A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = r2h, where r is the radius and h is the height of this cylinder.)

I need help finding the dr/dt only cause I already did the rest.

Answer is c .803 METERS

V= T*pi*R^2 , T is thickness

Noticew I didn't use your formula for volume.

dV/dt= PI*T*2R dr/dt+ PI*r^2 dT/dt
But dV/dt = 0 (there is no more oil).
so

dr/dt= r^2 / 2TR * dT/dt

so what is T when R=8m?

V=PI*r^2*T
T= 8m^3/(PI*64m^2)=1/PI*8 m

in cm...T=100/8PI= you do it.
now working all measurements in cm
dR/dt= r^2/2TR dt/dr
dr/dt= 800/[2(800/8PI)800] * .1cm/hr
check all that.
dr/dt should be in cm/hr

Best

Why did the oil slick join the circus? Because it heard there would be a lot of spills and thrills! Now, let's calculate the rate at which the radius of the slick is increasing.

Given:
Height, h = 0.01 m (converted from 0.1 cm)
Volume, V = 1 cubic meter
Radius, r = 8 meters

Using the volume formula for a cylinder, V = r^2h, we can express r in terms of h as:
r = sqrt(V/h)

Since we are given that h is changing with respect to time (t), we can differentiate the equation with respect to t:
dr/dt = (1/2)*(V/h^2)*(dh/dt)

Now let's substitute the given values and solve for dr/dt:

dr/dt = (1/2)*(1 cubic meter)/(0.01 m)^2)*(dh/dt)

dr/dt = (1/2)*(1 cubic meter)/(0.0001 square meters)*(dh/dt)

dr/dt = 5000*(dh/dt)

So, the rate at which the radius of the slick is increasing when the radius is 8 meters is 5000 times the rate at which the height is decreasing, dh/dt.

To find the rate at which the radius of the slick is increasing (dr/dt), we can use the relationship between the volume of the oil slick and the radius and height of the cylinder.

Given:
- The volume of the oil slick is 1 cubic meter,
- The thickness of the oil slick is decreasing at a rate of 0.1 cm/hr,
- The radius of the slick is 8 meters,

We need to find dr/dt, the rate at which the radius of the slick is increasing.

We can start by expressing the volume of the oil slick in terms of the radius and height of the cylinder, using the formula V = r²h.

Since the oil slick is very flat, we can consider the height (h) to be constant. Therefore, we can write:

V = r²h

Differentiating both sides of the equation with respect to time (t), we have:

dV/dt = d(r²h)/dt

Since h is constant, the derivative of h with respect to time is zero, so the equation simplifies to:

dV/dt = 2rh(dr/dt)

Now, let's substitute the given values into the equation:

dV/dt = 2 * 8 * h * (dr/dt)

We know that dV/dt = 1 cubic meter/hour, so:

1 = 16h(dr/dt)

Now we need to find the value of h, the constant height. The problem statement does not provide the value of h, so we won't be able to find the exact value of dr/dt. However, we can still find the relationship between dr/dt and h.

To find h, we can recall that the thickness of the oil slick is decreasing at a rate of 0.1 cm/hr, which is equivalent to 0.001 m/hr.

Since 1 cm = 0.01 m, we can express the thickness in meters:

0.1 cm/hr = 0.001 m/hr

We can consider the thickness as h in our equation. Therefore:

h = 0.001 m/hr

Now we can substitute this value back into our equation:

1 = 16 * 0.001 * (dr/dt)

1 = 0.016(dr/dt)

Finally, we can solve for dr/dt:

dr/dt = 1 / 0.016

dr/dt ≈ 62.5 meters per hour

Therefore, when the radius of the slick is 8 meters, the rate at which the radius is increasing is approximately 62.5 meters per hour.