Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s. How fast is the area of the spill increasing when the radius is 19 m?

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Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s. How fast is the area of the spill increasing when the radius is 19 m?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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This is a very simple problem: \[A=\pi r^2\] And you know that the change in radius dr/dt = 2 m/s
Now you have to find out what dA/dt is, when r = 19m To find this, take the derivative of Area: dA/dt = 2*pi*r*(dr/dt)
You need to know what is the change in area when r =19m and dr/dt = 2m/s so plug those values in: dA/dt = 2pi * 19m * 2m/s dA/dt = 76*pi (m^2/s) <=Your answer

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Other answers:

looks good , i concur

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