A tanker is spilling oil into the water resulting in an oil slick that is close to circular. At te time that the slick's diameter is growing at a rate of 8 ft/sec, the diameter is 250 feet. At what rate is the area of the oil sick increasing?
area = A = pi r^2
dA/dr = 2 pi r
dA/dt = 2 pi r dr/dt
which we did not really have to do because obviously the rate of change of area is the rate of change of radius times the circumference.
To find the rate at which the area of the oil slick is increasing, we need to differentiate the equation for the area of a circle with respect to time.
The area of a circle is given by the formula A = πr^2, where A is the area and r is the radius.
Since the diameter of the oil slick is given and we want to find the rate at which the area is increasing, we need to express the area in terms of the diameter.
We know that the diameter is growing at a rate of 8 ft/sec, so we can express the diameter as a function of time: d(t) = 250 + 8t, where d(t) represents the diameter at time t.
The radius of the circle is half of the diameter, so the radius can be expressed as r(t) = (d(t))/2 = (250 + 8t)/2 = 125 + 4t.
Now we can express the area of the circle as a function of time, A(t) = π(r(t))^2 = π(125 + 4t)^2.
To find the rate at which the area is increasing, we need to differentiate A(t) with respect to time:
dA/dt = 2π(125 + 4t)(4) = 8π(125 + 4t).
Therefore, the rate at which the area of the oil slick is increasing is 8π(125 + 4t) square feet per second.
To find the rate at which the area of the oil slick is increasing, we can use the derivative. The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.
However, we're given the rate at which the diameter is growing, not the radius. So we need to find a relationship between the diameter and the radius.
We know that the diameter is twice the radius, so we can write r = d/2, where r is the radius and d is the diameter.
Now, let's differentiate the formula for the area with respect to time (t) to find the rate at which the area is changing.
A = πr^2
Taking the derivative with respect to time (t) on both sides:
dA/dt = d/dt (πr^2)
Now, let's substitute the relationship between the radius and diameter:
r = d/2
Substituting this into the equation:
dA/dt = d/dt (π(d/2)^2)
Let's simplify this expression:
dA/dt = d/dt (π(d^2/4))
= (π/4) d/dt (d^2)
= (π/4) * 2d * dd/dt
= (π/2) d * dd/dt
We're given that dd/dt (the rate at which the diameter is growing) is 8 ft/sec. Also, we know that the diameter is 250 feet at the given time. So we can substitute these values into the expression:
d = 250 ft
dd/dt = 8 ft/sec
Now, we can calculate the rate at which the area is increasing:
dA/dt = (π/2) * 250 * 8
= 1000π
Therefore, the rate at which the area of the oil slick is increasing is 1000π square feet per second.