# Investigating the relationship between the derivative of the volume and the surface area of solids and regular polyhedra: tetrahedron case

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## Introduction

After having studied the topic of calculus in math class, I was keen to investigate examples about how derivatives and integrals relate to our physical world. I found several topics fascinating to investigate but there was one topic I was most intrigued about: how does the derivative of the volume of a solid relate to its surface area. This topic appealed to me when I first realized that the derivative of the volume of a sphere is equal to its surface area. I then tried to explore whether there was such an evident relationship in other common solids, such as pyramids or cones. In both these cases the derivative of the figure’s volume did not equal its surface area, nor it seem to have a clear relationship, which led me to reluctantly assume it might have been a coincidence the reason why it works for a sphere in the first place.

However, my interest in investigating this topic appeared again when exploring the figure of a torus as a solid of revolution. This solid was unknown yet fascinating for me, so I decided to delve deeper into its main features, such as its major radius and minor radius. Figure 1 shows a torus with major radius R and minor radius r.

I then decided to find out what the formula for the volume (V) and surface area (SA) of a torus was.

V = 2π2r2R

SA = 4π2Rr

I soon noticed that, keeping R constant -

$$\frac{dV}{dr}=4\pi^2Rr=SA$$

This was an intriguing ‘second coincidence’ where the derivative of the volume of a solid equals its surface area. I was convinced that these two cases were note mere coincidences, and that there must be specific reasons why the relationship holds. Unfortunately, after searching about this topic on the internet, I noted there was limited information that explored this intriguing idea.

This investigation is rooted in my desire to extend the exploration of the relationship between the derivative of the volume and the surface area to other solids and specifically regular polyhedra. Although a direct and evident initial relationship, as noted in the two previous examples, might not be observed, the goal of this investigation is to come up with different ways to express a solid’s volume such that by differentiating it, an exact or close relationship between the derivative of the volume and the surface area is obtained. The ultimate aim of this investigation is to explore the case of the regular polyhedron of a (regular) tetrahedron and come up with a new expression for its volume such that differentiating it is exactly equal to its surface area. This solid was chosen since, unlike a cube or a sphere, is complicated enough to not have a three-axes symmetry, yet simple enough to have a small number of plane faces. Thus, if successful, a generalization to all regular polyhedra could be stated, in addition to providing guidelines for a further exploration of this topic on other 3- Dimensional solids and even other mathematical areas.

## Sphere Case

First, I will explore the case of a sphere so as to gain insight and determine possible reasons why the derivative of the volume of a sphere equals its surface area. Figure 2 shows the formulae for the volume and surface area of a sphere with radius r.

A possible way to find an explanation for this observation is through the application of the definition of the derivative, as it will enable us to interpret the meaning of differentiating a sphere’s volume, as well as provide greater insight into how and why it results in exactly the surface area.

The definition of the derivative is as follows -

The derivative of f(x) with respect to x is expressed as3 f’(x) or $$\frac{df}{dx}$$ and is defined as -

$$f'(x)=\frac{df}{dx}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

When differentiating a sphere’s volume V with respect to its radius r and applying the definition of the derivative, we obtain the following equation -

$$\frac{dV}{dr}=\lim_{h\to0}\frac{v(r+h)-V(r)}{h }$$

$$=\lim_{h\to0}\frac{(\frac{4}{3}\pi(r+h)^3)-\frac{4}{3}\pi r^3}{h}$$

We can graphically represent the numerator of the equation above by presenting two spheres: one with radius r and the other one with radius r + h. The following diagram4 interprets and shows this -

For this reason, we can 'informally' rewrite the equation above as -

$$=\lim_{h\to0}\frac{{\text{Volume of sphere with radius (r + h) - Volume of sphere with radius r}}}{h}$$

Based on the diagram, we can quickly realize that the remainder of the subtraction in the numerator is exactly equal to the Surface Area of the sphere with radius r × h, as subtracting the volume of the sphere with radius r from the volume of the sphere with radius r + h results in the surface area of the former sphere increasing, in all directions, by exactly h units.

Therefore, we can simplify the equation above to -

$$=\lim_{h\to0}\frac{{\text{(Surface Area of sphere with radius r)}}×h}h{}$$

= limh→0 (Surface Area of sphere with radius r)

$$∴ \frac{dV}{dr}={\text{Surface Area of sphere with radius r}}$$

Hence, by applying the definition of the derivative for a sphere’s volume, we can observe that differentiating a sphere’s volume is exactly equal to the surface area because it can be represented as the sphere’s surface area increasing by h units in all possible directions, and thus, the numerator (when applying the definition of the derivative) being simplified (surface area) x h.

From this, we can make some general observations -

• Considering only the first component of the numerator, the application of the derivative of the volume of a sphere ‘created’ two spheres, one with radius (r + h) and another with radius r. Ultimately, the latter was subtracted from the former.
• Since the radius is equidistant to all parts of the Surface Area, the distance from any part of the surface area to the center of the first component of the numerator (Sphere with radius r + h) is equal to r + h, whereas that for the second component of the numerator (Sphere with radius r) is only r.
• For this reason, subtracting the volume of a sphere with radius r from the volume of a sphere with radius (r + h) results in exactly: (surface area of sphere with radius r) x h.
• An additional interesting observation is that the radius of a sphere meets all its surface area perpendicularly; at a 90-degree angle

With this in mind, we can suggest that perhaps the most important difference between a sphere and other solids for the derivative of the volume to be equal to the surface area is that, for a sphere, the volume is written in terms of a radius. Hence, the application of the definition of the derivative increases the distance from the surface area to the center by the exact same amount in all directions, and thus, solving for V(r + h) − V(r) exactly equals the (surface area of sphere) x h.

## Cube Case

Having already drawn the previous observations, I will now explore the case of a cube. This will enable me to compare this with the results obtained in the exploration of the case of a sphere, and thus, be ultimately able to establish some general guidelines for the case of a tetrahedron and other solids.

