Related rates: The volume of a cube is increasing at the rate of 40 mm^3/s. How fast is the surface area of the cube increasing when each edge is 10 mm?
for a cube of side s,
v = s^3
dv/dt = 3s^2 ds/dt
Thus, ds/dt = 2/15 mm/s
a = 6s^2
so
da/dt = 12s ds/dt
da/dt = 16 mm^2/s
To solve this problem, we need to find the relationship between the volume and the surface area of a cube, and then differentiate this relationship to find how the surface area changes with respect to time.
Let's start with the formula for the volume of a cube: V = s^3, where V is the volume and s is the length of one side (edge) of the cube.
Given that the volume is increasing at a rate of 40 mm^3/s, we can write this as dV/dt = 40 mm^3/s, where dV/dt represents the rate of change of volume with respect to time.
We are asked to find how fast the surface area of the cube is increasing when each edge is 10 mm. The formula for the surface area of a cube is A = 6s^2, where A is the surface area and s is the length of one side.
To find how fast the surface area is changing with respect to time, we need to differentiate the surface area formula with respect to time. Let's denote the rate of change of surface area with respect to time as dA/dt.
Now, we have V = s^3 and A = 6s^2, and we need to find dA/dt when s = 10 mm.
To find dA/dt, we can differentiate the surface area formula A = 6s^2 with respect to t, using the chain rule.
dA/dt = dA/ds * ds/dt
We know that dV/dt = 40 mm^3/s, which represents the rate at which the volume is changing with respect to time. Since V = s^3, we can differentiate both sides of this equation to find an expression for ds/dt in terms of dV/dt.
dV/dt = d/dt(s^3)
40 = 3s^2 * ds/dt
Simplifying this expression, we can solve for ds/dt:
ds/dt = 40 / (3s^2)
Now, we can substitute this expression for ds/dt back into the equation for dA/dt:
dA/dt = dA/ds * ds/dt
dA/dt = 12s * (40 / (3s^2))
dA/dt = 480s / (3s^2)
dA/dt = 160 / s
We know that s = 10 mm, so we can plug in this value to find:
dA/dt = 160 / 10
dA/dt = 16 mm^2/s
Therefore, when each edge is 10 mm, the surface area of the cube is increasing at a rate of 16 mm^2/s.