# Estimating volume of cube by differentiation

A cube with $10$ inch sides is coated with $0.2$ fiberglass. Use differentials to estimate the volume of the fiberglass shell.

I know that the formula for the volume of the cube is

$$V = x^3$$

and that by differentiating, it would be

$$\frac{\mathrm{d}V}{\mathrm{d}x} = 3x^2$$

That's correct way to start, then we have

$$V(10+\Delta x)=V (10)+V'(10)\cdot\Delta x\implies \Delta V=V'(10)\cdot\Delta x$$

Note also that we need to assume $\Delta x=2\cdot 0.2=0.4$.

Then we have

• $V'(10)=300$
• $\Delta x= 0.4$
• $\Delta V=120$

Note that we can also calculate by the area of the cube times the thickness of the coating and we obtain

• $\Delta V=A\cdot t=6\cdot 100 \cdot 0.2=600\cdot 0.2 = 120$
• thank you for your response- i don't understand why we should assume that the fiberglass is 0.2 x 2? – ComputerizedMyself1485 Mar 10 '18 at 19:38
• Because $x$ is the side and the coat is both sides, thus the side become $10+2\cdot .2=10.4$. – user Mar 10 '18 at 19:39
• @ComputerizedMyself1485 Also Arthur now has edited indeed, moreover the same result can be obtained by the surface area times the thickness. – user Mar 10 '18 at 19:40
• @ComputerizedMyself1485 if the cube has side 10 and we put 1 inch coat on all the surface the side becomes 12 not 11. Make a sketch if it is not clear. – user Mar 10 '18 at 19:50
• You are welcome! Note also that you can calculate by the surface and it is always good to have such kind of double check when possible. – user Mar 10 '18 at 19:54

Assuming "0.2 fiberglass" is $0.2$ inch thick, we are interested in $V(10.4)-V(10)$ (since the cost is applied to both sides). Considering the definition of $V'(x)$, we have $$V(10.4)-V(10)\approx V'(10)\cdot 0.4$$

• if 0.2 is the coat I think we should evaluate V(10.4). – user Mar 10 '18 at 19:34
• Thank you for your response- by that logic, that would mean that the coat is 600 inches, but the answer is 9 – ComputerizedMyself1485 Mar 10 '18 at 19:35
• There was a mistake in the solution, it's 120. thank you so much!! – ComputerizedMyself1485 Mar 10 '18 at 19:45
• @Arthur maybe at least a thanks should be given when a mistake is highlighted – user Mar 10 '18 at 19:48
• @gimusi You're right. Sorry. And thanks. – Arthur Mar 10 '18 at 19:52

That's right. So now you want to estimate $(10.4)^3-10^3$ by using that derivative. That is, you want to use $f(x+h)-f(x)\approx hf'(x).$

Your $3x^2$ is the rate of change of the volume with respect to the side of cube.

The linear approximation to the actual change is the rate of change times the change.

The change in the side of cube is $0.4\text { inch.}$

Thus the total change is $$3(10^2)(0.4)=120 \text { cubic inches}.$$