Starting at midnight, a 10 foot radius circular pond freezes inward from the outer edge at a rate of 4 inches per hour. How fast is the open area shrinking at when the radius is 9 feet? (Note: if I'd said "at 3am" instead of "when radius is 9 ft", this would just be a chain rule problem

y=csc^4 x - 2cot^2 x

y=x^2x+1

area = πr^2

d(area)/dt = 2πr dr/dt

given: dr/dt = -4 inches/hr
find d(area)/dt when r = 108 inches

d(area)/dt = 2π(108)(-4) inches^2/hr
= appr 2714.3 inches^2/h

if you had said 3:00 am, the time would have been 3 hrs, and the radius would have shrunk by 12 inches, making r = 9, the same as the first wording.

To find the rate at which the open area is shrinking when the radius is 9 feet, we can use the chain rule.

Let's first find the formula for the area of the circular pond as a function of the radius. The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.

Now let's differentiate the area function with respect to time, t, using the chain rule.

dA/dt = dA/dr × dr/dt

The rate of change of the area with respect to time is equal to the rate of change of the area with respect to the radius multiplied by the rate of change of the radius with respect to time.

Since the radius is decreasing at a rate of 4 inches per hour, we have dr/dt = -4 in/hr.

We need to find dA/dt when the radius is 9 feet, which is r = 9 ft.

Now let's find dA/dr, which is the derivative of the area function with respect to the radius.

A = πr^2
dA/dr = 2πr

Substituting r = 9 ft into the expression for dA/dr, we get dA/dr = 2π(9) = 18π ft^2.

Now substitute dr/dt = -4 in/hr and dA/dr = 18π ft^2 into the chain rule equation:

dA/dt = dA/dr × dr/dt
dA/dt = (18π ft^2) × (-4 in/hr)

To convert the units, we need to convert inches to feet:

1 ft = 12 inches
1 in/ft = 1/12 ft/in

So, we can convert the rate of change of the radius from inches per hour to feet per hour:

dr/dt = -4 in/hr × (1/12 ft/in) = -1/3 ft/hr

Substituting the values, we get:

dA/dt = (18π ft^2) × (-1/3 ft/hr)
dA/dt = -6π ft^3/hr

Therefore, the open area is shrinking at a rate of -6π square feet per hour when the radius is 9 feet.

To find the rate at which the open area is shrinking, we need to determine how the area is changing with respect to time.

Let's denote the radius as r, and let A represent the area of the pond at any given time.

We know that the pond is freezing inward at a rate of 4 inches per hour. This means that the radius is decreasing by 4 inches every hour, or in terms of time, dr/dt = -4 inches/hour.

The area of a circle is given by the formula A = πr^2. To determine how the area is changing with respect to time, we will take the derivative of the area with respect to time, dA/dt.

Using the chain rule, we have:

dA/dt = dA/dr * dr/dt

Applying the chain rule, dA/dr represents the derivative of A with respect to r, and dr/dt represents the rate at which the radius is changing with respect to time.

Taking the derivative of A = πr^2 with respect to r, we get:

dA/dr = 2πr

Now, substituting the given values, we can find dA/dt when the radius is 9 feet.

r = 9 feet
dr/dt = -4 inches/hour = (-4/12) feet/hour = -1/3 feet/hour

dA/dr = 2π(9) = 18π

Finally, we can calculate dA/dt:

dA/dt = dA/dr * dr/dt = 18π * (-1/3) = -6π

Therefore, when the radius is 9 feet, the open area is shrinking at a rate of -6π square feet per hour.