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December 9

Volume of cone circumscribed about a sphere

A student I'm tutoring had this problem for an exam. Somehow he got the right answer, but the way he did it doesn't seem to make sense. Here's the problem,

A right circular cone is to be circumscribed about a sphere of r=7. Find the height of the cone that has minimum volume.

He solved this problem by setting the volume of a cone equal to the volume of a sphere and plugging in 7 for r, and solving for h (height). But that assumes the volume of a cone and sphere are the same when a cone circumscribes a sphere, but how is that possible? Shouldn't the cone's volume always be greater? He got h=28 which is the correct answer. How else can this problem be solved? ScienceApe (talk) 21:05, 9 December 2013 (UTC)

This site gives a couple of alternative solutions, which result in h=4a (where h is the height of the cone, and a is the radius of the sphere. Equating the volume of the sphere, and the volume a cone where the base has the same radius as the sphere gives 4/3 π a^3=1/3 π a^2 h => 4a=h, the same expression. Note that the radius of the circumscribed cone is sqrt(2)a (easily see in the working of the second solution), so the cones are not identical, just the same height. It's possible your studet knew this identity, and used it, or there's some deeper principle I'm not spotting (or he just took a guess and has the luck of the gods). MChesterMC (talk) 10:19, 10 December 2013 (UTC)

I may be missing something, but isn't there only one way a right circular cone can be circumscribed around a given sphere? Sławomir Biały (talk) 14:41, 10 December 2013 (UTC)

Only if you choose the cone's height to base diameter ratio first. --CiaPan (talk) 08:07, 11 December 2013 (UTC)

"Right" only means that the cone's vertex is directly above the centre of its base (as opposed to a "slant" cone). So a right cone has two degrees of freedom (e.g. height and ratio of height to base diameter) and the condition that it has an inscribed sphere with a given radius only imposes one constraint, leaving one degree of freedom. Intuitively, you can have a tall thin cone with a base diameter only slightly larger than the sphere's diameter, or a short fat cone with height only slightly larger than the sphere's diameter, or anything in between. Gandalf61 (talk) 12:16, 11 December 2013 (UTC)

I thought "right" referred to the angle at the vertex, but apparently that is not the case. Sławomir Biały (talk) 13:35, 11 December 2013 (UTC)

See Cone lead, also Cone#Right circular cone. --CiaPan (talk) 09:49, 12 December 2013 (UTC)

I think the assumption is that the base of the cone is tangent to the sphere, as assumed in my solution below.--Jasper Deng (talk) 03:16, 15 December 2013 (UTC)

The solutions referred to by MChesterMC appear to me to be unnecessarily advanced, involving trigonometry and square roots and what have you.

The 6 variables are:

V: the volume of the cone

h: the height of the cone

A: the area of the base of the cone

R: the radius of the base of the cone

S: the side of the cone

a: the radius of the inscribed sphere

The 4 geometric equations are

0 = 3V − Ah = A − πR² = Sa − (h − a)R = S² − R² − h² .

Elimination of variables is done by multiplication and addition of polynomials. Elimination of A:

0 = (3V − Ah) + h(A − πR²)

0 = 3V − πhR²

Elimination of S:

0 = −(Sa + (h − a)R)(Sa − (h − a)R) + a² (S² − R² − h²)

0 = ((h − a)²− a²)R² − a²h² .

0 = (h² − 2ha)R² − a²h² .

0 = (h − 2a)R² − a²h .

Elimination of R²:

0 = (h − 2a)(3V − πhR²) + πh((h − 2a)R² − a²h) .

0 = 3(h − 2a)V − πa²h² .

Differentiation:

0 = 3(h − 2a)dV + 3(dh − 2da)V − 2πah²da − 2πa²hdh .

The radius a is constant, so da=0.

0 = 3(h − 2a)dV + 3Vdh − 2πa²hdh .

The condition of minimum volume is that dV=0 even when dh≠0. Chose dh=1.

0 = 3V − 2πa²h .

Elimination of V

0 = −(3(h − 2a)V − πa²h²) + (h − 2a)(3V − 2πa²h) .

0 = πa²h² − (h − 2a)2πa²h .

0 = πa²h(h − (h − 2a)2) .

0 = −h + 4a .

This latter equation is easily solved, h=4a, and when a=7 then h=28. Bo Jacoby (talk) 16:20, 12 December 2013 (UTC).

