AguillabChim3
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The radius of a circular oil slick expands at a rate of 2 m/min.

(a) How fast is the area of the oil slick increasing when the radius is 25 m
(b) If the radius is 0 at time , how fast is the area increasing after 4 mins

Respuesta :

As we knw that the dr/dt = 2
 And  The area of the circular slick is given by formul
 A = (pi)r^2.
Now lets take derivative: dA/dt=2πrdr/dt
As we knw that radius is 25
sodA/dt=2π(25)(2)=100π
which is equal to
=314.16 per minute this is how area is increasing
now for part b
 t = 4, the radius will have increased to 8 m.
   now we will put the values in formula dr/dt
so it will be.dA/dt=2π(8)(2)
=32π
=100.53 is the answer
hope it helps

Using implicit differentiation, it is found that:

a) The area is expanding at a rate of 314 m²/min.

b) The area is expanding at a rate of 100.5 m²/min.

The area of a circle of radius r is given by:

[tex]A = \pi r^2[/tex]

It's implicit derivative in function of time is:

[tex]\frac{dA}{dt} = 2\pi r \frac{dr}{dt}[/tex]

The radius of a circular oil slick expands at a rate of 2 m/min means that [tex]\frac{dr]{dt} = 2[/tex], thus:

[tex]\frac{dA}{dt} = 4\pi r[/tex]

Item a:

Radius of 25 m, thus [tex]r = 25[/tex].

The rate that the area is expanding is:

[tex]\frac{dA}{dt} = 4\pi (25) = 314[/tex]

The area is expanding at a rate of 314 m²/min.

Item b:

Radius starts at 0, expands at a rate of 2 m/min, thus, after 4 minutes, [tex]r = 8[/tex].

The rate that the area is expanding is:

[tex]\frac{dA}{dt} = 4\pi (8) = 100.5[/tex]

The area is expanding at a rate of 100.5 m²/min.

A similar problem is given at https://brainly.com/question/24861732