James Stewart was a monster
>>16748843Stewart's calculus will be remembered longer than Euclid's Elements, mark my words. Let the rudincels cope and seethe
>>16748843That's an interesting and challenging problem.I may try to solve it.There may be three cases.
>>16748856There are only two cases.
wild guess but h*sin(θ)
Total square area of Cone minus square area of a sphere with radius X divided by two. Total square area of Cone = YTotal square area of Sphere = ZRadius of sphere = X Examine Your Zipper, nerd.
Y - Z/2 = Answer
>
chatgpt solved it in 40 seconds. how long does it take you?
>>16748843At the time I thought such questions were dumb and needlessly contrived but the truth is that it was good they had problems like that.
INTERESTING CALC 2 CHALLENGE PROBLEM (by me)given a ketchup cup of radius R and height L, what angle [math]\theta[/math]should you unfold the cup to maximize that sweet red glop?
>>16749363i came up with this problem and worked out its solution years ago when i first learned that you could unfold the damn things.it's a problem i'd put on a final because it has a non-trivial solution and utilizes most concepts one has learned up to that point very nicely
>>16749363You filtered me. Then again my formal education only goes as as far as trig. All my calculus is self-taught. Without something linking the length of L in terms of R or vice versa I can get to a (probably wrong) derivative function pictured. There's probably some identities I could utilize to simplify things a bit more but I realized I probably already fucked up somewhere and it doesn't matter anyway because I'm gonna need to play this game where R/L > [whatever the fuck] and I just kinda said "fuck it."
>>16749553>>16749363Oh, and I probably vastly overcomplicated things anyway. But that's typical novice shit.
>>16749363You don't need that much ketchup.
>>16749553>>16749555it's important to sanity check things you deriveone obvious check is that theta=0 and theta=pi should result in 0 volumesince you only supplied what you think is the derivative, a quick glance shows the first term is tends to infinity at both theta=0 and theta=pi, which should instantly be throwing up red flags>>16749565keetchuuup
>>16749573Yeah, I'm pretty much positive I made an algebra error somewhere.To clarify my thought process:>n = theta - 90>area of cone section: (1/3)((pi)hk^2)-(1/3)((cot(n))/(R))(pi)(R^2) where h is the height of the whole cone and k is the radius of the circle opened by L's angle about R.>k = R+Lsin(n)>h = (L + Rcsc(n))sec(n)Everything else I did followed from there.(note: the derivative function I showed was for theta - 90 and not theta itself. but I'm drinking and can't be fucked to evaluate whether that matters)
>>16749573>>1674958190 as in 90 degrees btw. I didn't get far enough to care about radians vs degrees in this particular problem though obviously I should have stuck with radians (so (pi)/2)
>>16749582>90 as in 90 degreesof course. i roll over that automatically.i'm working out the problem at the moment for fun. i'll post it when i finish.
>>16749587My nigga in science. You posted this problem and don't know the solution? I would have figured you worked it out before posing it to us my melanated fren.You're good. I'm still learning myself. My current goal is wrapping my head around the Fourier transform, which I feel deliciously close to accomplishing.But what your problem requires is multivariable functions, which is cool, but very quickly becomes more "pain in the ass" than "interesting" in 90% of cases.
i've got the expression for the volume of the ketchup cup, but the derivative isn't as nice as i remember it being such that you can solve directly for the angle. i'll check it again tomorrow, but it's too late now to continue
Turns out I don't know calculus.
>>16748843R = infinity
>>16749005I solved it using intuition in about 3 seconds
Without spoiling yhe problem for me, can someone tell me if cutting this down to 2 dimensions is valid? I e. Triangle and circle?
>>16749823both are rotation volumes so it should be okay?calculate some areasI know a guy who went to econ school, he'd throw the thing in Excel and do a monte carlo on it
>>16750206the lengths people will go to not think are astounding
>>16750221numerical integration is a very respected field of mathematics
[math]r = \frac{h}{1+\frac{\cos2\theta}{\sin\theta}}[/math]
>>16749005chatgpt didn't solve itit copy pasted the answerthat's like saying google search solved it
>angle is more than 90>all the water falls outhuh?
Trig is chemistry tier mindnumbing
>>16750463trig is easy and useful and isn't a pile of misleading rules of thumb that give a skewed impression of reality
>>16748843It's icecream. It's a trick question you idiots.
