The volume of a melting cube is decreasing at a rate of 10cm^3/min. How fast is the surface area of the ice cube decreasing when the length of an edge is 30 cm?
To find the rate at which the surface area of the ice cube is decreasing, we need to differentiate the surface area formula with respect to time.
The volume of a cube with edge length "s" can be expressed as V = s^3.
Since the volume is decreasing at a rate of 10 cm^3/min, we can determine the rate of change as dV/dt = -10 cm^3/min.
Differentiating both sides of the equation with respect to time, we get:
dV/dt = d/dt (s^3)
-10 = 3s^2 * ds/dt
We can rearrange this equation to find ds/dt:
ds/dt = -10 / (3s^2)
Now, we know that the surface area of a cube can be expressed as A = 6s^2. Differentiating both sides of the equation with respect to time, we get:
dA/dt = d/dt (6s^2) = 12s * ds/dt
Substituting the value of ds/dt we found earlier:
dA/dt = 12s * (-10 / (3s^2)) = -40s / s^2 = -40 / s
Now, we can substitute the given value of s = 30 cm to find the rate at which the surface area is changing:
dA/dt = - 40 / (30 cm) = -4/3 cm^2/min
Therefore, when the length of an edge is 30 cm, the surface area of the ice cube is decreasing at a rate of 4/3 cm^2/min.
To find how fast the surface area of the ice cube is decreasing, we can use the chain rule for differentiation.
Let's assume that the edge length of the cube is "x" (in cm), and the volume of the cube is given by V = x^3 (in cm^3).
Since the volume is decreasing at a rate of 10 cm^3/min, we can express this as dV/dt = -10 cm^3/min.
To find dA/dt, the rate of change of the surface area, we need to differentiate the surface area formula of the cube, A = 6x^2, with respect to time (t).
Applying the chain rule, we have:
dA/dt = dA/dx * dx/dt
Now, let's differentiate A = 6x^2 with respect to x:
dA/dx = 12x
Substituting the given values:
dx/dt = -10 cm^3/min
x = 30 cm
dA/dt = (12x) * (-10) = -120x
Substituting x = 30 cm:
dA/dt = -120 * 30 = -3600 cm^2/min
Therefore, the surface area of the ice cube is decreasing at a rate of 3600 cm^2/min when the length of an edge is 30 cm.
V = x^3
dV/dt = 3x^2 dx/dt
10 = 3x^2 dx/dt
dx/dt = 10/(3x^2)
SA = 6x^2
d(SA)/dt = 12x dx/dt
dx/dt = d(SA)/dt / (12x)
d(SA)d/t / (12x) = 10/(3x^2)
d(SA)/dt = 120x/(3x^2)
which when x = 30
= 120(30)/2700 = 4/3 cm/min