The Calculus Problem Nobody Could Solve

[@DOROnoDORO]

as someone who hasn't taken a calculus class yet but did a decent amount of self studying, I'm quite proud of myself for being able so solve it :)

[@AliDicaprio]

Honestly, I wouldn’t wanna try to solve it, but when i saw ur comment, i felt it’s like a challenge cuz, me too, I didn’t study calculus, all my information just from self studying. So thank u, u made proud of myself too.

[@Jack_Callcott_AU]

@doro doro....That's good! I'm also quite proud of myself because I solved it too, but I have taken numerous calculus classes. 👁👍

[@anonamemous6865]

How do I feel that I did nothing?

[@The-Devils-Advocate]

@@anonamemous6865 you sought out useful review or novel material, depending on your level of math knowledge

[@3koningen]

Note: there is no trig in this book and that is a good thing. Second note: all books by Silverman are great. Especially his translations of Russian books. Last (for now): he also has a much larger calculus book; one of the best out there.

[@rubenvegas7926]

Why are people so against trig? Everything can be derived from the unit circle. Pretty damn convenient and easy if you ask me

[@cr1t1cs92]

Not only that Trig is actually quite handy too.

[@danielmrosser]

Arguably, the entire field of mathematics began with trigonometry. Without a millennia of human ingenuity focused on solving problems of trigonometry, modern civilization may not have developed beyond the Stone Age.

[@9WEAVER9]

SPECIAL FUNCTIONS (SILVERMAN IS RAD)

[@aomoussynonymous8712]

@@9WEAVER9 Hehehe Rad. So Silverman is a plane angle ?

[@nexusnexus9221]

I recently did a problem like this. The solution was to draw a graph, with a point. This causes you to think of two possible solutions and then it became easy. Thank you for showing us this.

[@martinepstein9826]

Paused at 3:36. Here is my solution The tangent line to f(x) = x^2 at the input c is the line through (c,c^2) with slope 2c y = 2c(x - c) + c^2 = 2cx - c^2 Now we plug in (x,y) = (1,-1) and solve for c -1 = 2c - c^2 c^2 - 2c - 1 = 0 c = 1 +- sqrt(2) c^2 = 3 +- 2 sqrt(2) The tangent lines are y = 2(1 + sqrt(2))x - 3 - 2 sqrt(2) y = 2(1 - sqrt(2))x - 3 + 2 sqrt(2)

[@violintegral]

I had the exact same solution. 👍

[@bubbotube]

You should have used (1,-1) as (x1,y1) to find the equation of the tangent lines

[@justintroyka8855]

I think what makes this problem hard isn't any specific knowledge or techniques you need to be aware of, but the fact that you aren't given that much information - it can be impossible to know where to start. This is why it's important for math students to not just learn specific knowledge or techniques, but general problem-solving skills that will serve them well in any situation. I was stumped for a little while, and then I decided to *name the point* on y=x^2 where the line was tangent. I named it (a,b), and that opened up the whole problem to be solved pretty easily. Naming your unknowns is an important problem-solving technique in math that can be useful anywhere.

[@galzajc1257]

We had that kind of problems in high school, also with other conics. But in this case you can also write intersection of line of slope a through (1,-1) with the parabola. a x - 1 - a == x^2 Then just solve for a, where discriminant is 0. a^2 - 4 a - 4==0 Then just place the solutions in a x - 1 - a. Another solution that comes to mind and is a bit more general is to just use point on parabola (t, f(t)), where t is a parameter. Then you can express the slope in two ways. f ' (t) =(f(t)-y0)/(t-x0)

[@davidellis1929]

Slightly simpler solution: Once we have the two points of tangency on the parabola, we know the slopes are twice their x-values. Use the common point (1,-1) with these two slopes to get a simpler equation for the two tangent lines.

[@lucien5112]

Yeah, I thought it was weird when he used the point on the parabola rather than the point given in the point-slope form equation

[@dobrojoeshive]

Using the point (1,-1) does result in an equation with fewer radicals, but when we deal with tangent lines, we're usually more interested in what's happening near the point of tangency than near some other point, so we tend to use the point-slope form centered around the point of tangency.

[@VeteranVandal]

Yep. This is the faster, more elegant method. The slow, rude method is shown in the vid... Of course I did it the dumber way, just as the channel did, but it's good to see people nailed the better method instead.