>>16748843What about when csc(theta) = 4?
p = volume(largest fully-submerged sphere)q = volume(largest half-submerged sphere)If sin(theta) = 2^(1/3) – 1, then p = q/2.
>>16748843Stewart was a publishing-industrial-complex shekels grifter.Boring, soulless exercises for computational midwits.
>>16748843case 1: wet topif r <= h/(csc(t) + 1) then v = v_1 = 4*pi*r^3/3case 2: dry top and wet equatorif h/(csc(t) + 1) <= r <= h*sin(t) then v = v_2 = pi*(r*(csc(t) + 2) – h)*(r*(csc(t) – 1) – h)^2/3case 3: dry equator and wet "tropic"if h*sin(t) <= r <= h*tan(t)*sec(t) then v = v_3 = v_2case 4: dry "tropic" (and wet bottom)if h*tan(t)*sec(t) <= r then v = v_4v_4 —> 0 as r —> 1/0
maximum:if r = h*csc(t)/(csc(t) + 2)/(csc(t) – 1) then v = 4*pi*h^3/3/(csc(t) + 2)^2/(csc(t) – 1)
>>16753844
>>16753535the probably with stewart is having 9000 editions of the same book whose only difference was different permutations of the same problems
>>16750327Oh, you're right.
>>16753845a = volume(displaced water)b = volume(cone)the ratio a/b is depictedexample:if theta = pi/4, then a/b ≈ 5/6
>>16749367>>16749363I think you don’t need calculus for this one, just trig it’s (big cone)-(smaller cone). Well unless you forget the formula for cone size, then you can use calculus ti get it
>>16749367>mfw glorified related rates problemis that all?
>>16753988the volume of the cup is[math]V(\theta) = \pi R^2 L \sin\theta - \pi R L^2 \cos\theta\sin\theta + \frac{1}{3}\pi L^3 \cos^2\theta \sin\theta[/math]which can be simplified in a variety of ways using trig identities, however, optimizing the derivative seems to have no closed form solution
>>16754016Hmm I think Theta overcomplicating the situation, you ca just make x,y which opey the Pythagorean theorem
>>16749363>>16754016i disagree. using this volume and h = L cos(t) and R = r + 2Lsin(t), I get[math]V(\theta) = \frac{4}{3}\pi L^3 \cos\theta \sin^2 \theta + 2\pi rL^2\sin\theta\cos\theta + \pi r^2L\cos\theta[/math]
>>16753993That's all that calculus is
>>16748843>>16748850>Let the rudincels cope and seetheTypical white-boy problem. Can't handle the intellectual rigor of the Chosen Ones, plays around with pedestrian problems instead.
>>16750443it was solved by humanity, we just accessed our memories
>>16754058suppose the truncated cone is aligned with the [math]z[/math]-axis and that the base of the cone with radius [math]R[/math] coincides with the [math]xy[/math]-plane.use the parameterization as described in >>16749363, with base radius [math]R[/math], side length [math]L[/math], and base angle [math]\theta[/math]the height of the truncated cone along the axis is [eqn]\Delta z = L \sin \theta[/eqn]the change in radius from the base to the top is [eqn]\Delta R = - L \cos\theta[/eqn]the radius as a function of [math]z[/math] is linear, and therefore[eqn]r(z,\theta) = R + z \frac{\Delta R}{\Delta z} = R - z \cot \theta[/eqn]integrating the radius as a solid of revolution along the axis gives the volume[eqn]V(\theta) = \int_0^{\Delta z} \pi r(z, \theta)^2 dz[/eqn]evaluate this integral[eqn]\begin{split}V(\theta) &= \pi \int_0^{\Delta z} [R - z \cot \theta]^2 dz \\&= \pi \int_0^{\Delta z} [R^2 - 2 R z \cot \theta + z^2 \cot^2 \theta] dz \\&= \pi R^2 \int_0^{\Delta z}dz - 2 \pi R \cot \theta \int_0^{\Delta z} z dz + \pi \cot^2 \theta \int_0^{\Delta z} z^2 dz \\&= \pi R^2 z \bigg|_0^{L \sin \theta} - 2 \pi R \cot \theta \frac{1}{2}z^2\bigg|_0^{L \sin \theta} + \pi \cot^2 \theta \frac{1}{3}z^3\bigg|_0^{L\sin\theta}\\&= \pi R^2 L \sin \theta - \pi R \cot \theta L^2 \sin^2 \theta + \pi \cot^2\theta \frac{1}{3} L^3 \sin^3\theta \\&= \pi R^2 L \sin \theta - \pi R L^2 \cos \theta \sin \theta + \frac{1}{3} \pi L^3 \cos^2 \theta \sin\theta\end{split}[/eqn]so[eqn]\boxed{V(\theta) = \pi R^2 L \sin \theta - \pi R L^2 \cos \theta \sin \theta + \frac{1}{3} \pi L^3 \cos^2 \theta \sin\theta}[/eqn]which, once again, can be simplified a variety of ways using trig identitiesfor instance, putting everything in terms of sines[eqn]V(\theta) = \pi R^2 L \sin\theta - \frac{1}{2} \pi R L^2 \sin 2 \theta + \frac{1}{3} \pi L^3 [1 - \sin^2\theta] \sin \theta[/eqn]all the powers of sine can be reduced to a single powers of sine (with multiples of [math]\theta[/math])