This whole problem is considerably simpler with Lagrange multipliers (provided one is familiar with basic partial differentiation). We are maximizing $V={\frac {1}{3}}\pi hr^{2}$ subject to the constraint (obtained via similar triangles - there is no way you can fail to observe this while also knowing the equations of the cone and sphere) $r^{2}(h-7)^{2}=49(r^{2}+h^{2})$. By Lagrange's theorem there exists a number $\lambda$ such that, at the desired point, $\lambda$ times the gradient of the constraint equals the objective function's gradient. In other words, with the constraint denoted as $g(r,h)=49h^{2}+14hr^{2}-r^{2}h^{2}=0$, we have:

$\nabla V=\lambda \nabla g(r,h)$ or

${\frac {\partial V}{\partial r}}=\lambda {\frac {\partial g(r,h)}{\partial r}}$

${\frac {\partial V}{\partial h}}=\lambda {\frac {\partial g(r,h)}{\partial h}}$

so we have the following system of three equations in r, h, and lambda (the Lagrange multiplier):

${\frac {2}{3}}\pi rh=\lambda (28hr-2rh^{2})$

${\frac {1}{3}}\pi r^{2}=\lambda (98h+14r^{2}-2hr^{2})$

$g(r,h)=49h^{2}+14r^{2}-r^{2}h^{2}=0$

To eliminate the pi and lambda we divide the first two equations to obtain

$2{\frac {h}{r}}={\frac {28rh-2rh^{2}}{98h+14r^{2}-2hr^{2}}}$

We cross-multiply:

$196h^{2}+28hr^{2}-4h^{2}r^{2}=28r^{2}h-2r^{2}h^{2}$

The second term on the left and the first on the right cancel and we obtain

$196h^{2}=2h^{2}r^{2}$

or $r=7{\sqrt {2}}$. We substitute this into the third equation and obtain

$49h^{2}+14(98)h-98h^{2}=0$

$h^{2}+28h-2h^{2}=0$

$h^{2}=28h$

or h=28 (we can discard h=0, of course).

The issue with setting ${\frac {1}{3}}\pi hr^{2}={\frac {4}{3}}\pi r^{3}$ is that the cone's radius must be greater than 7, or that would violate the definition of circumscribing. As expected, the cone's volume ${\frac {2744}{3}}\pi$ is greater than the sphere's ${\frac {1372}{3}}\pi$.--Jasper Deng (talk) 02:51, 15 December 2013 (UTC)

In addition, a single-variable calculus solution may be obtained by solving for r² in the constraint equation and substituting into the volume equation to obtain

$V={\frac {1}{3}}\pi h{\frac {49h^{2}}{h^{2}-14h}}={\frac {1}{3}}\pi {\frac {49h^{2}}{h-14}}$

This is differentiated and the derivative set equal to 0:

${\frac {dV}{dh}}={\frac {1}{3}}\pi ({\frac {98h}{h-14}}-{\frac {49h^{2}}{(h-14)^{2}}})=0$

${\frac {98h(h-14)-49h^{2}}{(h-14)^{2}}}=0$ so

$49h^{2}-14(98)h=0$

from which h=28 can be obtained, as above.--Jasper Deng (talk) 04:13, 15 December 2013 (UTC)

The method of Lagrange multipliers avoids solving for V in terms of h, which is cumbersome, and in my opinion is far less confusing than Bo Jacoby's solution (which is very hard to follow). The constraint in its original form is the triangle similarity relation ${\frac {r}{\sqrt {r^{2}+h^{2}}}}={\frac {7}{h-7}}$ which is readily obtained by constructing a diagram and noting that both are equal to the sine of the angle between the axis of the cone and its side.--Jasper Deng (talk) 04:59, 15 December 2013 (UTC)

Hi Jasper Deng. Thank you for your comments! The starting equations

0 = 3V − Ah = A − πR² = Sa − (h − a)R = S² − R² − h²

comes from this:

1. The volume of a cone is one third of the base area times the height, so 0 = 3V − Ah
2. The area of a circle is pi times the radius squared, so 0 = A − πR²
3. The right-angled triangle with hypotenuse h-a and cathetus a is similar to the right-angled triangle with hypotenuse S and cathetus R, so 0 = Sa − (h − a)R
4. The triangle with sides R,h,S is right-angled, so Pythagoras says that 0 = S² − R² − h²

The equations are manipulated without solving them and without substituting. Generally you cannot solve equations exactly. If you have two equations, say

0 = 3V − 2πa²h

and

0 = 3(h − 2a)V − πa²h²

the idea is to note that

0 = A(3V − 2πa²h) + B(3(h − 2a)V − πa²h² )

for arbitrary polynomials A and B, and then chose A and B such that the variable V is eliminated. That is A=h-2a and B=-1. Bo Jacoby (talk) 13:20, 15 December 2013 (UTC).

The problem is that you are not really allowed to introduce so many arbitrary parameters into the problem (such as your A and B). For example you cannot assume that dh=1 in your solution above, nor can you assume that their differentials are related like that (you can't do that unless all of them are (implicitly) functions of one variable such as t). My method, while involving substitution, avoids solving for V in terms of just one variable. It doesn't assume that you can solve for anything explicitly other than the objective function (if there's no closed-form solution, your method is equally unable to find it).--Jasper Deng (talk) 03:29, 16 December 2013 (UTC)

I merely introduced A and B in order to explain the method to you. dh is a free variable and, assuming dh≠0, I may set dh=1 for convenience. I didn't solve for V in terms of just one variable, but I rather eliminated V in order to reach the equation 0=-h+2a. The problem with several nonlinear equations in several variables is solved by you by substituting the solution of one equation into the other equations, and by me by eliminating the variables one by one. Your method involved fractions and square roots, while my method did not. So my method is more elementary, and also more general. You will benefit from appreciating that. Bo Jacoby (talk) 10:53, 16 December 2013 (UTC).

