If a snowball melts so that its surface area decreases at a rate of 2 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm.
Area=4PI(diameter/2)^2
dArea/dtime= 4PI*2 * diameter/2* ddiameter/dt*1/2 check that
dArea/dtime=2 PI diameter*ddiameter/dt
2=2PI*9*ddiameter/dt
solve
To find the rate at which the diameter decreases, we need to determine the relationship between the surface area and the diameter of the snowball.
The surface area of a snowball is given by the formula:
Surface area = 4πr^2,
where r is the radius of the snowball and π is a mathematical constant approximately equal to 3.14. Since the diameter of the snowball is twice the radius, we have:
Surface area = 4π * (d/2)^2,
where d is the diameter.
Given that the surface area decreases at a rate of 2 cm^2/min, we can rewrite this equation in terms of the rate of change of the surface area:
d(Surface area)/dt = -2 cm^2/min.
To find the rate at which the diameter decreases, we need to differentiate the equation with respect to time (dt):
d/dt [Surface area] = d/dt [4π*(d/2)^2].
Differentiating both sides of the equation with respect to time, we get:
d(Surface area)/dt = 2 * 4π * (d/2) * (1/2) * (dd/dt).
Simplifying the equation, we have:
-2 = 2 * 4π * (d/2) * (1/2) * (dd/dt).
Cancelling out the common factors, we get:
-2 = 2π * d * (dd/dt).
Now, we can plug in the given diameter (d = 9 cm) into the equation and solve for the rate at which the diameter decreases when the diameter is 9 cm:
-2 = 2π * 9 * (dd/dt).
Simplifying the equation further, we have:
-2π * 9 * (dd/dt) = -2.
Now we can solve for the rate at which the diameter decreases (dd/dt):
(9π * (dd/dt)) = 1.
dd/dt = 1 / (9π).
Therefore, when the diameter is 9 cm, the rate at which the diameter decreases is approximately equal to 1 / (9π) cm/min.