[@pacopascal3256]

This is the book that I learned calculus from when I was in high school. It's probably my favorite calculus book despite it not being as extensive with its topic coverage as other texts. Seeing you talk about this brings back allot of memories. In high school, my math teacher told me I should work on improving my analytic geometry. So, I was looking for problems to solve and making up my own. And one that stuck with me was finding where the tangent line of y = x^2 intersects the y axis. I remember using this book and the material I learned from it to solve it. The result (which is -y) at time got me excited. It stuck with me all these years. By the time I got to calculus in college, my understanding was much deeper than my peers. I was the only one in the class to solve the most difficult problem on the calc 1 final. Even the professor didn't anticipate the methodology in my solution. I credit this book for that.

[@valor36az]

Dear Professor TMS, I cannot thank you enough for the quality of your videos, it keeps people like myself who self study oriented.

[@TheMathSorcerer]

👍👍

[@fuoridisenno8226]

Hi math sorcerer! Great video as usual. Here in Italy this kind of problem is actually faced by students in the second year of high school (Liceo Scientifico, high school with scientific orientation), without using any calculus method. Here's how you can solve it without calculus: Lines: y = mx + q They have to meet the point (1;-1) ==> -1 = m*1 +q So m + q = -1 for both lines. You want the lines to meet the parabola, so basically finding points that respect the condition x^2 = mx + q (*) But those lines have to intersecate the parabola in only ONE point so we are looking for m and q such that there is only one solution for the (*) equation, and this happens in a second grade equation when the (b^2-4ac) term in the solving formula is 0. In this case, (b^2-4ac) is (m^2 + 4q). But we had found m + q = -1, so we finally have m^2 + 4(-1-m) = 0 which leads to 2 solutions for m, m1 = 2(1 + sqr(2)) and m2 = 2(1-sqr(2)), as you found. The rest is trivial. To be honest, I'm quite shocked by the fact that nobody in your classes was able to solve it, considering that I faced the very same kind of problem when I was 14 yo. But I guess it's all about the point of view you've learnt to look at the subject through.

[@30803080308030803081]

When you showed the problem, I paused the video, solved the problem in my notebook, then continued the video. The way you solved it was so different from the way I did. I kind of think my way was better. But I think you solved it how a math teacher would solve it.

[@donovanthompson9013]

Through some symbol moving, i found a general solution for function f(x) and point (x0,y0). you solve for x1 in the eq f(x1)+f'(x1)(x0-x1)=y0 and your line is y=mb+b where m=f'(x1) and b=f(x1)-f'(x1)x1 Derivation: given f(x) and (x0,y0), we look for eq y=mx+b st y0=mx0+b and there exists x1 st f(x1)=mx1+b => b=y0-mx0 =>f(x1)=mx1+y0-mx0 notice that for the line to be tangent, m=f'(x1) =>f(x1)=f'(x1)x1+y0-f'(x1)x0 =>f(x1)+f'(x1)(x0-x1)=y0 plugging back in to y0=mx0+b, we get b=f(x1)-f'(x1)x1 Using your problem, we solve for x in x^2+2x(1-x)=-1=>-x^2+2x+1=0 which has roots 1+- sqrt(2) solving for b as and plugging into y=mx+b gives your two solutions heres a desmos link with a couple examples:www.desmos.com/calculator/txcm00l92g

[@guyedwards22]

It actually strikes me as odd that not one student could get this, even first semester calc students. I started Calculus my junior year in High School and this problem would've been very straightforward to me even early on in the year. To be fair, our teacher was quite excellent.

[@javiergilvidal1558]

People are getting dumber by the minute!

[@wannabeactuary01]

I think that is the difference - how well taught the subject is and how good the textbooks are. Equations of lines is badly taught in general.

[@richardibarra4688]

@@javiergilvidal1558 I really don’t think that’s the case. I just think people forget the basics or they overthink the problem and they over complicate things.

[@afuzzycreature8387]

having just gone through this I would consider this to be a valid exam problem for a high schooler who just learned differentiation and knew point and intercept form. That nobody got it... no, I can't believe this. That just terrible on the instructor.

[@farmerjohn6526]

@@javiergilvidal1558 Society Protects Dumbness, otherwise evolution would wipe them all out.