Figure 3 shows a cube with side length S. The formula for the volume (V) and surface area (SA) of a cube, expressed in terms of its side length S, is -

V = S3

We can soon realize that -

SA = 6S2

$$\frac{dV}{dS}=3S^2=\frac{SA}{2}$$

Interestingly enough, although differentiating the volume of a cube with respect to its side length S does not exactly equal the surface area, it indeed equals half the surface area, which indicates that there must be a close relationship between the derivative of the volume of a cube and its surface area. Similar to when we were exploring the case of a sphere, we can apply the definition of the derivative for the volume of a cube to find a possible explanation for this -

$$\frac{dV}{dS}=\lim_{h\to0}\frac{V(S+h)-V(S)}{h}$$

$$=\lim_{h\to0}\frac{(S+h)^3-S^3}{h}$$

Once again, we can graphically interpret the equation above by presenting two cubes: one with side length s the other one with side length s + h. What it is pivotal to realize is that increasing the side length of a cube by h units is not the same as the distance from any part of the surface area to the center of the cube to have increased by h units, but rather -

Unlike a sphere, increasing the side length of a cube by h units can actually be represented as increasing the distance from the surface area to the center of the cube by h units but in only one direction per axis (if we consider a cube placed in a XYZ coordinate system as in the diagram) – or, alternatively, increasing the distance by $$\frac{h}2$$ units in both directions per axis. This is not what happens in the case of a sphere since, as the radius originates from the center of the solid and represents the distance from the surface area to the center, increasing the radius by h units results in increasing the distance from the surface area to the center by h units in all directions of the three axes.

From this, we can note that it is possible to rewrite the equation above of the derivative of a cube’s volume with respect to its side length by utilizing the concept of the triple integral, as we can exactly represent a cube’s increase in each of the three axes in the XYZ coordinate system. The advantage of using this method is that we can explicitly manipulate the endpoints of the triple integrals to analyze how the derivative of volume - surface area relationship changes accordingly, and thus determine what exact changes in the endpoints of the integrals are needed for the aimed relationship to hold.

The following is how we can rewrite the equation by using triple integrals -

$$\frac{dV}{dS}= \displaystyle\lim_{h\to0}\frac{∫^{x=S+h}_{x=0}∫^{y=S+h}_{y=0}∫^{z=S+h}_{z=0}dV- ∫^{x=S}_{x=0} ∫^{y=S}_{y=0} ∫^{z=S}_{z=0}}{h}$$

V(S + h) is equivalent to $$\displaystyle\int^{x=S+h}_{x=0}∫^{y=S+h}_ {y=0}∫^{z=S+h}_{z=0} dV$$ since, as observed in figure 5, it is the volume of a cube with side length S + h, which can be represented by a cube of side length S increasing its side length by h units in only one direction per axis.

We can now solve the limit -

$$=\displaystyle\lim_{h\to 0}\frac{\int^{x=S+h}_{x=0}\int^{y=S+h}_{y=0}\int^{z=S+h}_{z=0}{\text{dz dy dx}}-\int^{x=S}_{x=0}\int^{y=S}_{y=0}\int^{z=S}_{z=0}{\text{dz dy dx}}}{h}$$

$$=\lim_{h\to0}\frac{[(S+h-0][(S+h)-0][(S+h)-0]-[S-0][S-0][S-0]}{h}$$

$$=\lim_{h\to0}\frac{(S+h)^3-S^3}{h}$$

$$=\lim_{h\to0}\frac{(S^3+3S^2h+3Sh^2+h^3)-S^3}{h}$$

$$=\lim_{h\to0}\frac{h(3S^2+3Sh+h^2)}{h}$$

= limh→0 3S2 + 3Sh + h2

= 3S2 + 3S(0) + (0)2

= 3S2

$$∴ \frac{dV}{dS}= \frac{\text{Surface area of cube with side length S}}{h}$$

In this way, it can be deduced that the reason why the derivative of a cube's volume expressed in terms of its side length is only half its surface area is the fact that, after applying the definition of the derivative, the first part of the numerator, V(S + h), represents an increase of the cube’s side length by h units in only one direction per axis (or $$\frac{h}{2}$$ units in both directions). Therefore, based on this, we can predict that an increment in the side length of the first cube presented in the numerator when applying the definition of the derivative by h units in both directions per axis will lead to the solution of the limit being exactly equal to the surface area of the cube with side length S. If this is true, then we can determine specific changes needed to the original expression of the volume of a cube so that the aimed relationship is achieved.

We can test this assumption by manipulating the endpoints of the triple integral when applying the definition of the derivative for a cube's volume, specifically by letting the lower endpoints of the triple integrals (for the first component of the numerator) be − h -

$$\displaystyle\lim_{h\to0}\frac{\int^{x=S+h}_{x=-h}\int^{y=S+h}_{y=-h}\int^{z=S+h}_{z=-h}dV-\int^{x=S}_{x=0}\int^{y=S}_{y=0}\int^{z=S}_{z=0}dV}{h}$$

$$=\displaystyle\lim_{h\to0}\frac{\int^{x=S+h}_{x=-h}\int^{y=S+h}_{y=-h}\int^{z=S+h}_{z=-h}\,dz\, dy\, dx-\int^{x=S}_{x=0}\int^{y=S}_{y=0}\int^{z=S}_{z=0}\,dz \,dy \,dx}{h}$$

$$=\lim_{h\to0}\frac{[(S+h)-(-h)][(S+h)-(-h)][(S+h)-(-h)]-[S-0][S-0][S-0]}h{}$$

$$=\lim_{h\to0}\frac{[S+2h]^3-S^3}{h}$$

$$=\lim_{h\to0}\frac{(S^3+6S^2h+12Sh^2+8h^3)-S^3}{h}$$

$$=\lim_{h\to0}\frac{h(6S^2+12Sh+8h^2)}{h}$$

= 6S2 + 12Sh + 8h2

= 6S2 + 12S(0) + 8(0)

= 6S2

∴ = Surface area of cube with side length S

Thus, it can be observed that our assumption holds. From this we can conclude that, at least for the case of a cube, for the derivative of its volume to be equal to its surface area, the application of the definition of the derivative should lead to -

• An equidistant increment of the first component of the numerator by h units in both directions of all three axes

If this does not hold, then the derivative of the volume will not result in exactly the surface area.

Comparing this to the case of a sphere, although applying the definition of the derivative for a sphere’s volume can also be deemed as leading to an increment of the first component of the numerator by h units in both directions per axis, it is much more accurate to indicate that it is an increment by h units in every possible direction of the XYZ coordinate system. This makes sense if we associate (although it is theoretically not totally correct) a sphere to a regular polyhedron with “infinite faces”. Alluding to this perception, we can explain why in the case of a sphere it is needed that the increment by h units is observed in all possible directions of the XYZ system, whereas for a cube it is only necessary for this increment to be in six directions (both directions per axis). There seems to be a very close relationship between the number of faces of a solid and the number of directions (as well as the specific directions) in which the increment by h units should be observed.

