In calm waters, the oil spilling from the ruptured hull of a grounded tanker spreads in all directions. Assuming that the polluted area is circular, determine how fast the area is increasing when the radius of the circle is 60 ft and is increasing at the rate of 1 2 ft/sec?

Question

Mathematics

Guest

29 January, 19:08

In calm waters, the oil spilling from the ruptured hull of a grounded tanker spreads in all directions. Assuming that the polluted area is circular, determine how fast the area is increasing when the radius of the circle is 60 ft and is increasing at the rate of 1 2 ft/sec?

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Answers (1)

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Callie Wolfe · Answer 1

2260.8 ft ^ 2 / s

Step-by-step explanation:

To solve the problem we first know that:

dr / dt = 6 ft / s

r = 60 ft

Now they ask us for dA / dt, the change of the area with respect to time, by chain rule we have to:

dA / dt = dA / dr * dr / dt

we have that A = pi * r ^ 2

d (pi * r ^ 2) / dr

deriving the above we are left with:

2 * pi * r

replacing:

dA / dt = (2 * pi * r) * 6

dA / dt = 2 * 3.14 * 60 * 6

dA / dt = 2260.8 ft ^ 2 / s

therefore the area change per second is 2260.8 ft ^ 2 / s