[@perplexedon9834]

I haven't done calc in a few years or beyond grade 12, though I was pretty good. I managed this really easily in a couple of minutes by setting up 3 equations: y=x^2, m=2x, y=m(x-1)-1 From m=y'; y-y1=m(x-x1) 3 equations 3 unknowns, and 1 and 2 easily substitute into 3. Solve 3 with the quad formula, sub into 2 and then back into 3. So pretty similar to how you did it. I attribute me finding this easy to playing around with graphing calculators a lot, practicing generalising equations for geometric relationships (like tangent lines. You can even do it without any calculus: y=m(x-1)-1 y=x^2 Sub 2 into 1: x^2=m(x-1)-1 Rearrange and solve for X with quad formula: x=[m±√(m^2-4(m+1))]/2 For the line to be tangent to x^2, m must give a single simultaneous solution. Therefore the b^2-4ac=0. m^2-4m-4=0 gives m=2±2√2, sub back into 1 and you're done.

[@fuoridisenno8226]

this is exactly how it's done here in Italy in second year of high school

[@JR-uc3nk]

I bought this book because of you! One of my favorites!

[@zgennaro]

Wow, it took me about an hour and several failed attempts but to my surprise I actually got it. The messy algebra made me doubt but when I graphed my results… Eureka! So proud of myself!

[@TheMathSorcerer]

nice:)

[@paul21353]

Alternative solution. Line through (1,-1) with slope m: y+1=m(x-1) Intersect line with y=x^2: x^2+1=m(x-1). We want tangents, so set D=0 in this quadratic equation and you get m=2+sqrt(2) or m=2-sqrt(2) and this gives the equations of the two tangents as required. No derivative needed.

[@picassodilly]

It struck me that once you’d found the two x values for the slopes, it would be simpler to just jump to the point/ slope equations using (1,-1) as your point instead of doing the extra math to find your y values.

[@enalaxable]

a bit easier to find the 1- parametric line from (1,-1), intersection with x^2, and require it to have one double solution for each line. A tad less pencil used, but it's a general way to find tangent lines, without derivatives.

[@fesosorpro1106]

One simpler way to solve the problem is to use the property of discriminant and the tangency of line. First state the general eqn of lines passing through (1,-1), then form a system of eqn with y=x². Very soon eliminate y and form a quadratic on x. Because of the tangency of the line the discriminant must be zero, hence by equating the discriminant of the quadratic with zero we end up with a new quadratic on the slope of line, which gives the value of the slope once solved.

[@timschumacher]

Certainly agree with this. Set the tangent as y=mx+c as (1,-1) is on the tangent, get c=-1-m, so the tangent is y=mx-1-m Solve simultaneously with y=x² to get x²-mx+1+m=0 This quadratic has the discriminant of 0 (this is the crucial step) So m²-4m-4=0 which then gives the two yangent gradients.

[@chazzbunn7811]

Cool problem. I want to share my solution because it's a little different from the video's, I started with this: y - f(c) = f'(c)(x - c), which is a slope equation tangent to f. Then I did this: y - c^2 = 2c(x - c), because f(c) = c^2 and f'(c) = 2c. Then I plugged in the point (1, -1) for x and y, -1 - c^2 = 2c(1 - c). That equation yields the same quadratic as in the video.

[@tomasgarau5338]

If you want to make life a bit easier for your self when finding the y intercept, you don't even need the y coordinates you calculated - you can just use the point (1,-1)! For example sub (x,y) = (1,-1) into y = mx + c for -1 = 2(1+sqrt2) 1 + c c = -3 -2sqrt2 which gives y = 2(1+ sqrt2)x - 3 - 2sqrt2

[@willyh.r.1216]

All college math-learners should expect to solve non-standard math problems. Otherwise, test = mechanical repetition with different numbers and letters.

[@Noam_.Menashe]

I used only modern methods (at least what I learned), I called the point on the parabola A (x_0,x_0^2), called (1,-1) B, equated the slope to the derivative, solved it, checked, and done.

[@JoeTaxpayer]

Fun problem. The kind of problem that a student may look at, and since they haven’t seen a nearly identical problem, might lack the imagination to get the answer as simple as it may seem once the process is shown to them. The other thing I have learned from you? It doesn’t take all kinds of expensive software and video editing tools to make a brilliant math video. Your videos are perfect just as they are, pencil and paper still rules! (and, yes the Dixon Ticonderoga cannot be beat. I bought a box of Amazon brand pencils at a great price. Worst purchase of my life. I left all of them on the table in the math office. I’ll stick with the best.)

[@TheMathSorcerer]

thanks for your comment:)

[@williamguest3887]

Another solution would be to assume the equation of a tangent to the parabola at x=a is y=mx + c. You can obtain m using differentiation, and therefore c by observing that there should be exactly one solution to finding the intersection between the line and the parabola. Plugging in x=1, y=-1 gives the two solutions pretty neatly.

