If the volume of an expanding cube is increasing at the rate of 7m^3/sec,how fast is it surface area increasing when the surface area is 24cm^2.
To find the rate at which the surface area is increasing, we need to find the derivative of the surface area with respect to time.
Let the side length of the cube be denoted by x. The volume of the cube is given by x^3, and the surface area is given by 6x^2.
We are given that the volume is increasing at a rate of 7 m^3/sec, so we have dV/dt = 7. Taking the derivative of both sides of the volume equation, we get:
dV/dt = d/dt (x^3) = 3x^2 (dx/dt)
Since we are given that dV/dt = 7, we can solve for dx/dt:
7 = 3x^2 (dx/dt)
dx/dt = 7 / (3x^2)
Now, we are given that the surface area is 24 cm^2, which is equal to 0.0024 m^2. So we have:
6x^2 = 0.0024
x^2 = 0.0004
x = 0.02 (since side length cannot be negative)
Now we can use the value of x to find the rate at which the surface area is increasing:
dx/dt = 7 / (3(0.02)^2)
dx/dt = 7 / (3(0.0004))
dx/dt ≈ 58333.33 m^2/sec
Therefore, the surface area is increasing at a rate of approximately 58333.33 m^2/sec when the surface area is 24 cm^2.
To find the rate at which the surface area of the cube is increasing, we can use the relationship between the surface area and the volume of a cube.
1. Let’s start by converting the given surface area from cm^2 to m^2. Since 1 m^2 = 10,000 cm^2, the surface area is:
24 cm^2 = 24/10,000 m^2 = 0.0024 m^2
2. Let's denote the side length of the cube as s (in meters). Since all sides of a cube are equal, we have:
Surface Area (SA) = 6 * s^2
Volume (V) = s^3
3. We are given that the volume is increasing at a rate of 7 m^3/sec. So, dV/dt = 7.
4. We need to find ds/dt, the rate at which the side length is changing when the surface area is 24 cm^2. To do this, let's differentiate the volume equation with respect to time (t):
dV/dt = d(s^3)/dt
5. Using the power rule of differentiation, we have:
7 = 3s^2 * ds/dt
6. Rearranging the equation, we get:
ds/dt = 7 / (3s^2)
7. Now, let's find the value of s when the surface area is 0.0024 m^2 by substituting the values into the surface area equation:
0.0024 = 6s^2
8. Solving for s, we have:
s^2 = 0.0024 / 6
s^2 = 0.0004
s = √(0.0004)
s ≈ 0.02 m
9. Finally, substitute the value of s into the equation for ds/dt to find the rate at which the surface area is increasing:
ds/dt = 7 / (3 * (0.02)^2)
10. Calculating ds/dt:
ds/dt = 7 / (3 * 0.0004)
ds/dt = 7 / 0.0012
ds/dt ≈ 5833.33 m^2/sec
Therefore, the surface area of the cube is increasing at a rate of approximately 5833.33 m^2/sec when its surface area is 24 cm^2.