An offshore oil well is leaking oil and creating a circular oil slick. If the radius of the slick is growing at a rate of 4 miles/hour, find the rate at which the area is increasing when the radius is 7 miles. (The area of a disc of radius r is A = πr2.)
Area = πr^2
d(area)/dt = 2πr dr/dt
given: dr/dt = 4
find: d(area)/dt when r = 7
d(area)/dt = 2π(7)(4) = 56π miles^2 / hour
Thank you!
Hmmm. Assuming that in precal you have not yet learned to take a derivative, consider that the area is increasing at a rate equal to the change in radius times the circumference. That is, if r changes very little, you are just adding a new, very thin circle to the area.
That of course, would be (2πr cm)*(4 cm/hr) = 56 cm^2/hr
To find the rate at which the area is increasing, we need to differentiate the area function with respect to time. Let's break down the problem into smaller steps:
Step 1: Find the equation for the area of the oil slick in terms of the radius.
The area of a disc is given by the formula A = πr^2, where r is the radius of the slick.
Step 2: Differentiate both sides of the equation with respect to time (t).
dA/dt represents the rate of change of the area with respect to time.
Step 3: Apply the chain rule to differentiate.
Since both A and r are functions of time, we need to use the chain rule to differentiate A with respect to t.
Now let's solve the problem using these steps:
Step 1: Equation for the area of the oil slick:
A = πr^2
Step 2: Differentiate both sides with respect to time (t):
dA/dt = d/dt(πr^2)
Step 3: Apply the chain rule:
dA/dt = 2πr(dr/dt)
Now we need to substitute the given information into the equation.
Given information:
dr/dt = 4 miles/hour (rate at which the radius is growing)
r = 7 miles (radius)
Substitute the values into the equation:
dA/dt = 2π(7)(4)
Simplify the expression:
dA/dt = 56π
Therefore, the rate at which the area is increasing when the radius is 7 miles is 56π square miles per hour.