the radius of the circle after 2 hours? What is the radius of the circle after 2.5 hours? Use the result of part (c) to determine the area of the circle after 2 hours and 2.5 hours. Determine a function that represents area as a function of time, t. Use the result of part (e) to determine the area of the circle after 2 hours and 2.5 hours. Compute the average rate of change of the area of the circle from 2 hours to 2.5 hours. Compute the average rate of change of the area of the circle from 3 hours to 3.5 hours. Based on the results obtained in parts (g) and (h), what is happening to the average rate of change of the area of the circle as time passes? If the oil tanker is 150 yards from shore, when will the oil spill first reach the shoreline? (1 yard = 3 feet). How long will it be until 6 miles of shoreline is contaminated with oil? (1 mile = 5280 feet).

Solution

I am attempting the circled parts:

c) Circle at rate of 2.2 feet /hr

After 2 hrs radius = 2.2 x2 = 4.4 feet

After 2.5 hrs radius = 2.2 x 2.5 = 5.5 feet

d) Area for 4.4 feet = pi(4.4)^2 = Area for 5.5 feet = pi(5.5)^2 = 60.82 feet^2

Area for 5.5 feet = pi(5.5)^2 = pi(5.5)^2 = 95.03 feet^2

e) After time ,t, hrs radius = 2.2xt = 2.2t

So, Area = pi(radius)^2 = pi(2.2t)^2 =15.2t^2

g) Avg rate of change of area from 2hrs to 2.5 hrs

Avg rate of change ( 2---> 2.5 hrs) = ( 15.2(2.5)^2 -15.2(2)^2 )/( 2.5 -2)

= 68.4

h) Avg rate of change ( 3---> 3.5 hrs) = ( 15.2(3.5)^2 -15.2(3)^2 )/( 3.5 -3)

= 98.8

i) Based on g , h we see that average rate of change of area increases with time.