In the case of a sphere, as it can be deemed to have “infinite faces” (although this is theoretically not totally correct), there would be expected to be an increment by h units in “infinite directions” or, more clearly, in all possible directions of the XYZ coordinate system. In the case of a cube, as it has only six faces, an increment by h units is only required for six directions, of which are both directions of each of the three axes if, and only if, each of the cube’s plane surfaces are parallel to one axis (or we could conversely say perpendicular to two axes). If this is not the case (for instance, if a cube is placed in such a way that its plane surfaces are neither parallel nor perpendicular to any axis), it would make no sense to require an increase by h units in both directions per axis. Instead, to hold the observations previously mentioned, each plane face should “grow”, by h units, in an outwards direction that is perpendicular to the face.

In the case of a sphere, as it can be deemed to have “infinite faces” (although this is theoretically not totally correct), there would be expected to be an increment by h units in “infinite directions” or, more clearly, in all possible directions of the XYZ coordinate system. In the case of a cube, as it has only six faces, an increment by h units is only required for six directions, of which are both directions of each of the three axes if, and only if, each of the cube’s plane surfaces are parallel to one axis (or we could conversely say perpendicular to two axes). If this is not the case (for instance, if a cube is placed in such a way that its plane surfaces are neither parallel nor perpendicular to any axis), it would make no sense to require an increase by h units in both directions per axis. Instead, to hold the observations previously mentioned, each plane face should “grow”, by h units, in an outwards direction that is perpendicular to the face.

This is why expressing a solid’s volume in terms of an “internal radius” – we will later better define what is meant by internal radius – that is equidistant and touches perpendicularly all the faces of the solid is crucial for the aimed relationship to be achieved. The internal radius must touch each face perpendicularly because, by doing so, no matter how the solid is placed in the XYZ coordinate system, the first component of the numerator – when applying the definition of the derivative to this solid’s volume expressed in terms of this internal radius – can be graphically represented as increasing in an outwards direction that is perpendicular to each plane face by h units.

It is now important to clarify the concept of a solid’s internal radius that, although is not a formal mathematical concept, it will be constantly used in the following sections of this investigation. By considering the characteristics of a sphere’s radius and making general observations about it, we can come up with generic guidelines that will enable us to extrapolate the concept of a radius – which, for all the other solids explored in this investigation, will be referred to as their internal radius – to other solids. These generic guidelines indicate both what is meant by a solid’s internal radius and the specific characteristics of it that should hold -

• The internal radius of a solid originates at exactly its centroid – the centroid is defined as the “arithmetic mean position of all the points” in a figure or solid
• A solid’s internal radius must be equidistant and touch all of its the faces perpendicularly
• For this reason, the internal radius of a solid can only touch each of its faces once; it can only touch each face at one specific position (this is because it is not mathematically possible for a line segment that originates at the centroid of a solid to touch one of its plane faces perpendicularly at several position)

The only unknown up to this point is the specific position at which a solid’s internal radius must touch each of its faces. However, as we know that a solid’s internal radius must originate from its centroid and must touch each of its faces perpendicularly, we can work out the specific position of each face at which the radius must touch them.

For the case of a cube, we can soon realize that, if we were to determine its internal radius, the only possible way that the previous observations hold is by it touching specifically the centroid of each of its plane faces, as otherwise the radius originating in the centroid of the cube will not be able to touch each face perpendicularly. Moreover, we can note that, in this case, all of the previous observations and conclusions apply, since the radius would also conveniently be equal to half the cube’s side length, which is – as can be deducted from the manipulation of the triple integral – the necessary relationship for the derivative of a cube’s volume expressed in terms of its internal radius to be exactly equal to its surface area. Figure 5 presents a cube with its internal radius R -

If we express the volume V and surface area SA of this cube in terms of its internal radius R, we get -

$$V=(2R)^3=8R^3$$

SA = 6(2R)2 = 24R3

We can now clearly observe that, confirming our original assumption, the derivative of the volume of the cube expressed in terms of its internal radius is indeed exactly equal to the surface area -

$$\frac{dV}{dR}=\frac{d(8R^3)}{dR}$$

= 24R2

$$= 24(\frac{S}{2})^2$$

= 6S2

$$\frac{dV}{dR}$$ = Surface Area of cube with internal radius R and side length S

This can be explained by the fact that, since the cube’s volume is expressed in terms of its internal radius, applying the definition of the derivative lead to -

$$\frac{dV}{dR} = \lim_{h\to0}\frac{V(R+h)-V(R)}{h}$$

$$= \lim_{h\to0}\frac{8(R+h)^3-8R^3}{h}$$

The first component of the numerator, V(R+h), can now be represented by a cube whose internal radius has increased by h units. Unlike the original case (the cube’s volume being expressed in terms of its side length) where the distance between the surface area and the center of the first component of the numerator increased by only h units in only one direction per axis, the application of the derivative of the cube’s volume V with respect to its internal radius R leads the cube represented by the first component of the numerator, V(R+h), to grow, by h units, in an outwards perpendicular direction to each face.

Therefore, we can also represent the equation above using the concept of triple integral, but this time considering the lower endpoints for the first component of the numerator to be - h and the upper endpoints to be 2R + h -

$$\frac{dV}{dR}=\lim_{h\to0}\frac{\displaystyle\int^{x=2R+h}_{x=-h}\int^{y=2R+h}_{y=-h}\int^{z=2R+h}_{z=-h}\,dz \,dy \,dx-\int^{x=2R}_{x=0}\int^{y=2R}_{y=0}\int^{z=2R}_{z=0}\,dz \,dy \,dx}{h}$$

$$=\lim_{h\to0}\frac{[(2R+h)-(-h)][(2R+h)-(-h)][(2R+h)-(-h)]-[2R-0][2R-0][2R-0]}{h}$$

$$= \lim_{h\to0}\frac{(8R^3+24R^2h+24Rh62+8h^30-8R^3}{h}$$

$$\lim_{h\to0}\frac{h(24R^2+24Rh+8h^2)}{h}$$

limh→0 24R2 + 24Rh + 8h2

= 24R2 + 24R(0) + 8(0)

= 24R2

∴ $$\frac{dV}{dR}$$ = Surface area of cube with internal radius R

In this way, the results obtained for the case of a cube show that, in order that differentiating a cube’s volume to be exactly the surface area, it is essential to express the cube’s volume in terms of an internal radius R, as otherwise applying the definition of the derivative for the cube’s volume will not result in the distance between the surface area and the center of the first component increasing by h units in an outwards perpendicular direction to each face. For this reason, it can be assumed that, if we are able to express any solid’s volume in terms of its internal radius, differentiating it should result in exactly the solid’s surface area. This is because applying the definition of the derivative in these conditions will lead the distance between the surface area and the center of the first component of the numerator to increase by h units in an outwards perpendicular direction to all its plane faces.

