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Use differentials to estimate the amount of ice in cubic inches that covers a 3 ft cube if the ice is $\frac{1}{2}$ inch thick.

Since it is a cube, I believe the equation should be:

$$ y = x^3 $$

Take the derivative:

$$ y' = 3x^2 $$

Plug in 3:

$$ y = 3*3^2 $$

Then plug in 3 + .5 (1/2 inch)

$$ y = 3 * (3.5)^2 $$

Resulting in:

36.75 - 27 = 9.75 inches cubed

Is this correct, or did I make a mistake somewhere along the way?

EDIT

$ y = 3 * (3)^2 * \frac{1}{12} $

Resulting in: 2.25 inches cubed. (seems more reasonable).

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It doesn't look correct. You should use the formula $\Delta y\approx f'(3)\cdot\Delta x$, where $f(x)=x^3$ and $\Delta x=0.5$. (Here, $\Delta y$ is the volume of the ice.) –  David Mitra Dec 12 '12 at 17:21
    
@DavidMitra You mean like this: $y = 3*(3)^2 * (0.5)$ Leaving me with 13.5 inches cubed? –  StrugglingWithMath Dec 12 '12 at 17:24
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@David: $\Delta x$ should be $1$ because the side length of the iced-over cube includes a layer of ice on both sides of it. –  Henning Makholm Dec 12 '12 at 17:26
    
Oops, yes; I made two errors in the previous comment. $\Delta x$ is $1$ inch, or $1/12$ feet. So $\Delta y\approx3\cdot3^2\cdot(1/12)$. –  David Mitra Dec 12 '12 at 17:28
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Yes, but it's $\color{maroon}{\Delta y} \approx 3\cdot3^2\cdot{1\over12}$, not $y=\cdots$. $y$ is the volume of the cube. The volume of the ice is the change in volume of a cube (with no ice) from $x=3$ to $x=3+1/12$. –  David Mitra Dec 12 '12 at 17:56

1 Answer 1

up vote 1 down vote accepted

The problem states that you should use differentials. If $y = x^3$, then $dy = 3x^2 \cdot dx$, where $dx = \Delta x$. (Consequently, if $dx$ is small, $dy \approx \Delta y$). Also, $1/2$ inch is not the same as $0.5$ feet. Use $dx = 1/24$ feet to remain consistent with units.

$$ \Delta y \approx dy = 3x^2 \cdot dx = 3(3)^2\left(\frac{1}{24}\right) = \frac{9}{8}. $$ The other thing to remember is that for a cube, the total surface area should involve all 6 faces. In fact the differential of $x^3$ only accounts for three of the faces (the three faces that are "growing" as $x$ grows). Therefore, the total amount of ice should be $\approx \frac{9}{4}$ cubic feet.

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Thank you for correcting me on some obvious mistakes (I feel dumb for not thinking about those). I noticed a user above commented that since there is .5 inch on each side, should it stay 1/12? Which would account for all 6 faces then? –  StrugglingWithMath Dec 12 '12 at 17:33
    
That's true... Using $dx = 1/12$ would account for all sides of the cube. Then $dy = 3(3)^2(1/12) = 9/4$, as expected. –  Shaun Ault Dec 13 '12 at 2:37

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