If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate at which the diameter decreases when the diameter is 11 cm.
a = pi d^2
da/dt = 2pi d dd/dt
You have da/dt and d, so find dd/dt
To find the rate at which the diameter decreases, we need to relate the surface area and the diameter of the snowball. The surface area of a sphere is given by the formula:
Surface area = 4πr^2
Where r is the radius of the sphere. Since the diameter is twice the radius, we can write:
Surface area = 4π(d/2)^2
Surface area = 4π(d^2/4)
Surface area = πd^2
Now, we need to find the rate at which the diameter decreases when the surface area decreases at a rate of 3 cm^2/min. To do this, we differentiate the equation with respect to time (t):
d(Surface area)/dt = d(πd^2)/dt
Now we can apply the chain rule:
d(Surface area)/dt = 2πd(d/dt)
Given that d(Surface area)/dt = -3 (since the surface area is decreasing at a rate of 3 cm^2/min), we can solve for the rate at which the diameter decreases (d(diameter)/dt):
-3 = 2πd(d(diameter)/dt)
Now we can plug in the initial diameter value (d = 11 cm) to find the rate at which the diameter decreases when the diameter is 11 cm:
-3 = 2π(11)(d(diameter)/dt)
Simplifying the equation:
-3 = 22π(d(diameter)/dt)
Finally, we solve for d(diameter)/dt:
d(diameter)/dt = -3/(22π)
So, the rate at which the diameter decreases is approximately -0.043 cm/min when the diameter is 11 cm. Note that the negative sign indicates that the diameter is decreasing.