[@othman31415]

Nice problem! another approach is to consider the difference: y=mx-m-1 is the equation of all (non-vertical ) lines passing through (1,-1), one of those line will be a tangent iff it intersects with x->x² in exactly one point which means that the equation x²-mx+m+1=0 would have only one solution thus its discriminant should be equal to zero i.e delta= m²-4m-4=0, solving this equation will determine all the tangents.

[@pyrotas]

Same as I thought (and wrote elsewhere not having seen this comment). For such a problem there is no need to introduce derivation and indeed I think that solving it in a more "naive" but "constructive" way as the one proposed by you goes a long way into teaching intuition in geometry :)

[@othman31415]

@@pyrotas I agree "the more elementary the better" that's my math motto. But if we replace the function x->x^2 in the problem by say.. x->exp(x) then we can't escape derivatives.

[@pyrotas]

@@othman31415 sure thing! However, if I had been faced with that particular problem (with a quadratic, i.e.) I would have approached it in the simplest way, regardless it was a calculus class or what the teacher was expecting from me to do :D Yet again, I was kind of a trouble-maker when I was a student :D

[@Matlockization]

The reason why people get destroyed is because there is not enough work done on the cross pollination of newer and earlier work. New maths topics appear to be independent of each other in module form. Which unless they are shown how to do, will not see that the direct connection is doable.

[@mitchellm4197]

Bro thank you so much for recommending this book. I've been grinding Algebra, Trig, and Geometry for a while and still have some work to do, but I was eager to learn Calculus and starting reading this book. I love it. such a good read so far. Cheers!

[@TheMathSorcerer]

awesome!

[@michaelzumpano7318]

I paused the video at the start and worked through the problem. I made a couple arithmetic mistakes but got back on track when the answers didn’t look reasonable. This was a great exercise - more please :-)

[@TheMathSorcerer]

❤️❤️

[@ndotl]

Professorship 101: On the midterm and/or final, assign one problem that is not completely covered by the standard text. Eventually, I was able to recognized when I encountered said problem. From my perspective, it was a way of determining if the student really understood the concepts as opposed to understanding computation mechanics.

[@brianmccormick8328]

Surprised no one solved this. It's straight forward.

[@carultch]

When I saw the title, I thought this was going to be about either a currently unsolved problem in mathematics, or a famous problem that took centuries to solve. Rather than just a simple application of tangent lines.

[@The1RandomFool]

I took a slightly different approach and used point-slope form to find the points at which the tangent lines touch the function, being x^2+1 = 2*x*(x-1). It's essentially the same, though. I'm surprised nobody could solve it though, even with the current state of the education system. It doesn't seem hard.

[@marionfelty7247]

When I was a senior in high school (1977), without the use of calculus (only analytical geometry), I DESIGNED a program for my state-of-the-art TI-58 programmable calculator that would solve this same type of problem in general. You just imputed the coefficients of the quadratic equation, and then the (x,y) coordinates of the point. The program then tested the point for how many solutions that were available. For example: if the point is outside the quadratic-2 solutions, on the curve-1 solution, inside the curve on the axis of symmetry-2 solutions, and inside the curve but not on the axis of symmetry-0 solutions. Once this was known it would calculate the tangent line equations for how many were possible. Do that when you are only 17 years old. 😎

[@yamomwasthebomb7159]

wow im so impressed

[@marionfelty7247]

@@yamomwasthebomb7159 Yes, of that, I'm sure. lol

[@kevinfoflygen1627]

Thanks for the problem! I enjoyed seeing your solution, which was somewhat different than mine. I wrote g(x) = mx+b for the tangent lines. For one of the tangent lines, say g_1(x) = m_1*x+b_1, the slope m_1 = f'(x_1) = 2x_1, where x_1 is the point of tangency with f(x). Observing that f'(x) = 2x, I saw that the y-intercept b = -f(x) = -x^2. Thus g_1(x) = 2x_1*x-x_1^2. In particular, g_1(1) = -1 = 2x_1-x_1^2 ---> x_1^2-2x_1-1 = 0 ---> x_1 = 1+/-sqrt(2). From this point, my solution was the same as yours. The main differences, I guess, were that (1) is used geometry to find b, and (2) I didn't need to find y explicitly.

