[dag12]

ok, I will help. I will edit this post as I go along.

1. F = Km1m2/(s^2)
dF/ds = -2Km1m2/(s^3)
ds = dF / [-2Km1m2/(s^3)]

In the limit of small changes, we approximate the differentials dF and ds as ΔF and Δs, respectively:
Δs = ΔF / [-2Km1m2/(s^3)]

Now simply substitute the value of s, as well as ΔF. ΔF is given by -0.11F.
You will need to evaluate F for s = 8m using the formula given.

2.A simple use of the Pythagorean theorem yields:
P = 4*l / sqrt(2), where P is the perimeter and l is the length of the diagonal.
dP = 4/sqrt(2) * dl
Using the same method above,
ΔP = 4/sqrt(2) * Δl
Since we want the percent change of the P, we want ΔP/P, which is given by:
ΔP/P = (4/sqrt(2) * Δl) / (4*l / sqrt(2))
ΔP/P = Δl /l

Since the diagonal changes from l = 26cm to 26.351, Δl = 0.351.
So the answer is simply:
ΔP/P = 0.351 / 26
ΔP/P = 0.0135
Expressed as a percent, this would be 1.35%.

Note that this is consistent with the real answer, which can be found by evaluating:
((4*26.351/sqrt(2))-(4*26/sqrt(2)))/(4*26/sqrt(2))

3. Letting the length of the side of the cube be x, we have
V = x^3
A = 6*x^2
dV/dx = 3x^2
dA/dx = 12x
Dividing one by the other (which mathematicians will tell you to be wary of, but the details of that are unnecessary here) we have
dV/dA = x/4
Hence under the small error approximation,
ΔA = ΔV*4 / x
Note that this expression is dimensionally consistent, as V has units of length^3, and dividing by x [length] will give units of length^2.
ΔV = (546.612-512) cm^3
and x = cube root[512]

Substitution yields
ΔA = 17.31 cm^2

This is approximately equal to the real answer, which can be found by evaluating:
6*[(546.612)^(1/3)]^2-6*[(512)^(1/3)]^2 = 17.12 cm^2