Ultimately, from the exploration of both the sphere and the cube case, we can define a solid’s internal radius as an equidistant line segment that originates in the centroid of the solid and extends to touch, perpendicularly, the centroid of each of the solid’s plane faces. It is important to notice that this definition applies only for solids that have no curved faces, as for the latter the definition should be slightly modified since the idea of the centroid of a curved surface can be initially unclear. However, if we express the volume of a solid with curved surfaces in terms of an internal radius such that the previous observations hold, we should also expect that by differentiating it we obtain the solid’s surface area.

### Testing for a Cylinder

If the previous conclusions and observations are true, and if they hold for a given solid, then, regardless of the nature of the solid, we should be able to express its volume in such a way that the derivative of it equals the surface area. In order to check the validity of this assumption before relying on it to work for a tetrahedron, I will first apply the previous generic guidelines for the case of a cylinder, as it is a solid that, similar to a tetrahedron (though unlike a cube or a sphere), does not have a three-axis (x-axis, y-axis, and z-axis) symmetry. Moreover, as it has a curved surface it can enhance the conclusion that, regardless of the nature of the solid, if we can express its volume in terms of an internal radius, we should be able to relate the derivative of the volume with the surface area. Ultimately, testing for the case of a cylinder will be also useful because its volume is expressed in terms of two variables, the radius (r) and the height (h), which is why it might initially seem that there is no direct relationship between the derivative of its volume and its surface area.

We can use partial derivatives to check whether there is a clear relationship between the (partial) derivative of the volume of a cylinder and its surface area. This will be useful, as it will enable us to determine whether treating one variable as a constant while differentiating the other one can lead to any relationship with the surface area.

Partial derivatives of (r, h) -

(r, h) = πr2h

$$\frac{dV}{dr}=2\pi rh$$

$$\frac{dV}{dh}=\pi r^2$$

Interestingly enough, a close relationship with the surface area can be noted. As a matter of fact, $$\frac{dV}{dr}$$ provides exactly the surface area of a hollow cylinder (the surface area of its circular surface), whereas $$\frac{dV}{dh}$$ results in exactly half the surface area of the top and bottom faces (bases) – it results in the surface area of only one base. These results can be thoroughly explained by the analysis drawn from the previous exploration for the case of a sphere and cube. $$\frac{dV}{dr}$$ equals exactly the surface area of the circular surface of a cylinder, as by applying the definition of the derivative, we get -

$$\frac{dV}{dr}=\lim_{a\to0}\frac{V(r+a)-V(r)}{a}$$

$$=\lim _{a\to0}\frac{\pi(r+a)^2h-\pi r^2h}{a}$$

Which can be ultimately simplified to -

$$= \lim_{a\to0}\frac{\,Surface \,area \,of \,the \,circular \,surface \,of \,a \,cylinder × a}{a}$$

$$∴ \frac{dV} {dr} = \,Surface \,area \,of \,the \,circular \,surface \,of \,a \,cylinder$$

The result of $$\frac{dV} {dh}$$ , on the other hand, is exactly the surface area of one of its bases, as similar to a cube expressed in terms of its side length, applying the definition of the derivative for $$\frac{dV}{dh}$$ will result in the graphical increase of h by a units in only one direction – or an increase of h by $$\frac{a}{2}$$ units in both directions. As with a cube, if we rewrite the height of a cylinder in terms of a new variable x such that, by applying the definition of the (partial) derivative to a cylinder’s volume with respect to x, the distance between the centroid of each of the circular bases and the center of the first component of the numerator increases by a units in an outwards perpendicular direction both, we should obtain the exact surface area of the both circular bases. For this to happen the relationship 2x = h must hold. We can check this through the following equations -

V(r, x) = πr2(2x)

$$\frac{dV}{dx}=2\pi r^2$$

$$\,∴ \frac{dV}{dx}=2×\frac{dV}{dh}=\,Surface \,area \,of \,the \,two \,bases \,of \,a \,cylinder$$

These results are revealing, as they suggest that we can use the findings encountered in this exploration to find the surface area of a solid’s specific component (such as either the surface area of the circular surface of a cylinder or the surface area of its bases) even if the volume of this solid is expressed in terms of multiple variables (through partial derivatives). This can turn out to be a very useful observation, as it can enable us to find the surface area of a specific component of any given solid by simply applying the definition of the derivative for the volume of the surface area. If necessary, we could manipulate this equation and determine what changes to the original equation – or to only a variable of a multivariable equation – would be necessary for the aimed relationship to be achieved.

In order to express a cylinder’s volume in such a way that differentiating it gives exactly the surface area (including both its circular surface and its two bases), it would be necessary to express it in terms of its internal radius (R) such that it touches the centroid of the two bases and that of the circular surface. Unfortunately, this cannot be possible for cylinders with any dimensions, as the relationship h = 2r must hold so that the internal radius is equidistant and touches the centroid of both bases and that of the circular surface. This outlines a limitation of the methodology employed, as it can only work when specific conditions are previously met.

Notwithstanding this limitation, the guidelines established in Section 3. help us easily determine the specific conditions that must be met to be able to express a solid’s volume in terms of an internal radius such that differentiating will exactly equal the solid’s surface area. For the case of a cylinder, the only conditions that must be met for this to happen is that the following relationship holds -

If we now write the volume (V) and the surface area (SA) of this cylinder, in terms of its internal radius, we obtain -

V = π(R)2(2R)

V = 2πR3

SA = 2π(R) + 2π(R)2

SA = 4πR2 + 2πR2

SA = 6πR2

We can now quickly realize that -

$$\frac{dV}{dR}=6\pi R^2$$

$$\,∴ \frac{dV}{dR}=\,Surface \,area \,of \,cylinder$$

This is particularly conclusive, as any other type of cylinder (cylinders with radii and heights that have no relationship whatsoever) will not provide the same results, which suggests that it is the application of the observation and guidelines stated in this investigation that enable the expected results to happen. Thus, these three examples (sphere, cube, and cylinder case) demonstrate that the methodology is reliable enough to apply it to fulfill the original objective of this investigation: rewrite the volume of a regular tetrahedron in such a way that differentiating it equals the surface area of the tetrahedron.

