The volume of a cube is increased from 729 cubic centimetres to 849.285 cubic centimetres.
Use differentials to determine:
a) by approximately how many centimetres the side length of the cube increases ?
B) by approximately how many square centimetres does the surface area of the cube increase ?
Again?
http://www.jiskha.com/display.cgi?id=1413057054
ya i just didn't get the second part because its volume not the surface area and i got a big number I'm not sure about it
a = area = 6 x^2
da/dx = 12 x
so
da = 12 x dx
To solve this problem using differentials, we can start by understanding the formulas for the volume and surface area of a cube.
The volume of a cube is given by the formula: V = s^3, where s represents the side length of the cube.
The surface area of a cube is given by the formula: A = 6s^2, where A represents the surface area and s represents the side length.
a) Let's find out approximately how many centimeters the side length of the cube increases.
We are given that the volume of the cube increased from 729 cubic centimeters to 849.285 cubic centimeters. So, we can set up the equation:
849.285 = s^3
To determine the approximate increase in the side length, we can take the derivative of both sides of the equation with respect to s:
dV/ds = 3s^2
Then, we can use differentials to approximate the change in volume, dV, for a small change in side length, ds:
dV ≈ 3s^2 ds
Substituting the values, we have:
20.285 ≈ 3s^2 ds
Now we can solve for ds:
ds ≈ (20.285 / (3s^2))
Let's assume the starting side length is s₀ and the increase in side length is ds. So, the new side length would be s₀ + ds.
To find the approximate increase in side length, we can plug in the values of s₀ and ds into the equation:
Δs ≈ s₀ + ds - s₀
Now you can substitute the appropriate values and calculate the increment in the side length of the cube.
b) Next, let's find out approximately how many square centimeters the surface area of the cube increases.
The surface area of the cube is given by the equation: A = 6s^2
To find the increase in surface area, we can differentiate both sides of the equation with respect to s:
dA/ds = 12s
Again, we'll use differentials to approximate the change in surface area, dA, for a small change in side length, ds:
dA ≈ 12s ds
Now, we can substitute the values and solve for ds:
ΔA = dA = 12s₀ ds
To find the approximate increase in surface area, substitute the appropriate values for s₀ and ds into the equation.
Using these approaches, you can approximate the increase in side length and surface area of the cube.