[@sgkhk2397]

In my school teacher gave me a very beautiful problem..I solved it today ..... . If from the points (h, 2-5h)where h≠1 two distinct tangents can be drawn to the curve y = x ^ 3 - 3x ^ 2 - ax + b then find the value of(a + b) (h, a, b are real numbers) please make a video on this question ....

[@stanleythompson5321]

This guy already has the answer for the problem and back track down on writing, i really appreciate his explanation.

[@futuure]

I think my method might be a more straightforward alternative: We know any tangent to the graph in point a on the real axis verifies y = (x-a)f’(a) + f(a) (E) where f:x->x^2 We’re looking for a such that for y = -1 and x = 1 this remains verified. This leads to y = (x-a)2a + a^2 i.e. when inputting x and y as said: a^2 - 2a - 1 = 0. We know this has 2 real solutions (discriminant etc.) which are a1 = 1 + sqrt(2) and a2 = 1 - sqrt(2). So the tangents at these two x coordinates verify the proposition. Inputting them back into (E) it gives us the corresponding equations: y1 = 2(1+sqrt(2))x - 3 - 2sqrt(2) and y2 = 2(1-sqrt(2))x - 3 + 2sqrt(2)

[@coolguy7676]

I am an engineering graduate who took Calc 1 several years ago. It felt really nice to solve this problem because these were the kind of puzzles that made me get interested in maths. I self study analysis and abs algebra so it feels "cute" to get to see a problem which has numbers in it after so much time haha. ❤

[@englematics]

This problem can easily be done without Calculus! Write equation of line with slope m that passes through (1,-1). Then solve for intersection between line and parabola. This will result in a quadratic equation and tangent lines will occur when the discriminant is zero, i.e. a quadratic in m yielding the two possible slopes.

[@dneary]

A small improvement: instead of using the point (1+\sqrt{2}, 3+2\sqrt{2}) to calculate the equation of the line, you make the math a bit easier by using the other point you know is on the line: (1,-1). In both cases, you get the equations of the lines to be: y = (2+\sqrt{2})x - (1+2\sqrt{2}) or y = (2-\sqrt{2})x - (1-2\sqrt{2})

[@amritsagar3506]

you can find the chord of contact by using the formula T=0 taught to us in junior high school coordinate geometry and then solve it with the curve to find the point of contact i.e (y+y1)/2=xx1 (y-1)/2=x on simplifying you get y=2x+1 On solving it with the curve y=x^2 you get 2x+1=x^2 x^2-2x-1=0 On Solving this quadratic you get x=1+√2 and x=1-√2 from this we can calculate respective slopes by dy/dx=2x therefore 1st slope is 2+2√2 and 2nd slope is 2-2√2 and since both lines passes through 1,-1 by slope point point the combined equation can be written as (y+1)=(2±2√2)(x-1)

[@lightyagami1752]

I haven't watched your video beyond 3:44, but I solved your problem in less than a minute. y = 2(1+√2)x - 3 - 2√2 and y = 2(1-√2)x - 3 + 2√2 Very direct and easy question. Just let the point of tangency on the parabola be (α,α²) and since the gradient of the tangent line is 2α, its equation is y = 2αx - α² after simplification. Now just substitute x = 1, y = -1 and solve the quadratic!

[@skybird8231]

I solved it only by using the fact that the tangent lines from (1,-1) to f(x) only intersects once per line, and are the ONLY lines to do so (the two answers). This means that if solved by m(x-x1)=y-y1 -> m(x-1)-1=y=x**2 ->x**2-mx-m-1=0 Then use quadratic formula, but write c as (m-1). This gives x=m+-sqrt(m**2-4*1*(m-1))/2. Because we know they only intersect once, the squareroot has to be equal to zero (in the end we get two m values, which also only gives two x values). m**2-4m+4=0 -> m=2+-2*sqrt(2). Then we can just proceed to create the lines with the given point (1,-1) and get the two answers.

[@dalitlegreenfuzzyman]

Haha I just commented the same solution and was like…. Has no one else noticed that this takes NO calculus? Then I scrolled through the comments to see yours haha

[@ultravioletgaming4217]

You can also create a system of equations with y=x^2 and the equation of a line passing through (1, -1) with slope f'(x), or y=2x(x-1)-1. Then solve the system of equations for x to get the same values. You then don't need to find the y values of the points the lines touch x^2, as you can put them right into y=2x(x-1)-1, since you can use any point on the line for point-slope form.