No, I could've easily solved my system by elimination too. Volume is always derivable by integration over three dimensions and thus it is never necessary to first solve for V. I'm still not convinced that your arbitrary parameters can be manipulated like that. Your implicit differentiation is fundamentally imprecise (what are you differentiating with respect to? That h is a free variable is patently false, because it is a function of whatever variable you differentiated with respect to, unless it is h, in which case the other differentials should have a dh under them or the dh should still stay dh; you didn't at all address my points). The method of Lagrange multipliers also eliminates the dependence on other variables cleanly and furthermore, is strictly true, while you potentially introduce extraneous solutions with the extra parameters. This method might be better in another situation, but when you know a strict objective function subject to one or more constraints in the form of $g(\mathbf {x} )=0$, Lagrange multipliers are the more fundamental way. Just because I have a square root doesn't mean my method is somehow more complicated (in fact I even squared both sides when presenting the solution above, so it's completely irrelevant). I don't care about there being square roots or fractions if they can be eliminated. If I have a more precise method to use that avoids introducing arbitrary confusing parameters, there's no reason not to use it.--Jasper Deng (talk) 11:21, 16 December 2013 (UTC)

1. Surely you could solve the problem by elimination, but you didn't, and you object against it.
2. When the volume of a cone is computed by integration, the result is that 0=3V-Ah.
3. Which 'arbitrary confusing parameters' are you talking about? My variables are (V,A,R,S,h,a) and their differentials.
4. The rules of differentiation used are d(x+y)=dx+dy and d(x·y)=x·dy+dx·y, no matter which variable I am differenting with respect to.
5. h is not a free parameter, dh is.
6. That dV/dh=0 can be written dV=0, dh=1.
7. Lagrange multipliers are not wrong, but not necessary either.
8. 'Just because I have a square root doesn't mean my method is somehow more complicated'. No, but it is more advanced. My method can be used by someone who hasn't been taught square roots or fractions.

Did you understand my explanation, or do you still find the calculation difficult to follow? Bo Jacoby (talk) 23:06, 16 December 2013 (UTC).

1. Why use elimination when in this situation it's easier to just solve it using substitution?
2. That I do not dispute. But it's not the point of my argument, which is that if you have $V=...$ the method of Lagrange multipliers is always applicable, and because of 3D integration it's always possible to get such a formula (if you can't get one in closed form, you won't get a form suitable for making polynomials either).
3. The polynomials A and B can't just be anything.
4. The rules for differentials do not apply if the variables are not functions of the same parameters.
5. Also, it is improper to use ordinary differentials in a multivariate situation (if you want to avoid solving for V in terms of one variable).
6. Per the above, dh isn't; also differentials are infinitesimally small so you can only assign a finite value to the quotient of two differentials or the integral of a differential.
7. Strictly nothing but differentiation is necessary but you want the easiest solution, right?
8. No-one is taught calculus without first knowing the basic operations of square rooting and geometry. Basic differential calculus courses that give these kinds of problems do not omit related rates problems, which require one to look at the geometry of the situation.

I would be extremely surprised if someone came to calculus without first knowing the most basic geometry and arithmetic concepts of square roots and fractions (in fact the derivative is fundamentally a limit of a fraction, so you cannot claim that someone who knows differentials doesn't at least also know fractions).--Jasper Deng (talk) 23:21, 16 December 2013 (UTC)

1. Is your objection, that you did not understand the elimination method, still valid?
2. I don't get your point.
3. The equations 0=A·0+B·0 is true for any A and B.
4. Really?
5. Really?
6. Differentials may be interpreted as formal variables to which finite values may be assigned.
7. I wanted an elementary solution. You don't find it easy. I do.
8. My students are taught calculus without square rooting.

I wouldn't be surprised if you get surprised. Bo Jacoby (talk) 08:06, 17 December 2013 (UTC).