## Methodology for Other Solids

Based on the previous analyses for the three solids explored, we can now come up with a step-by- step methodology to determine whether it is possible to express the volume of more complicated solids in terms of their internal radius and, if so, ultimately show that differentiating it will result in exactly the surface area of the solid.

Before presenting the established methodology, it is pivotal to understand that, since the centroid is defined as the “arithmetic mean position of all the points” 14 in a figure or solid, we can find the exact coordinates of the centroid of any plane figure or solid of which we know the specific coordinates of its vertices by simply calculating the arithmetic mean of the components (X, Y, and Z coordinate component) of all vertices of the figure or solid. This will be extremely useful when finding the exact coordinates of both the centroid of a regular tetrahedron and the centroid of each of the tetrahedron’s plane faces.

The following is the step-by-step methodology for more complicated solids with plane faces -

• Place the solid in an XYZ coordinate system such that the origin $$\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$$ is the coordinates of one of the vertices of the solid
• Determine the coordinates of all the other vertices of the solid
• By using the method of calculating a centroid through the arithmetic mean of the coordinate components of all vertices, calculate the centroid of the solid and that of each of the solid’s plane faces
• Determine and write, in vector notation, all the possible vectors that represent the solid’s internal radius by originating in the solid’s centroid and extending to the centroid of each of the solid’s plane faces
• Check whether the magnitude of all the vectors representing the solid’s internal radius is the same, as this would mean the solid’s internal radius is equidistant
• Check, through the use of the dot product of vectors formula, whether the vectors representing the solid’s internal radius touch each of the solid’s faces perpendicularly
• If all the previous holds, express the volume and surface area of the solid in terms of the established internal radius (we can use the magnitude of the vectors previously found)
• Finally, differentiate the volume of the solid expressed in terms of its internal radius and check whether it equals the solid’s surface area

This methodology will be specifically followed for the case of a regular tetrahedron, as it is very straightforward and takes all the important factors determined throughout this investigation into account. Moreover, although there are alternative methodologies that could be followed, such as inscribing a sphere in a regular tetrahedron with the conditions that the center of the sphere is the centroid of the solid and that the sphere touches each of the solid’s plane faces once (in this way, the radius of this sphere will be exactly the internal radius of the tetrahedron), these are difficult to thoroughly carry out without knowing additional information about the solid. Nevertheless, it should also be noted that the limitation of this methodology is that it only works if we consider solids with no curved surfaces, as finding the specific position at which the solid’s internal radius should touch them can be more challenging. For this reason, this methodology can be extremely useful for the case of a tetrahedron but needed to be slightly adapted for the case of a cone, for example. Notwithstanding, this methodology might still provide insightful observations when trying to apply it to solids with curved or circular surfaces.

## Tetrahedron Case

A tetrahedron is a polyhedron that has four faces of which all of them are triangles. In the case of a regular tetrahedron, all its four faces are equilateral triangles with the same side length. For this investigation, a regular tetrahedron was selected so as to be able to draw further conclusions about what is needed for the derivative of a solid to be exactly equal to its surface area. Moreover, as the tetrahedron is only symmetrical upon one axis, applying the previously stated methodology to a tetrahedron will show whether some possible initial assumptions one could have, such as the importance of both symmetry in all axes and the nature of the solid per se, are correct or not.

The volume and surface area of a regular tetrahedron are the following -

$$V=\frac{\sqrt{2}a^3}{12}$$

$$SA = \sqrt3a^2$$

The diagram labels the vertices of the regular tetrahedron. The following are the coordinates of these vertices -

$$A:\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

$$B:\begin{pmatrix} \frac{\sqrt3a}{2} \\ \frac{a}{2} \\ 0 \end{pmatrix}$$

$$C:\begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix}$$

$$D :\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{3} \end{pmatrix}$$

With this, we are able to calculate the centroid of the tetrahedron (which we will refer to as point O) and the centroid of each of its four triangular faces (points E, F, G, and H).

The following are the coordinates of these additional points -

$$O :\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{12} \end{pmatrix}$$

$$E :\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ 0 \end{pmatrix}$$

$$F:\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}$$

$$G:\begin{pmatrix} \frac{\sqrt3a}{18} \\ \frac{a}2{} \\ \frac{\sqrt6a}{9} \end{pmatrix}$$

$$H: \begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{2a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}$$

We can now determine the four vectors that will represent the internal radius of the regular tetrahedron -

$$\vec {OE}: E-O=​​​​​​​​\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ 0 \end{pmatrix}-\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{6} -\frac{\sqrt3a}{6}\\ \frac{a}{2} -\frac{a}{2}\\ 0-\frac{-\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} 0\\ 0\\-\frac{\sqrt6a}{12}\end{pmatrix}$$

$$​\vec {OF}:F-O=\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}-\begin{pmatrix} \frac{2\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{2\sqrt3a}6 -\frac{\sqrt3a}{6}\\ \frac{a}3 -\frac{a}2\\ \frac{\sqrt6a}{9}-\frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{18} \\ -\frac{a}{6} \\ \frac{\sqrt6a}{36} \end{pmatrix}$$

$$\vec {OG}:G-O=\begin{pmatrix} \frac{\sqrt3a}{18} \\ \frac{a}{2} \\ \frac{\sqrt6a}{9} \end{pmatrix}-\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{2\sqrt3a}{18}-\frac{\sqrt3a}{6} \\ \frac{a}{2}-\frac{a}{2} \\ \frac{\sqrt6a}{9}-\frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{-\sqrt3a}{9} \\ 0 \\ \frac{\sqrt6a}{36} \end{pmatrix}$$

$$\vec {OH}:H-O=\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{2a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}-\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{2\sqrt3a}{9}-\frac{\sqrt3a}{6} \\ \frac{a}{3}-\frac{a}{2} \\ \frac{\sqrt6a}{9}-\frac{\sqrt6a}{12} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{18} \\ \frac{a}{6} \\ \frac{\sqrt6a}{36} \end{pmatrix}$$

