The radius in meters of a circular oil slick increases at a rate of (12 - 0.3t), where t is measured in hours. Which of the following expressions would calculate the rate at which the area of the oil slick is increasing when the radius is 100 meters?
dr/dt = (12 - .3 t)
a = pi r^2
da/dt = 2 pi r dr/dt
da/dt = 2 pi r (12-.3 t)
when r = 100
da/dt = 200 pi (12 - .3 t)
To calculate the rate at which the area of the oil slick is increasing when the radius is 100 meters, we need to find the derivative of the area with respect to time.
The area of a circle can be calculated using the formula A = πr^2, where A is the area and r is the radius.
In this case, the radius is a function of time: r(t) = 100 meters.
To find the rate at which the area is increasing, we need to find the derivative of A with respect to t: dA/dt.
Using the chain rule for differentiation, we can express the derivative of the area as follows:
dA/dt = dA/dr * dr/dt,
where dA/dr is the derivative of the area with respect to the radius and dr/dt is the derivative of the radius with respect to time.
The derivative of the area with respect to the radius is given by:
dA/dr = 2πr.
The derivative of the radius with respect to time is given by:
dr/dt = 12 - 0.3t.
Now, we can substitute the values into the expression for dA/dt:
dA/dt = dA/dr * dr/dt
= 2πr * (12 - 0.3t).
Substituting r = 100 meters:
dA/dt = 2π(100) * (12 - 0.3t)
= 200π(12 - 0.3t).
Therefore, the expression that calculates the rate at which the area of the oil slick is increasing when the radius is 100 meters is:
200π(12 - 0.3t).