1. That was never my objection. I said it's confusing and imprecise, not incomprehensible.
2. The argument about not being able to solve for V therefore does not apply.
3. You can't do that in the sense that it introduces extraneous solutions and in the sense that if you choose A and B to keep the whole equation balanced, you are no longer guaranteed that both terms are 0. Therefore any roots you find may purely be due to your A and B.
4. No they may not, unless you can show otherwise: Differentials are only related by such equations in the limit as their size approach zero (the radioactive decay law, for example, is only integrable when we allow the change in time to approach 0). The definition of differentiability can be stated as requiring every change $\Delta y$ (at a point x=c) to be expressed in the form $\Delta y={\frac {dy}{dx}}\Delta x+g(x-c)$ where $\lim _{x\to c}g(x-c)=0$ (assuming your ordinary differentiation is valid, which it is not strictly, per below), and you basically are ignoring g (the equation assuming g = 0 is only valid when it is 0, presumably in the limit as x approaches c). This is a requirement that's easily extended to the multivariable case. I will be convinced if you can strictly show that the function g drops out regardless of what the parameter's value is.
5. That solution is not elementary when you have implicit differentiation. It's certainly not what comes to the calculus student's mind. I am perfectly capable of doing it but I don't feel like the OP benefits from it (it likely is not exactly what his teacher wants, and without a good understanding of the implicit differentiation is bound to encourage errors).
6. You've got to be kidding me. They are not ready to solve geometric problems if they don't know their... geometry (basic algebra is a prerequisite for calculus, and that includes square roots - you'd teach extremely well if you could somehow teach someone calculus who doesn't have a full understanding of what a function is (that's what algebra's for)). Besides, they can't ever go without knowing that the power rule is true for rational exponents (including the square root). What if the objective function were not an integer power of the parameters?

The variables in the situation are functions of more than one variable (even in polynomial form V is an implicit function of both r and h) unless a constraint is applied to reduce it to a one-variable situation, which would not be applicable in either of our solutions.--Jasper Deng (talk) 09:19, 17 December 2013 (UTC)

Looking at your differentiation in further detail, I see that you forgot that V is still a function of a radius variable (that of the cone), even if that variable is not actually visible (dV also has to be 0 regardless of a change in the radius variable, which you don't strictly consider, since R does affect h under the constraints placed (by your argument I could've easily chosen dh=2, and suddenly I get something else)). If you did intend to consider V's dependence on multiple variables (eliminating its parameters' appearances in the expression is not sufficient), please be more clear about that, because I assume ordinary differentiation when I see d rather than the partial symbol. I can see where you're trying to apply the product rule, but V does not depend only on h and a. You have to insert a ${\frac {\partial V}{\partial R}}$ term somewhere to indicate dependence on R.

Finally, though, let's take a look at your polynomials again. Our methods are fundamentally the same if you notice that you are merely expressing V as a function of more variables than just R and h ($V(A,a,h,R,S)={\frac {Sa-(h-a)R+Ah}{3}}$; because you incorporate the sphere's dimensions) and applying more constraints than I did. Your differentiation steps just happen to be equivalent to taking the gradients of all the constraints, except you seem to run into a mess when you try to move all the differentiation to one step (but clearly you probably don't like manipulating the polynomials after taking the gradients of all the constraints that your polynomials impose). I merely placed the differentiation step where it belongs.--Jasper Deng (talk) 09:52, 17 December 2013 (UTC)

But I will note that only the univariate solution conclusively proves it to be a minimum without performing the complicated (for functions of 3 variables and above, that is, as we would have to do for the Lagrangian function) second partial derivative test (the univariate case is easily treated with the first derivative test).--Jasper Deng (talk) 10:24, 17 December 2013 (UTC)

This discussion is derailed. You wrote on my talk page: "Even I have trouble following your solution to the problem because I don't see where you get all the starting equations". When writing "Even I" you seem to suggest that you yourself are smarter than other readers, which may or may not be the case. I explained how I got the starting equations from the geometry of the problem, and asked: "Is your objection, that you did not understand the elimination method, still valid?", and you answered: "That was never my objection. I said it's confusing and imprecise, not incomprehensible". Don't be the teacher while you are confused and are having such trouble following the solution. Be the student. Ask questions rather than argue. Bo Jacoby (talk) 08:24, 18 December 2013 (UTC).

I should've been clearer - I thought I did imply that I now comprehend where you got the equations (to your credit, I wrote my other replies pretty late at night when I did not have the entire discussion in mind). Your differentiation step was what really confused me (No, I'm not a teacher, but someone who insists on very strict arguments when it comes to mathematics, especially calculus). I talked with other experts in mathematics and they concur with me that your "I can set dh=1" argument is patently false because you are not looking for a linear approximation but an exact solution, and the differential equation is not valid if you set any differential to a finite value (and I also argue that it isn't even valid in the first place because you did not consider the dependence on R).

I'm sorry if I come as overly combative here, but I do believe that your solution needs to be corrected before I can learn anything from it (yes I do have a habit of jumping from point to point, so let me summarize: Thanks for clarifying where you got the polynomials and how you approached it, but to me you are getting a few basic differentiation rules just outright wrong (I know all the mathematics behind it so when I see a new method, I seek to verify its validity). When I said "Even I", I implied that if I could not understand it, the OP could not either (sorry for my arrogance, though).