>>16753552v_4 = pi*w^2*(r – w/3)w = r – sqrt[r^2 – (h*tan[t])^2]
>>16749363This ketchup sub-thread shouldn't be ITT.It should be its own thread in /sci/.Things related to the OP's problem can be investigated, instead of introducing an entirely different problem.In other words, "not all has been said and done".
>>16748864>There are only two cases.Here are three cases:L: The sphere doesn't fit into the cone.M: The sphere fits into the cone, but is only partially submerged.S: The sphere is fully submerged.S = smallM = mediumL = largeWhich "two cases" do you mean?S and M?Or M and L?
>>16754710>V(t)>Integrates w.r.t. zRetard alert. Should be your first indicator that you're wrong. Not to mention that your volume can be negative for specific values of L & R.
>>16755719S&M. It obviously cannot be L.
>>16755769>Not to mention that your volume can be negative for specific values of L & Rgo ahead and try, i dare you to post a combination of [math]L,R\in \mathbb{R}^+[/math] could take to make my equation negative within [math]\theta \in [0, \pi)[/math]
>>16755769also, one limit of integration is a constant (namely 0) and the other is a function of [math]\theta[/math], so the integrated quantity is only a function of [math]\theta[/math]. integrated quantities don't depend on the dummy variables of integration
>>16755816oops, my bad, i meant [math]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/math]
oops, my bad again, i REALLY meant [math]\theta\in[0,\pi][/math]shouldn't try to pay attention to more than one thing at a time...
>>16755816>>16755822>>16755834i accept your concession that you've not properly checked the limits and constraints on your function. you should feel very embarrassed rn
>>16755839>he can't come up with two numbers that back up his claim
itt: bunch of sperglords with too much math education and not enough innate talent.
>>16748850And none of my school books are based upon Stewart or rudin instead using proprietary calculus.
>>16755774>It obviously cannot be L.The bottom of every sphere -- and that includes large spheres of course -- is submerged.
>>16753552h = 7t = theta = pi/4The blue curve is v_2 = v_3.The yellow curve is Re[v_4].The left side of the graph is at r = r_12 = h/(csc(t) + 1).If r_12 < r < r_23, then only the blue curve is valid and v = v_2.If r = r_23 = h*sin(t), then the blue and yellow curves intersect.If r_23 < r < r_34, then only the blue curve is valid and v = v_3.The maximum of the blue curve is in this interval.Thus the sphere is less than half submerged.If r = r_34 = h*tan(t)*sec(t), then the blue and yellow curves are tangent.If r_34 < r, then only the yellow curve is valid and v = v_4.The right side of the graph is at r = h/(csc(t) – 1).
>>16757412h = 7t = theta = pi/9The blue curve is v_2 = v_3.The yellow curve is Re[v_4].The left side of the graph is at r = r_12 = h/(csc(t) + 1).If r_12 < r < r_23, then only the blue curve is valid and v = v_2.The maximum of the blue curve is in this interval.Thus the sphere is more than half submerged.If r = r_23 = h/csc(t), then the blue and yellow curves intersect.et cetera
>>16756577sorry but proof by "just look at it" excludes that possibility
>>16758191You're actually disputing, that the bottom of every sphere is submerged?Oh my word.You're saying, that a large sphere floats on top of the water?That it doesn't sink a little bit, until it's securely held in place by the cone's circle?Oh my word.
