Sketching the change in a cube's volume

[wisredz]

Hi all,
I've come by an interesting while studying. Here it goes

The volume V=x^3 of a cube of with edges of length x increases by an amount \Delta V when x increases by an amount \Delta x. Show with a sketch how to represent \Delta V geometrically as the some of the volumes of

(a) Three slabs of dimensions x by x by \Delta x
(b) Three bars of dimensions x by \Delta x by \Delta x
(c) One cube of dimensions \Delta x by \Delta x by \Delta x

The differential formula dV=3x^2*dx estimates the change in V with three slabs.

Well that is kinda interesting right? Why is it so? I think the rest (3 bars and a cube) is the error in the estimate. It it right?

 

[HallsofIvy]

Yes, that's true and I agree that it is interesting!

 

[M98Ranger]

Hi there,

Yes, you are correct. The rest of the components (3 bars and 1 cube) represent the error in the estimate for the change in volume. This is because the differential formula dV=3x^2*dx is an approximation and not an exact measurement. By adding the extra components, we are accounting for any potential errors in the estimate.

To better understand this concept, let's take a look at the sketch. As we can see, the original cube has a volume of x^3. When we increase the length of one of the edges by \Delta x, the resulting cube has a volume of (x+\Delta x)^3. This increase in volume can be divided into three different components - the three slabs, three bars, and one cube.

The first component is the three slabs, each with dimensions of x by x by \Delta x. These slabs represent the change in volume due to the increase in length of one edge. The second component is the three bars, each with dimensions of x by \Delta x by \Delta x. These bars represent the change in volume due to the increase in length of two edges. Lastly, the third component is the one cube with dimensions of \Delta x by \Delta x by \Delta x. This cube represents the change in volume due to the increase in length of all three edges.

So, by adding these three components together, we are accounting for any potential errors in the estimate dV=3x^2*dx. This is why the differential formula is often represented as an approximation, as it does not take into account the potential errors in measurement.

I hope this helps clarify the concept for you. Keep up the good work with your studies!