I will try to learn something from the solution as soon as you can prove its soundness, but I'm firmly convinced that the differentiation step is just outright wrong. I know that you know Euler's formula, from another discussion on the reference desk, so I know you are capable of answering my arguments in a succinct way (To your credit, someone not taught to solve for V in terms of one variable likely will take this approach).--Jasper Deng (talk) 20:44, 18 December 2013 (UTC)

Speaking with another expert, while I'm still not convinced that you've expressed V as a function of just one variable (and really the steps for "eliminating" A are just redundant), you're not choosing dh=1 because if you divide both sides of the equation by dh, the dh coefficients vanish. But really, V has hidden dependencies on other variables - you still need to consider things like ${\frac {\partial V}{\partial R}}{\frac {dR}{dh}}$, eliminating parameters does not eliminate variables (your V is a function of 5 variables, so ${\frac {dV}{dh}}=\nabla V(A,a,R,S,h)\cdot {\frac {d<A,a,R,S,h>}{dh}}$, so ${\frac {\partial V}{\partial h}}=0$ is a necessary but not a sufficient condition; $dV=\nabla V\cdot <dA,da,dR,dS,dh>$ (you have to use chain rules for all variables except h)).

I'm not going to be condescending and "give you advice", but it's my firm opinion that if you want to do it using polynomial operations, you must carry out the above differentiation step first (use implicit differentiation as necessary to obtain derivatives of R and the other variables with respect to h) before doing any polynomial operations. --Jasper Deng (talk) 23:01, 18 December 2013 (UTC)

Thank you!

Do you agree that 0=3(h−2a)V−πa²h² follows from the geometry equations?

As the radius a is constant, implicit differentiation with respect to h gives 0=3(h−2a)dV/dh+3V−2πa²h . Do you agree?

Formal multiplication by dh gets rid of the fraction. So I wrote 0=3(h−2a)dV+3Vdh−2πa²hdh . Don't be alarmed. The meaning is unchanged. It is only a matter of notation.

Now set dV/dh=0 in the first equation, or dV=0 and dh=1 in the second equation. In both cases we get 0=3V−2πa²h . Do you agree?

Differentials like dV and dh are not infinitesimals (whatever that means). They are formal variables, defined axiomatically by the rules for differentiating sums and product, and the rule that da=0 if a is constant, and that dV=0 if V is minimum. (If V is maximum, then -V is minimum, and so d(-V)=0, but d(-V)=-dV by the above axioms, and so dV=0 too). Bo Jacoby (talk) 00:09, 19 December 2013 (UTC).

It is dangerous to think of differentials as anything but infinitesimals (infinitesimals were replaced by $\lim _{\Delta \to 0}$ but we still use the Leibniz notation for them; you can't just take off the limit operator (which is effectively what you're doing here)), for in various other situations it results in an invalid equation because of the "error" term above (the one I said whose limit must be 0), unless you are doing an error approximation (take Riemann sums for example). The student who does not understand that will be tempted, for example, to set dV=3 (if the student does not realize that dV must effectively be 0 as dh-->0). I unfortunately cannot agree with the implicit differentiation on the grounds that you haven't isolated V as a function of h alone.--Jasper Deng (talk) 01:13, 19 December 2013 (UTC)

Don't worry, there is no danger! The definition of the differential quotient of y=f(x) is f'(x)=lim_t→x((f(t)−f(x))/(t−x)). This is also written dy/dx=lim_Δx→0(Δy/Δx), but not dy/dx=(lim_Δx→0Δy)/(lim_Δx→0Δx). Don't write dy=lim_Δx→0Δy or dx=lim_Δx→0Δx. The differentials are not infinitesimals. Avoid this historical confusion. The differentials dy and dx are free variables except that they must satisfy the relation dy=(lim_Δx→0(Δy/Δx))dx. From the definition follows the algebraic rules d(x+y)=dx+dy and d(x·y)=dx·y+x·dy. As these rules are sufficient for differentiation of polynomials of several variables, you will not need the definition any more. Bo Jacoby (talk) 15:10, 19 December 2013 (UTC).

But you cannot claim, for example, that you can obtain the exact value of a function using Euler's method. You can only obtain the exact value of an antiderivative (for example) in the limit as the differential approaches 0 (or the Riemann sum does not work). My textbook explicitly says (in general) that differentiability means $\Delta f(\mathbf {x} )=\nabla f(\mathbf {x} )\cdot \Delta \mathbf {x} +g(\Delta \mathbf {x} )$ (x is a vector, i.e. in a multivariate domain) where $\lim _{\Delta \mathbf {x} \to \mathbf {0} }g(\Delta \mathbf {x} )=0$, so when you omit $g(\Delta \mathbf {x} )$, you get an equation that is only valid when $\Delta \mathbf {x} \to \mathbf {0}$, i.e. when the differential is infinitesimally small.--Jasper Deng (talk) 19:49, 19 December 2013 (UTC)

I am not using Euler's method. Note that the result h=4a is exactly correct. No approximation was made. The functions that are differentiated are polynomials, and the differentiation is exact. No term is omitted. Bo Jacoby (talk) 21:42, 19 December 2013 (UTC).

