Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 7.5 mi^2/hr. How rapidly is radius of the spill increasing when the area is 8 mi^2?

Question

Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 7.5 mi^2/hr. How rapidly is radius of the spill increasing when the area is 8 mi^2?

I tried using A=(pi)r^2
and dA/dt=2(pi)r(dr/dt)

and plugging in 7.5=2(pi)8(dr/dt)

but the answer is incorrect. Please help!

Answer 1

HOW DID YOU USED R=1.60 CM

Answer 2

To solve this problem, you're correct in using the formula A = πr^2 to represent the area of the oil spill, where A is the area and r is the radius.

The given information states that the area increases at a constant rate of 7.5 mi^2/hr. This means that dA/dt = 7.5.

You want to find how rapidly the radius (dr/dt) is changing when the area is 8 mi^2. Substituting the given values into the equation, you have:

dA/dt = 2πr(dr/dt)
7.5 = 2πr(dr/dt)

Now you need to find the radius (r) when the area is 8 mi^2. Use the formula for the area of a circle:

A = πr^2
8 = πr^2

Solve for r:

r^2 = 8/π
r^2 = 2.546...
r ≈ √2.546
r ≈ 1.595 mi (rounded to three decimal places)

Now substitute the known values into the equation:

7.5 = 2π(1.595)(dr/dt)

Divide both sides by 2π(1.595) to solve for (dr/dt):

(dr/dt) = 7.5 / (2π(1.595))
(dr/dt) ≈ 0.739 mi/hr (rounded to three decimal places)

Therefore, the radius of the oil spill is increasing at a rate of approximately 0.739 mi/hr when the area is 8 mi^2.

Answer 3

To solve this problem, you correctly started with the formula for the area of a circle, A = πr^2. Next, you differentiated both sides of the equation with respect to time (t) using the chain rule, which gives you dA/dt = 2πr(dr/dt).

Let's break down the steps to find the correct answer:

1. Given: dA/dt = 7.5 mi^2/hr (rate at which the area increases)
2. Find: dr/dt (rate at which the radius is increasing) when A = 8 mi^2

Now, substitute the given values into the equation dA/dt = 2πr(dr/dt):

7.5 = 2π(8)(dr/dt)

Simplify:

15π(dr/dt) = 7.5

Divide both sides by 15π:

(dr/dt) = 7.5 / 15π

Simplify further:

(dr/dt) = 0.1 / π

So, when the area is 8 mi^2, the rate at which the radius is increasing is approximately 0.1/π mi/hr.

Note: When you calculated the derivative correctly, the issue may have arisen during the calculation of the value. Be sure to check your calculations and use the approximation for π when necessary.

Answer 4

pi r^2 = 8

so
r = 1.60 miles

dA/dt = 2 pi r dr/dt yes, correct
so
7.5 = 2 pi (1.6) dr/dt

dr/dt = .748 miles/hr

You used the area, 8 for the radius R