2
$\begingroup$

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).

  • 0
    @DavidMitra Thanks again for the correction, seems to be a habit for me just using y.2012-12-12

1 Answers 1

1

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.

  • 0
    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.2012-12-13