But I want to let you know that letting a differential equal something arbitrary (note that the correct result is not obtained unless you set dh=1) results in disaster in general. For example, $\int dy=\int 93xdx$ is not valid if the differentials were of finite value, because the Riemann sums assume the differentials are infinitesimally small.--Jasper Deng (talk) 00:33, 20 December 2013 (UTC)

The correct result is obtained by chosing dh≠0. The minimum condition dV/dh=0 is equivalent to dV=0 when dh≠0. The value dh=1 was chosen for convenience. dh=2 would also do the trick, but is an unnecessary complication. You are confusing the differential dx with the difference Δx. Bo Jacoby (talk) 01:14, 20 December 2013 (UTC).

They are fundamentally the same. In fact, the d in dx was chosen to replace delta. My textbooks say the choice of d requires that it be infinitesimal (and they are very recent). In fact a derivative is not even formally the quotient of two numbers (or variables; it is a shorthand for the limit that defines the derivative), and the fact that it obeys multiplication and division rules is only a convenient coincidence. If differentials are to be useful for Riemann sums, i.e. $\int dx=x+C$, they must be infinitesimal (you cannot set dx to a finite value in that situation because the limit is being taken).--Jasper Deng (talk) 01:28, 20 December 2013 (UTC)

Why choose 'd' in dx to replace 'Δ' if they are fundamentally the same? No sir, they are fundamentally different. Your objection was treated earlier in this thread. Listen and learn: The differentials dy and dx are chosen such that the differential quotient dy/dx is equal to lim_Δx→0(Δy/Δx). To realize that x+C is an integral of dx, just note that d(x+C)=dx+dC=dx+0=dx. The theories of Riemann and Lebesgue are not needed in order to integrate polynomials. You are using a sledgehammer to crack a nut. And integration was not used to find the cone of minimum volume circumscribed a fixed sphere, so you are off topic anyways. Bo Jacoby (talk) 09:43, 20 December 2013 (UTC).

I never said you used integration here. But if you want to claim that differentials can be treated like that, you must show that such a definition is consistent with all uses of differentials ($\int dx=x+C$does not work if dx = 5, because then you'd add infinitely many 5's; furthermore when it comes to definite integrals, a finite value of dx would only be an approximation). Otherwise I'm not going to believe a word of what you say. In fact, my textbook explicitly states that "The differentials of the independent variables dx and dy are $dx=\Delta x,dy=\Delta y$" (and I have no reason to doubt this textbook, which is among the very best I've read).--Jasper Deng (talk) 17:58, 20 December 2013 (UTC)

I talked with a(nother, he is the fourth expert I have consulted) teacher who teaches this (he has a degree in mathematics) and when I told him about the notion of differentials being finite, he called that notion absurd, even for this problem. Maybe what you're getting at is to just define two variables a and b such that ${\frac {a}{b}}={\frac {dy}{dx}}$, but they can't be called differentials, because that is inconsistent with their use in Riemann sums.--Jasper Deng (talk) 19:56, 20 December 2013 (UTC)

Please acknowledge what you do understand before you change the subject. Can you now follow my solution to the problem and see that it produces the correct result? I don't mind explaining Riemann integration to you, but let's take one step at a time. Bo Jacoby (talk) 23:04, 20 December 2013 (UTC).

Take my work at Talk:Fresnel integrals for example. Just because it comes out with the expected result doesn't mean the method itself is valid. I imagine you're going to argue that the integral sign implies a limit as the change in the variable of integration goes to 0 (as integration does do) but I do not consider that a sufficient argument to validate your definition of a differential. The book I use considers "the differential of an independent variable to be its increment", where "increment" is the numerator or denominator of the difference quotient of a partial derivative. The use of "increment" here should not act to imply that differentials are finite, because like I said above, the "error" term only vanishes in the limit, not at any ordinary value of the increment; the book also discusses Newton's method in the multivariate case, only approximately equating a differential and an increment in the equations it uses. If you can't agree on that definition, this discussion is over (and I'm not inclined to change it, as I have spoken with other experts who have had decades of experience with this and while one of them said that your idea of differentiation is correct, they unanimously agree with me that you cannot set a differential to a finite value. If we wanted a finite change, we use delta instead of d, or if we want a quotient that doesn't require taking the instantaneous value of a difference quotient, don't use the term differential for that).--Jasper Deng (talk) 00:20, 21 December 2013 (UTC)

Do you agree that the result was correct? Do you agree on the definition dy/dx=lim_Δx→0(Δy/Δx)? Do you accept your expert's explanation: just define two variables a and b such that a/b=dy/dx? Do you mind me using the (otherwise unused) names dy and dx instead of a and b just to avoid the extra variable names a and b? Are you able to follow my calculation? Do you recognize your attitude problem? Bo Jacoby (talk) 09:21, 21 December 2013 (UTC).

