An oil spill is increasing such that the surface covered by...

An oil spill is increasing such that the surface covered by the spill is always circular. Find the rate at which the area $A$ of the surface is changing with respect to the radius $r$ of the circle at
(a) any value of $r$
(b) $r=500 \mathrm{ft}$

Key Concepts

Power Rule

The power rule is a basic rule in calculus used to differentiate functions of the form f(x) = x?, where the derivative is given by f'(x) = n·x??¹. This rule is particularly useful when differentiating the area of a circle with respect to the radius, as the area expression involves a squared term.

Circle Geometry

The area of a circle is determined by its radius using the formula A = ?r². This fundamental relationship in geometry connects a one-dimensional measure (radius) to a two-dimensional measure (area), forming the basis for many applications involving circular shapes.

Differentiation

Differentiation is a core concept in calculus that involves computing the rate at which one quantity changes with respect to another. In the context of geometric relationships, it allows us to determine how the area of a shape changes as its defining parameters, such as the radius of a circle, change.

Transcript

00:01 The area of a circle is equal to pi times the radius squared.

00:04 So we're going to find the instantaneous rate of change function for our area, which i'm going to denote as just a prime, and that's just equal to the limit as h goes to 0 of a of r plus h, minus a of r, all divided by h.

00:24 So we use the same equation we would use if we are looking for the instantaneous velocity from a displacement function.

00:31 So this means that a prime, or the instantaneous rate of change of our area, is equal to the limit as h goes to 0 of pi times r plus h squared, minus pi times r squared, all divided by h.

00:49 So i'm going to expand this factor here.

00:52 So we have this is equal to the limit as h goes to zero of pi, times r squared plus 2 r h not 2 r h plus h squared and then this is minus pi r squared and divided by h so now i'm going to distribute this pi through this factor here so we have this is equal to the limit as h goes to zero of pi r squared plus 2 pi r h plus pi r h plus pi r h plus pi h squared plus pi h squared minus pi r squared divided by h.

01:33 And so what we can do now is we can cancel out this pi r squared with this negative pi r squared in the numerator.

01:40 And then we can also factor out in h.

01:43 So this will be equal to the limit as h goes to zero of h multiplied by 2 pi r plus pi times h divided by h.

01:55 And again all i did was cancel out to two factors that i could in the numerator and then factor out in h of what was remaining...

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