>>16758502>You're actually disputing, that the bottom of every sphere is submerged?nope
>>16758502Are you Chinese?
>>16759005>Are you Chinese?Believe it or not, I'm a descendant of Chinggis Khan."Genghis Khan ([...]; c.1162 – August 1227), also known as Chinggis Khan, was the founder and first khan of the Mongol Empire."And I enjoy East Asian food.And I love rice.
>>16760143>love ricemegalomaniac
>>16760143Ask me how I figured it out..
>>16758191>proof by "just look at it">>16758988>nopeThe definition of a large sphere is:r >= h Sec[θ] Tan[θ]The amount of sinkage is:w = r – Sqrt[r^2 – (h Tan[θ])^2]How's that for more rigor?
>>16760259Imagine being this autistic. Nigga just use your minds eye. It can't be the large sphere.
>>16748843Now shucks you just listen here one second mister!
>>16748843well. yeah he was a literal homosexual. they are all child molesters
>>16760259it doesn't need rigor, just look at it, this isn't complicated
>>16760143
The whole thing, starting with this post:>>16758191>sorry but proof by "just look at it" excludes that possibilityand ending with this post:>>16760560>it doesn't need rigor, just look at it, this isn't complicatedwas a misunderstanding, on my part.Now I finally understand, that you're the speaker of "just look at it".At first, I thought, that you meant, that I'm the speaker thereof.Or that you were being critical of my lack of rigor.All of the posts pertaining to this matter can be deleted now.
>>16750443That's not how it works you braindead techbro.
>>16761006You are retarded kek
>>16755774>S&MA small sphere is fully submerged.Thus the volume of the displaced water is equal to the volume of the small sphere.The volume of a small sphere is a strictly increasing function of the radius.Thus only the largest small sphere is a candidate, for displacing the most water.And the largest small sphere is equal to the smallest medium sphere.Thus small spheres are of no interest!
>>16761825>You are retarded"I know you are, but what am I?"
>>16748843Let the cone have height hand semi-vertical angle t, so the rim radius is R = h tan(t)Put a sphere of radius a on the axis. It touches the cone along a circle. If the center of the sphere is at height z_0 above the apex, tangency gives[math]a=z_0\sin\theta\quad\Rightarrow\quad z_0=\frac{a}{\sin\theta},\qquadz_t=z_0\cos^2\theta=\frac{a\cos^2\theta}{\sin\theta},[/math]where z_t is the height of the circle of tangency.For any ball that can pass through the rim we have [math]a\le R[/math], and then[math]z_t=\frac{a\cos^2\theta}{\sin\theta}\le\frac{R\cos^2\theta}{\sin\theta}=h\cos\theta<h,[/math]so the intersection (water displaced) is the cone from z=0 to z=z_t plus the spherical segment from z=z_t to z=h.Computing that volume V(a) and simplifying,[math]V(a)=\frac{\pi}{3}\Big(z_t^{\,3}\tan^2\theta + [\text{spherical segment }(z_t\to h)]\Big)\quad\Rightarrow\quad\frac{dV}{da}=\frac{\pi}{\sin^3\theta}\big(a\sin^2\theta-a+h\sin\theta\big)^2\ge0.[/math]Thus V(a) is non-decreasing in a on the feasible interval [math]0<a\le R[/math]; it is strictly increasing there (the derivative vanishes only at [math]a=h\sin\theta/\cos^2\theta\ge R[/math], i.e., outside the feasible set).Therefore the spill is maximized by taking the largest sphere that can be inserted--i.e., one whose radius equals the cup’s rim radius:[math]\boxed{\,a_{\max}=h\tan\theta\, }.[/math]
why isnt the answer just hte circumfernence of the sphere because that is what is touching hte water
The following aren't quotes.They're paraphrases.>>16749005>r = (h Sin[θ])/((1 – Sin[θ]) (1 + 2 Sin[θ]))Correct.>>16750327>r = h/(1 + Cos[2 θ]/Sin[θ])Correct.>>16753844>r = (h Csc[θ])/((Csc[θ] + 2) (Csc[θ] – 1))Correct.>>16762410>r = h Tan[θ]Incorrect!And "you're late to the party" pal.Please find the error(s) in your post, and post a correction.So that we don't have to.
>>16760347The correct James Stewart is depicted.
>>16762410(z_0)/sec(θ) = (z_t)/cos(θ)
