If a snowball melts so that its surface area decreases at a rate of 4 cm2/min, find the rate at which the diameter decreases when the diameter is 12 cm.
To find the rate at which the diameter is decreasing, we first need to establish a relationship between the surface area and the diameter of the snowball.
The surface area of a sphere is given by the formula: A = 4πr^2, where r represents the radius.
Since the diameter is twice the radius (d = 2r), we can rewrite the equation as: A = 4π(d/2)^2 => A = πd^2.
Now, we can differentiate both sides of the equation with respect to time to find the rate of change of the surface area:
dA/dt = d(πd^2)/dt => dA/dt = 2πd(dd/dt).
Given that the surface area is decreasing at a rate of 4 cm^2/min, we have:
dA/dt = -4 cm^2/min.
When the diameter is 12 cm, we can substitute this value into the equation:
-4 = 2π(12)(dd/dt).
Simplifying further:
-4 = 24π(dd/dt).
To find the rate at which the diameter is decreasing (dd/dt), we divide both sides of the equation by 24π:
dd/dt = -4 / (24π).
Simplifying this expression further:
dd/dt ≈ -0.053 cm/min.
Therefore, when the diameter is 12 cm, the diameter is decreasing at a rate of approximately 0.053 cm/min.
To find the rate at which the diameter decreases, we first need to establish a relationship between the surface area and the diameter of the snowball.
The surface area of a sphere can be given by the formula:
A = 4πr²
where A is the surface area and r is the radius of the sphere. Since we are given the diameter, we need to convert it to radius by dividing it by 2. Therefore, the formula can also be expressed as:
A = 4π(d/2)²
Simplifying further:
A = πd²/2²
A = πd²/4
Now, we know that the surface area decreases at a rate of 4 cm²/min. This means that the rate of change of surface area with respect to time (dA/dt) is -4 cm²/min (negative sign indicates decrease).
Next, we need to find the rate at which the diameter decreases when the diameter is 12 cm. Let's call the rate of change of diameter with respect to time as dd/dt.
To relate the rate of change of diameter with the rate of change of surface area, we can differentiate the surface area equation with respect to time:
dA/dt = d/dt(πd²/4)
Now, let's differentiate each term individually:
dA/dt = (1/4)d/dt(πd²)
dA/dt = (1/4)(2πd)(dd/dt)
dA/dt = (π/2)dd/dt
We already know that dA/dt = -4 cm²/min, so substituting this value into the equation, we have:
-4 = (π/2)dd/dt
Solving for dd/dt, we get:
dd/dt = -8/π
Therefore, when the diameter is 12 cm, the rate at which the diameter decreases is approximately -8/π cm/min.
a = 4 pi r^2 = pi d^2
da/dt = 2 pi d dd/dt
Now just solve
-4 = 2 pi (12) dd/dt