if a snowball melts so that its surface area decreases at a rate f 1cm^2/min, find the rate at which the diameter decrease when the diameter is 10cm?
SA = 4πr^2
d(SA)/dt = 8πr dr/dt
when d(SA)/dt = -1 , and r = 5
-1 = 8π(5) dr/dt
dr/dt = -1/(40π)
dD/dt = -1/(20π) or appr -.016 cm/min
(the negative sign indicates that the diameter is decreasing)
To find the rate at which the diameter decreases when the diameter is 10 cm, we can use the formula for the surface area of a sphere, which is given by:
A = 4πr^2
Where A is the surface area and r is the radius (which is half the diameter).
Since we want to find the rate at which the diameter decreases, we need to find the derivative of the surface area with respect to time.
First, let's express the surface area in terms of the diameter instead of the radius. Since the radius is half the diameter, we have:
A = 4π(r^2) = 4π[(d/2)^2] = πd^2
Now, let's take the derivative of the surface area with respect to time (t):
dA/dt = d/dt (πd^2)
To find dA/dt, we can differentiate each term separately. The derivative of π is 0 since it is a constant, and the derivative of d^2 is 2d (using the power rule).
dA/dt = 0 + 2πd(d/dt)
Now, we need to determine the value of (d/dt), which is the rate at which the surface area decreases, given as 1 cm^2/min. So we substitute that value into the equation:
dA/dt = 2πd(d/dt)
1 = 2πd(d/dt)
Solving for (d/dt), the rate at which the diameter decreases, we divide both sides of the equation by 2πd:
(d/dt) = 1 / (2πd)
Now we can substitute the given diameter of 10 cm into the equation to find the rate at which the diameter decreases when it is 10 cm:
(d/dt) = 1 / (2π(10))
(d/dt) = 1 / (20π)
(d/dt) = 1 / (20 × 3.14)
(d/dt) ≈ 0.008 cm/min
Therefore, the rate at which the diameter decreases when the diameter is 10 cm is approximately 0.008 cm/min.