The a and b quotient thing was not an expert's explanation (they all reject the entire notion; that ${\frac {dy}{dx}}=\lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$ is not disputed, but you're not going to get away with claiming that ${\frac {dy}{dx}}={\frac {\Delta y}{\Delta x}}$ for any arbitrary finite $\Delta x$, because $\Delta y$ is strictly a change in the function y, not a change that results from just varying $\Delta x$ in the differential equation (we're not doing a linear approximation of any sort)), but my conjecture on what you were trying to get at. In math there's no room for "oh, the differentials are as if it were just one finite variable divided by another so we can call any numerator/denominator pair whose quotient is the derivative differentials"; if it doesn't meet the definition it does not qualify, period. It is either a differential or it is not. There's nothing wrong with the attitude of insisting on mathematic rigor especially when it comes to calculus. It is only a coincidence that you do not run into major trouble in this particular problem (I read your steps carefully before starting this discussion), because your steps happen to correspond to the general formal steps that must be taken. To say a=0 in my conjecture is equivalent to saying ${\frac {dV}{dh}}=0$ because 0=0, but that's a coincidence. If we had been searching for a nonzero value for the derivative, the a/b form is just wrong - you'd quickly find that you cannot manipulate the two separately because their ratio is fixed in that case, and furthermore, it is meaningless to even use it because we could be wanting the derivative to be equal to any arbitrary real number, and any real number can be expressed as the quotient of two others (what makes a/b so special? Nothing. You can't even get a valid change in the dependent variable using that).--Jasper Deng (talk) 11:42, 21 December 2013 (UTC)

I am not claiming that dy/dx=Δy/Δx for any arbitrary finite Δx. I don't know how you got that misunderstanding. I use the undisputed definition of dy/dx, a limit of fractions, not a fraction of limits. There exist numbers, dy and dx, such that dy/dx is the fraction that it formally seems to be. These differentials are finite, they are not infinitesimal. Many text books define differential quotients without defining differentials. Bo Jacoby (talk) 20:42, 21 December 2013 (UTC).

Don't call them dy and dx, because I still insist that they must be infinitesimal, you're not making a convincing argument that they are finite (as I have repeatedly shown you cannot apply that definition to other uses of differentials; furthermore you cannot use your definition with partial derivatives' difference quotients). (When it comes to calculus textbooks, there is only one text I trust out of those available to me, the one I'm referring to here, the others often don't even get the strict definition of a limit right). For example, if you treat the differentials in a Jacobian matrix as identical and take the determinant, the equating of all the differentials with the same symbol (such as $\partial z$) leads to the entire determinant to equal 0 because all the terms appear to have the same numerator and denominator, an obviously invalid result (you would argue that each partial derivative is a quotient of a unique pair of variables, but your erroneous use of differential symbols in place of those variables' symbols would lead to this kind of fallacy). Also, in physics, anything but an infinitesimal is contrary to any of the applications of the derivative in it.--Jasper Deng (talk) 22:07, 21 December 2013 (UTC)

You cannot be helped when you insist in misunderstanding what I write. Partial differentiation is written like this: dz=(δz/δx)dx+(δz/δy)dy. My application was d(3(h − 2a)V − πa²h²)=3(h − 2a)dV + 3(dh − 2da)V − 2πah²da − 2πa²hdh . This is standard procedure. Bo Jacoby (talk) 06:29, 22 December 2013 (UTC).

You just admitted that your definition does not work in the most general case (and please indent your comments). So you're basically saying that you're using some variables a, b, and c, and don't want to actually make the distinction between differentials and the a/b convention (Don't wrongly accuse me of misunderstanding partial differentiation; I wrote the partial derivative notation above using vector notation, the Jacobian determinant is completely different ($\partial V$ is not necessarily called a "differential", as the book does not define them as such, but you can clearly see that you're going to get fallacies with your notion of a differential)). If four experts concur with me, as well as other students, there's no way it could be "standard". The standard procedure is to not multiply by dh and treat each derivative strictly as one quantity, not the quotient of two others (also, your notion that you could teach calculus to people w/o square roots was called even more absurd by the fourth expert I asked). I'm not interested in trying to convince you anymore, math is very strict about its rules (the quotient of two numbers being equal to the derivative does not mean they are differentials just because they "look sorta like them") and if you don't want to play by its rules, I don't think you have any premise to try to convince me of anything. I hate being so argumentative over what's such a small deal after all , but I cannot accept a method that introduces an idea that's patently wrong. By the way, if you think differentials are defined by some obscure axioms, please read my text on them (the requirement for their quotient to be a derivative does not work ummodified in the multivariate case, since there's no single difference quotient possible - $df(\mathbf {x} )=\nabla f(\mathbf {x} )\cdot d\mathbf {x}$; differential equations relate infinitesimal increments consistent with the notion that integration is the exact inverse of differentiation and as I said your "differentials" are patently incompatible with definite integration, because the proof of the fundamental theorem only proves it for infinitesimal differentials). I'm doing it the way Newton and Leibniz did it (they didn't have an idea of the epsilon-delta definition but "finite differentials" has no place).--Jasper Deng (talk) 06:53, 22 December 2013 (UTC)

