At what rate, in cm/s, is the radius of the circle increasing when the radius is 5 cm if oil is poured on a flat surface, and it spreads out forming a circle and the area of this circle is increasing at a constant rate of 5 cm2/s?

Question

Samantha Adkison · Accepted Answer

You can think that both your area #A# and radius #r# are function of time so basically:
#A(t)=pi*[r(t)]^2# (area of the circle) deriving you get that your area changes and your radius changes as: #(dA(t))/dt=2pir(t)(dr(t))/dt# But: #(dA(t))/dt=5(cm^2)/s#
Considering the instant when #r(t)=5cm# you get:
#5=2pi*5(dr(t))/dt# you can simplify #5# and take #2pi# to the side dividing;
and finally the rate of change of your radius will be:
#(dr(t))/dt=1/(2pi)=0.16(cm)/s#

Emily Ammerman · Answer

undefined To find the rate at which the radius of the circle is increasing when the radius is 5 cm, given that the area of the circle is increasing at a constant rate of 5 cm²/s, we can use the formula for the rate of change of the area of a circle with respect to its radius.

Let $ A $ be the area of the circle and $ r $ be its radius.

Given:
$$ \frac{dA}{dt} = 5 \, \text{cm}^2/\text{s} $$

Using the formula for the area of a circle:
$$ A = \pi r^2 $$

Differentiating both sides with respect to time:
$$ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $$

We can solve for $ \frac{dr}{dt} $:
$$ 5 = 2\pi \times 5 \times \frac{dr}{dt} $$
$$ \frac{dr}{dt} = \frac{5}{10\pi} $$
$$ \frac{dr}{dt} = \frac{1}{2\pi} $$

So, the rate at which the radius of the circle is increasing when the radius is 5 cm is $ \frac{1}{2\pi} $ cm/s.