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I am very confused when it comes to related rates. I am not comprehending how I need to go about solve these questions. An example problem is ...

The radius of a circular oil slick expands at a rate of 6 m/min.

(a) How fast is the area of the oil slick increasing when the radius is 20 m? (b) If the radius is 0 at time t=0, how fast is the area increasing after 3 mins?

I know that I would need to use the area of a circle formula and then differentiate implicitly but everything is still very confusing. Any guidance is appreciated.

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Let's look at part $(a)$:

First, write what you want as an expression:

  • We want to know the rate at which the area is increasing, that is, ${dA\over dt}$ where $A$ represents area (measured in meters$^2$).

Second, write what you know with equations.

  • The radius expands at a rate of $6$ m/min - that is, the ${dR\over dt}=6$, where $R$ represents the radius (measured in meters) and $t$ represents time (measured in seconds).

  • The radius is $20$ meters - that is, $R=20$.

  • Finally, the area and the radius of the slick are related: $A=\pi R^2$.

So to recap, we know ${dR\over dt}=6$, $R=20$, and $A=\pi R^2$; and we want to know ${dA\over dt}$. The problem is basically how to get from $A$ to $R$.

This is where the expression $A=\pi R^2$ comes in: we have ${dA\over dR}=2\pi R$. Do you see how to combine this fact with something we already know to figure out a general expression for ${dA\over dt}$? (Hint: chain rule . . .)


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The radius of a circular oil slick expands at a rate of 6 m/min. $$ \frac{dr}{dt} =6$$

(a) How fast is the area of the oil slick increasing when the radius is 20 m? (b) If the radius is 0 at time t=0, how fast is the area increasing after 3 mins?

$$ A = \pi r^2 $$ a) find $\frac{dA}{dt}$ when $r=20$

b) find $\frac{dA}{dt}$ when $r=6(3)=18$

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