A spherical snowball melts in such a way that its surface area decreases at a rate of 1 cm2/min. Find the rate at which its diameter is decreasing when the diameter is 10 cm
A = 4 pi r^2
r = D/2
so
A = 4 pi D^2/4 = pi D^2
dA/dD = 2 pi D
dA/dt = 2 pi dD/dt
1 = 2 pi dD/dt
dD/dt = 1/(2pi)
To find the rate at which the diameter of the snowball is decreasing, we need to use the relationship between the surface area and the diameter of a sphere.
The surface area (A) of a sphere is given by the formula: A = 4πr², where r is the radius of the sphere.
Since we need to find the rate at which the diameter is decreasing, we will work with the diameter (d) instead of the radius. The diameter is twice the radius, so we have: d = 2r.
Differentiate the equation with respect to time (t) to get the rate of change:
dA/dt = 8πr(dr/dt)
We know that dA/dt = -1 cm²/min (since the surface area is decreasing at a rate of 1 cm²/min), and we want to find the value of dr/dt when d = 10 cm.
Substituting the given values into the equation:
-1 cm²/min = 8π(10 cm)(dr/dt)
Simplifying further, we have:
-1 = 80π(dr/dt)
Now, solve for dr/dt:
dr/dt = -1 / (80π)
Calculating the value, we get:
dr/dt ≈ -0.004 cm/min
Therefore, the rate at which the diameter is decreasing when the diameter is 10 cm is approximately -0.004 cm/min.