The magnitudes of these vectors are the following -

$$\vec {|OE|}=\sqrt{(0)^2+(0)^2+(-\frac{\sqrt6a}{12})^2}=\frac{\sqrt6a}{12}$$          units

$$\vec {|OF|}=\sqrt{(\frac{\sqrt3a}{18})^2+(-\frac{a}{6})^2+(\frac{\sqrt6a}{36})^2}=\sqrt{\frac{a^2}{108}+\frac{a^2}{36}+\frac{a^2}{216}}=\sqrt{\frac{a^2}{24}}=\frac{a}{2\sqrt6}=\frac{\sqrt6a}{12}$$          untis

$$\vec {|OG|}=\sqrt{(-\frac{\sqrt3a}{18})^2+(0)^2+(\frac{\sqrt6a}{36})^2}=\sqrt{\frac{a^2}{27}+\frac{a^2}{216}}=\sqrt{\frac{a^2}{24}}=\frac{a}{2\sqrt6}=\frac{\sqrt6a}{12}$$         units

$$\vec {|OH|}=\sqrt{(\frac{\sqrt3a}{18})^2+(\frac{a}{6})^2+(\frac{\sqrt6a}{36})^2}=\sqrt{\frac{a^2}{108}+\frac{a^2}{36}+\frac{a^2}{216}}=\sqrt{\frac{a^2}{24}}=\frac{a}{2\sqrt6}=\frac{\sqrt6a}{12}$$          units

These results are conclusive, as in fact, all the vectors representing the solid’s internal radius have the same magnitude ($$\frac{\sqrt6a }{12}$$ units), and hence, it was shown that the requirement of being equidistant is true for the tetrahedron’s tentative internal radius (represented by the four aforementioned vectors).

It is important to also check whether the vectors representing the solid’s internal radius meet the centroid of the face they touch perpendicularly. We can do this by first determining vectors that are parallel to each of the solid’s faces and then using the dot product of vectors between the vectors representing the tetrahedron’s internal radius and the vectors parallel to the face they touch. Specifically, if we obtain that all the dot product of vectors is 0, we can then assert that the angle between the vectors representing the tetrahedron’s internal radius and the vectors parallel to each of the solid’s faces is equal to 90°, and hence, the vectors representing the solid’s internal radius meet the centroid of the face they touch perpendicularly. The simplest way to find a vector parallel to a solid’s face is by generating a vector that connects to known coordinates of the face, such as the centroid and a vertex of the face. The following are the vectors that were found in order for each one of them to be parallel to specifically one of the tetrahedron’s faces -

$$\vec {EA}:A-E=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ 0 \end{pmatrix}=\begin{pmatrix} -\frac{\sqrt3a}{6} \\ -\frac{a}{2} \\0 \end{pmatrix}$$               (Parallel to face ABC)

$$\vec {FD}:D-F= \begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{3} \end{pmatrix}-\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{6}-\frac{2\sqrt3a}{9} \\ \frac{a}{2}-\frac{a}{3} \\ \frac{\sqrt6a}{3}-\frac{\sqrt6a}{9} \end{pmatrix}=\begin{pmatrix} -\frac{\sqrt3}{18} \\ \frac{a}{6} \\ \frac{2\sqrt6a}{9} \end{pmatrix}$$          (Parallel to face ABD)

$$\vec {GD}:D-G=\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{3} \end{pmatrix}-\begin{pmatrix} \frac{\sqrt3a}{18} \\ \frac{a}{2} \\ \frac{\sqrt6a}{9} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{6}-\frac{\sqrt3a}{6} \\ \frac{a}{2}-\frac{a}{2} \\ \frac{\sqrt6a}{3}-\frac{\sqrt6a}{9} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{9} \\ 0 \\ \frac{2\sqrt3a}{9} \end{pmatrix}$$        (Parallel to face ACD)

$$\vec {HD}:D-H=\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{3} \end{pmatrix}-\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{2a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}=\begin{pmatrix} \frac{\sqrt3a}{6}-\frac{2\sqrt3a}{9} \\ \frac{a}{2}-\frac{2a}{3} \\ \frac{\sqrt6a}{3}-\frac{\sqrt6a}{9} \end{pmatrix}=\begin{pmatrix} -\frac{\sqrt3a}{18} \\ -\frac{a}{6} \\ \frac{2\sqrt6a}{9} \end{pmatrix}$$         (Parallel to face BCD)

Now, to show that the tentative internal radius touches the centroid of each of the tetrahedron’s faces perpendicularly, we can use the dot product between the vector touching the centroid of a face and the vector parallel to that face. Since there are four vectors representing the solid’s internal radius, we have four dot products -

$$\vec {OE}.\vec{EA}=\begin{pmatrix} 0 \\ 0 \\ -\frac{\sqrt6a}{12} \end{pmatrix}.\begin{pmatrix} -\frac{\sqrt3a}{6} \\ -\frac{a}{2} \\ 0 \end{pmatrix}=(0)(-\frac{\sqrt3a}{6})+(0)(\frac{a}{2})=(-\frac{\sqrt6a}{12})(0)=0$$

$$\vec{OF}.\vec{FD}=\begin{pmatrix} \frac{\sqrt3a}{18} \\ -\frac{a}{6} \\ \frac{\sqrt6a}{36} \end{pmatrix}.\begin{pmatrix} -\frac{\sqrt3a}{18} \\ \frac{a}{6} \\ \frac{2\sqrt6a}{9} \end{pmatrix}=(\frac{\sqrt3a}{18})+(-\frac{\sqrt3a}{18})+(-\frac{a}{6})(\frac{a}{6})+(\frac{\sqrt6a}{36})(\frac{2\sqrt6a}{9})=-\frac{a^2}{108}-frac{a^2}{36}+\frac{a^2}{27}=0$$

$$\vec{OG}.\vec{GD}=\begin{pmatrix} -\frac{\sqrt3a}{9} \\ 0 \\ \frac{\sqrt6}{36} \end{pmatrix}.\begin{pmatrix} \frac{\sqrt3a}{18} \\ 0 \\ \frac{2\sqrt6a}{9} \end{pmatrix}=(-\frac{\sqrt3a}{9})(\frac{\sqrt3a}{9}+(0)(0)+(\frac{\sqrt6a}{36})(\frac{2\sqrt6a}{9})=-\frac{a^2}{27}+0+\frac{a^2}{27}=0$$

$$\vec{OH}.\vec{HD}=\begin{pmatrix} \frac{\sqrt3a}{18} \\ \frac{a}{6} \\ \frac{\sqrt6a}{36} \end{pmatrix}.\begin{pmatrix} -\frac{\sqrt3a}{18} \\ -\frac{a}{6} \\ \frac{2\sqrt6a}{9} \end{pmatrix}=(\frac{\sqrt3a}{18})(-\frac{\sqrt3a}{18})+(\frac{a}{6})(-\frac{a}{6})+(\frac{\sqrt6a}{36})(\frac{2\sqrt6a}{9})=-\frac{a^2}{108}-\frac{a^2}{36}+\frac{a^2}{27}=0$$

Since all the dot products equal 0, it has been shown that the tentative internal radius of the tetrahedron also meets the requirement of touching the centroid of all of the solid’s plane faces perpendicularly. This is because of the following formula -

$$Cos\theta=\frac{v.w}{|v||w|}$$

Hence, if v · w equals 0, then the angle between these two vectors (θ) must be 90° (since θ ≤ 180°).