Your equation $df(\mathbf {x} )=\nabla f(\mathbf {x} )\cdot d\mathbf {x}$ and my equation dz=(δz/δx)dx+(δz/δy)dy mean the same thing: partial differentiation of a dependent variable with respect to several independent variables. So it does work in the multivariate case. If differentials dy and dx are tiny then they approximate differences Δx and Δy, otherwise they don't. It's no big deal. Bo Jacoby (talk) 07:59, 22 December 2013 (UTC).

That I don't dispute. But the "differentials are finite" notion does not work for the multivariate case, let alone the single-variable. You can define $a=\nabla f(\mathbf {x} )\cdot <b,c,d...>$, but once again <b,c,d...> is not dx and the difference quotient that defines the total derivative in general uses only the magnitude of dx→0 (so you cannot assign it a finite value, as that says nothing about the direction of dx). The reference to the Jacobian is an example of fallacies based on ${\frac {a}{b}}{\frac {c}{d}}={\frac {a}{d}}{\frac {b}{c}}$: using this, it could be erroneously concluded that ${\frac {dy}{dx}}{\frac {du}{dv}}={\frac {dy}{dv}}{\frac {du}{dx}}$, when neither u or y have any dependence on x and v respectively, leading to the absurdity that the product of those derivatives must always be 0 (or more strictly, the notion that it's even possible to take ordinary derivatives of y with respect to v or u with respect to x, when there is no known pair of functions relating those pairs in that way).--Jasper Deng (talk) 09:15, 22 December 2013 (UTC)

You ment to write ${\frac {a}{b}}{\frac {c}{d}}={\frac {a}{d}}{\frac {c}{b}}$ ? When neither u or y have any dependence on x and v respectively, then dy=adx+0dv=adx and du=0dx+bdv=bdv and there is no problem. a=δy/δx and b=δu/δv are partial differential coefficients. Bo Jacoby (talk) 10:35, 22 December 2013 (UTC).

Yeah I did. But 1 - neither of our definitions of "differential" cover the "numerator" or "denominator" of partial derivatives, 2 - that leads to the absurdity that the products of the (partial) derivatives on the left must necessarily equal 0, or the absurdity that the "numerator" of a partial derivative does not change meaning when the "denominator" is changed. In short, they don't completely obey rules we can take for granted with regular variables.--Jasper Deng (talk) 20:00, 22 December 2013 (UTC)

dx is a differential. δx is not a differential. There is no problem. Bo Jacoby (talk) 21:45, 22 December 2013 (UTC).

What if the product on the left is a product of two ordinary derivatives? I can take the product of two unrelated derivatives just like any other pair of functions. You can see that you can't consider the left side to be the product of two quotients of "finite" differentials, even though arguably the ordinary derivative is the quotient of a pair of related differentials. This is also what I meant by saying that differentiation rules do not apply if the functions aren't functions of the same parameters.--Jasper Deng (talk) 22:01, 22 December 2013 (UTC)

I don't understand. Can you please give a specific example? Does your objection apply to the problem of the minimum cone, or are you at last admitting that my solution is correct? Bo Jacoby (talk) 05:26, 23 December 2013 (UTC).

Like I said your solution is based on this wrong notion of a "finite differential", and so like my solution to the Fresnel integrals' limits, is bad even though it returns the correct result.--Jasper Deng (talk) 16:21, 23 December 2013 (UTC)

The idea of infinitesimal differential is based on the old erroneous definition (see above) dy/dx=(lim_Δx→0Δy)/(lim_Δx→0Δx), such that dy=lim_Δx→0Δy and dx=lim_Δx→0Δx where dy and dx are infinitely small without actually being zero. This idea was already criticized by George Berkeley calling the infinitely small differentials ghosts of departed quantities. In the modern definition dy/dx=lim_Δx→0(Δy/Δx) the differentials dy and dx are not infinitely small, they are not infinitesimals. My point is that you do not need to solve equations, just manipulate equations. The differentials dx and dy are then nontrivial solutions to the homogenous equation 0=lim_Δx→0(dx⋅Δy-dy⋅Δx), (the trivial solution being dx=dy=0). So I solved the cone problem without using fractions or root signs, and my students have learned differential calculus without relying on solving equations. You may learn from this if you are inclined to learning rather that arguing and protesting. Bo Jacoby (talk) 22:44, 23 December 2013 (UTC).