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(a) If $ A $ is the area of a circle with radius $ r $ and the circle expands as time passes, find $ dA/dt $ in terms of $ dr/dt. $(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spills increases at a constant rate of $ 1 m/s $, how fast is the area of the spill increasing when the radius is $ 30 m? $

a) $\frac{d A}{d t}=2 \pi r \times \frac{d r}{d t}$b) 60$\pi \frac{m^{2}}{8}$

04:17

Alex L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Oregon State University

Harvey Mudd College

University of Nottingham

Boston College

Lectures

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Yes, in part A, we're told, is the area of the circle of radius R and circle. It stands as time passes and ratifying d a d t. In terms of DRC's well, it is. The area of a circle of radius R A is equal to high on swear and therefore thinking there isn't on both sides with respective t. We have that already today, with respect to t is equal to buy the chain rule. We have two times high art times. Derivatives are with respect to teach and so we have DVT in terms of the Arctic highest, then in part B or call to suppose that oil spills from a ruptured tanker and spreads in a circular Tatham. We're told the radius of the oil spills increases at a constant rate of one meters per second and rest how fast the area of the still is increasing and the radius is 30 m. In other words, this is part A except for yeah, changing the radius drd t. This is one million per second and your radius is 30 m always. And so we have that Yes, the rate at which the area of the spill is changing d a d t. Well, this is two times pi times 30 times one, which is 60 pie and the human. It's are well, the area is in square meters and the time is in second. So square meters per second to work it's

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