Because all the characteristics – expressed in Section 3. “Cube Case” – that a solid’s internal radius must have to be deemed as such are true for the case of a tetrahedron, we can contend that the four vectors $$\vec{OE}$$ ,  $$\vec{OF}$$, $$\vec{OG}$$, and $$\vec{OH}$$  can be confidently stated to be the internal radius of the regular tetrahedron’s. The magnitude of these vectors, as they are all the same, is equal to the length of the tetrahedron’s internal radius R, which therefore holds the relationship -

$$R=\frac{\sqrt6a}{12}$$

$$∴a=\frac{12R}{\sqrt6}$$

With this relationship in mind, we can ultimately give a final answer to the original inquiry of this investigation by rewriting the volume and the surface area of a tetrahedron in terms of its internal radius, and then checking whether differentiating this new expression for the volume of a tetrahedron equals its surface area -

$$V=\frac{\sqrt2a^3}{12}=\frac{\sqrt2(\frac{12R}{\sqrt6})^3}{12}=8\sqrt3R^3$$

$$SA=\sqrt3a^2=\sqrt3(\frac{12R}{\sqrt6})^2=24\sqrt3R^2$$

Once we have written the volume and the surface area of the tetrahedron in terms of its internal radius, we only need to differentiate the newfound expression for the volume -

$$\frac{dV}{dR}=24\sqrt3R^2$$

$$∴ \frac{dV} {dR} = Surface area of regular tetrahedron$$

Thus, it has been shown that expressing the volume of a regular tetrahedron in terms of its internal radius R and differentiating it results in exactly the tetrahedron's Surface Area. This is particularly conclusive, as it indicates that, regardless of the nature of the solid, as long as the methodology in Section 5. can be seamlessly followed and the characteristics (expressed in Section 3.) that a solid’s internal radius must have to be deemed as such hold, then we must be able to express a solid's volume (in terms of an internal radius) such that differentiating this new expression of the solid’s volume exactly equals the solids surface area.

## Conclusion

In this way, through inductive reasoning (by taking into consideration all the previous examples and specifically the case of a regular tetrahedron), I can confidently indicate that, for the five platonic solids (the five existents regular polyhedra), it is possible to express their volume in terms of an internal radius such that differentiating it will result in exactly the solid’s surface area. Furthermore, this investigation also sets the ground for further exploration about other types of solids besides regular polyhedra, as observed in the cylinder’s case. An interesting discovery was that partial derivatives, in the case a solid’s volume is expressed in multiple variables, can provide revealing results, as they can be used to find the surface area of specific faces and surfaces of a solid rather than the total surface area of a solid. In this way, it would be intriguing to extrapolate this analysis to irregular solids, as well as for solids of revolution, and establish further guidelines that can help either express a solid’s volume such that differentiating it equals the solid’s surface area or apply partial derivatives and the definition of the derivative to determine the surface area of specific faces or surfaces of the solid rather than the solid’s total surface area.

Thus, despite the specificity of this investigation, we can observe that it can be of great theoretical interest since it can lead to further discoveries and conclusions in broader areas of mathematics. For instance, we can note that the explored throughout this mathematical investigation also applies, and can be extended, to the second dimension – although now relating a figure’s area and perimeter. In particular, similar to the sphere case in the third dimension, the derivative of a circle’s area is conveniently exactly equal to its perimeter. This suggests that conclusions similar to the ones drawn for 3D solids can be suggested for 2D figures (namely that 2D figures whose area can be rewritten in terms of an internal radius and later differentiated will ultimately result in the figure’s perimeter). For this reason, it can also be expected that similar conclusions should be drawn for figures in the fourth dimension (relating hypervolume and surface volume), which implies that this topic can serve to better understand the nature of different dimensions.

For example, take the case of the first dimension. It is difficult to imagine any clear way in which this research can be extended to it, as a line (the only ’1-Dimensional figure’) is only expressed in terms of its length or distance (or in terms of other similar quantities). Therefore, differentiating a line’s length (which can be defined as: “final position” − “initial point") can be deemed to have no mathematical significance, as it is like differentiating the subtraction of two first-degree variables with no coefficients. However, interestingly enough, these ‘first-degree variables’ are two points with a specific position (which relates to even one dimension lower, to the concept of the 0th dimension: points with a position but no dimensions whatsoever) and a specific coordinate location: the starting and ending 1-Dimensional positions of a line. This suggests that applying the definition of the derivative to a line’s length or distance should relate, in a way or another, to the specific 1- Dimensional coordinate locations of the ending and starting points making up the line, in spite of ultimately being equal to 0. However, as with the case of a volume being expressed in terms of two variables, if we rewrite the length of a 1-Dimensional line in terms of its midpoint (which is similar to the concept of internal radius for 2 and 3D solids) and then we differentiate it, we should specifically be able to obtain a result relating the starting and ending position of a line. This is how this topic can serve to both better understand and study the nature of different dimensions and better comprehend the meaning of the derivative.

After all, throughout this investigation I learned the huge significance of interpreting the process of differentiating (such as by applying the definition of the derivative) to make relevant observations and determine aspects that need to change for a certain result to be obtained. This is because only differentiating an expression and obtaining a final result omits several mathematical concepts and numerical values that turned out to be pivotal to answer my initial inquiry. In this way, by ‘manually’ applying the definition of the derivative, we can expect to find hidden implicit answers and ideas about specific mathematical queries, which ultimately converts the idea of differentiating in a much more powerful tool than only obtaining a final numeric or algebraic answer.

## Bibliography

Mendes Disconzi, Marcelo. “Math 155A Fall. 13 Examples Section 5.2..” DisconsiNet. PDF File. http://www.disconzi.net/Teaching/MAT155A-Fall-13/files/Examples_5.2.pdf.

“Python Math: Calculate surface volume and area of a cylinder.” w3resource. August 19, 2022. https://www.w3resource.com/python-exercises/math/python-math-exercise-5.php.

https://en.wikipedia.org/wiki/Centroid.

http://hawaii.hawaii.edu/math/Courses/Math100/Chapter2/Text/Text2202.htm.

https://samjshah.com/2018/04/19/for-a-sphere-why-is-the-derivative-of-the-volume-the-equation-for-the-surface-area/

## Appendix

### Appendix A -

Calculating the coordinates of the vertices of the regular tetrahedron

Vertex A -

This vertex starts at the origin; hence, its coordinates are $$\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$$

Vertex C -

This vertex only extends a units in the y-axis from the origin. Hence, its coordinates are $$\begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix}$$

Vertex B -

The y component of this vertex is exactly the midpoint of vector $$\vec{AC}$$ , which is $$\begin{pmatrix} 0 \\ \frac{a}{2} \\0 \end{pmatrix}$$

Since triangle $$A \frac{\vec {AC}}{2}B$$ is a 30-60-90 triangle, we can find the x component of this vertex through similarity (let the midpoint of $$\vec{AC}$$ be point X) -

$$\frac{\vec{|AX|}}{\vec{|XB|}}=\frac{1}{\sqrt3}$$

$$\frac{(\frac{a}{2})}{\vec{|XB|}}=\frac{1}{\sqrt3}$$

$$\vec{|XB|}=\frac{\sqrt3a}{2}=\,x \,component \,of \,vertex \,B$$

Hence, the coordinates of vertex B are $$\begin{pmatrix} \frac{\sqrt3a}{2} \\ \frac{a}{2} \\0 \end{pmatrix}$$

Vertex D -

The x and y components of the coordinates of this vertex are exactly the same as the x and y coordinates for the centroid of triangle ABC (point E), as well as those of the centroid of the tetrahedron. Hence, the coordinates of vertex D are $$\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ height \end{pmatrix}$$.

The height of the tetrahedron was encountered by using Pythagoras for right triangle AED -

$$\vec{|AD|^2}= \vec{|AE|^2}+\vec{|ED|^2}$$          where $$\vec{|ED|}$$ = height

$$\vec{|AD|}=a$$

$$\vec{AE}=\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ 0 \end{pmatrix}$$

$$\vec{|AE|}=\sqrt{(\frac{\sqrt3a}{6})^2+(\frac{a}{2})^2}=\sqrt{\frac{4a^2}{12}}=\frac{\sqrt3}{3}a$$

$$height=\sqrt{a^2-(\frac{\sqrt3}{3}a)^2}=\sqrt{\frac{2}{3}a^2}=\frac{\sqrt6a}{3}$$

Hence, the coordinates of vertex D are $$\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{3} \end{pmatrix}$$

### Appendix B -

Determining the coordinates of the centroid of the regular tetrahedron and the centroid of each of its faces

To determine the coordinates of the centroid of the regular tetrahedron (O), we only need to obtain the arithmetic mean of each component of the coordinates of the solid’s vertices. Hence -

Arithmetic mean of the x component of the coordinates of all vertices -

$$\frac{0+\frac{\sqrt3a}{2}+0+\frac{\sqrt3a}{6}}{4}=\frac{\sqrt3a}{6}$$

Arithmetic mean of the y component of the coordinates of all vertices -

$$\frac{0+\frac{a}{2}+a+\frac{a}{2}}{4}=\frac{a}{2}$$

Arithmetic mean of the z component of the coordinates of all vertices -

$$\frac{0+0+0+\frac{\sqrt6a}{3}}{4}=\frac{\sqrt6a}{12}$$

Hence, the coordinates of the centroid of the regular tetrahedron are $$\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ \frac{\sqrt6a}{12} \end{pmatrix}$$

Similarly, to determine the centroid of a given plane face of the regular tetrahedron, we only need to obtain the arithmetic mean of each component of the coordinates of the vertices forming that specific triangular plane face.

Centroid of triangular plane face ABC (point E) -

Arithmetic mean of the x component of the vertices A, B, and C -

$$\frac{0+\frac{\sqrt3a}{2}+0}{3}=\frac{\sqrt3a}{6}$$

Arithmetic mean of the y component of the vertices A, B, and C -

$$\frac{0+\frac{a}{2}+a}{3}=\frac{a}{2}$$

Arithmetic mean of the z component of the vertices A, B, and C -

$$\frac{0+0+0}{3} =0$$

Hence, the coordinates of the centroid of the triangular plane face ABC are $$\begin{pmatrix} \frac{\sqrt3a}{6} \\ \frac{a}{2} \\ 0 \end{pmatrix}$$

Centroid of triangular plane face ABD (point F) -

Arithmetic mean of the x component of the vertices A, B, and D -

$$\frac{0+\frac{\sqrt3a}{2}+\frac{\sqrt3a}{6}}{3}=\frac{2\sqrt3a}{9}$$

Arithmetic mean of the y component of the vertices A, B, and D -

$$\frac{0+\frac{a}{2}+\frac{a}{2}}{3}=\frac{a}3$$

Arithmetic mean of the z component of the vertices A, B, and D -

$$\frac{0+0+\frac{\sqrt6a}{3}}{3}=\frac{\sqrt6a}{9}$$

Hence, the coordinates of the centroid of the triangular plane face ADC are $$\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}$$

Centroid of triangular plane face ACD (point G) -

Arithmetic mean of the x component of the vertices A, C, and D -

$$\frac{0+0+\frac{\sqrt3a}{6}}{3}=\frac{\sqrt3a}{18}$$

Arithmetic mean of the y component of the vertices A, C, and D -

$$\frac{0+a+\frac{a}{2}}{3}=\frac{a}{2}$$

Arithmetic mean of the z component of the vertices A, C, and D -

$$\frac{0+0+\frac{\sqrt6a}{3}}{3}=\frac{\sqrt6a}{9}$$

Hence, the coordinates of the centroid of the triangular plane face ACD are $$\begin{pmatrix} \frac{\sqrt3a}{18} \\ \frac{a}{2} \\ \frac{\sqrt6a}{9} \end{pmatrix}$$

Centroid of triangular plane face BCD (point H) -

Arithmetic mean of the x component of the vertices B, C, and D -

$$\frac{\frac{\sqrt3a}{2}+0+\frac{\sqrt3a}{6}}{3}=\frac{2\sqrt3a}{9}$$

Arithmetic mean of the y component of the vertices B, C, and D -

$$\frac{\frac{a}{2}+a+\frac{a}{2}}{3}=\frac{2a}3$$

Arithmetic mean of the z component of the vertices B, C, and D -

$$\frac{0+0+\frac{\sqrt6a}{3}}{3}=\frac{\sqrt6a}{9}$$

Hence, the coordinates of the centroid of the triangular plane face BCD are $$\begin{pmatrix} \frac{2\sqrt3a}{9} \\ \frac{2a}{3} \\ \frac{\sqrt6a}{9} \end{pmatrix}$$