[@vibhu91]

I don’t really understand why was this a particularly hard problem? It was really straightforward for me. I have attached my solution below for everybody’s reference. I prefer my solution because it is more structured than one shown in the video, albeit they are the same thing in essence: Sol: you have three pieces of information 1. A point (1,-1) which passes through some straight lines 2. The same lines are tangents to a parabola 3. The graph of the parabola shows that the point is outside the parabola such that there are two such lines in question Suppose the the point where the line is tangent to the parabola is given by x=t, hence y=t^2 Hence, these points pass through the same straight lines, so the must satisfy the general eqn of a straight line aX + bY +c = 0 where a,b,c are some unknowns The slope of this line is -a/b But the line is tangent to the parabola, whose slope at (t,t^2) is given by 2t Hence we have three unknowns a,b,c and three equations a-b+c=0 at+bt^2+c=0 -a/b = 2t Solve and you will get the equations to both the lines. Very simple and solved under 2 mins.

[@RichardJohnson_dydx]

This is not a difficult problem but one I have not practiced in a long time. I recognized what to do but I could not remember how to do it.

[@blakeh95]

Interesting, I came at it kind of the other way around. I started with point-slope from of y-y1 = m(x-x1). We have y=x^2, m=2x, and (x1,y1) = (1,-1). This gives x^2-(-1)=2x(x-1) => x^2+1=2x^2-2x => x^2-2x-1 = 0. From there, I solved for the x values like you did. Once you have those, I went to y=mx+b form, with m now known for the tangency point (it's the solutions to the quadratic) and (x,y) = (1,-1) again [since the equation is for the entire tangent line, any point works--we don't have to use the point of actual tangency]. Then you can solve for the 2 values of b. I checked the point-slope forms you had at the end, and our results were equivalent.

[@wowZhenek]

Not even sure what is hard about this problem tbh ... tangent line formula is pretty much something everyone has to know: y = f(x0) + f'(x0)*(x-x0) where x0 is the point of intersection. Plug in (1;-1) and you get -1 = x0^2 + 2*x0*(x - x0) which leads to the solution.

[@timotejkochjar9225]

Exactly. I was so questioning if I solved it correctly when I wrote it down on paper because he made it sound like it was super difficult.

[@ivolva]

As you can see from pervious comments no Calculus needed here. Just consider y=x^2 and y=k*x+b. Tangent line crossing parabola in 1 point only, so for this point we have x^2=k*x+b and we want a discriminant of this equation equal zero. That give us a slope k. We can find b from condition tangent line is passing point (1,-1).

[@ingobojak5666]

A nicer form of the final solution is y = 2 * (1 +/- sqrt(2)) * (x - 1) - 1. This form naturally results if one calculates as follows: [A] Line equation y = m * x + b. [B] point (-1, 1) on line, - 1 = m * 1 + b, so b = - m - 1 and y = m * (x - 1) - 1. [C] at tangent point (x_t, y_t) slope equals derivative m = 2 * x_t and on parabola y_t = x_t^2, thus x_t^2 = 2 * x_t * (x_t - 1) - 1. This gives quadratic equation x_t^2 - 2* x_t - 1 = 0 with solutions x_t = 1 +/- sqrt(2). The final form given in the beginning results from substituting the result for m.

[@tiger12506]

I think you'll find that what trips up students is the mixing of y and x between the very distinct cases of f(x), and the function for the tangent line. They spend a good part of the algebra career trying to learn that variables are equivalent to numbers and that equal things are equal, but in this problem, you have distinct meanings for (x,y) in each case so that y_f(x) != y_tangent etc.

[@jimschneider799]

@8:20 - since 1 +/- sqrt(2) is the root of x^2 - 2 x - 1 = 0, you can solve for x^2, giving x^2 = 2 x + 1, so (1 +/- sqrt(2))^2 = 2 (1 +/- sqrt(2)) + 1 = 3 +/- 2 sqrt(2). Also, @10:00 - why not use (1, -1) for the point? You'd get the same line, but a vastly simpler equation.

[@emmynoether2198]

You could just use a parameter (t,t^2) for the curve find the slope, write the tangent equation and substitute the point to get a quadratic in t.

[@rtheben]

That’s really strange, we have this kind of problems in all books of Italian high schools, it’s considered an algebra conic problem not calculus, you ave to solve for a slope so the line passing through the given point intersects the parabola in only one point ( impose one solution for the system, you come up with imposing the discriminant of a quadratic equation to be zero)

[@jewulo]

You know that you are dealing with a very serious mathematics teacher when they get high just on the sniff of pencil-lead.

[@aaronw6195]

I remember doing a few problems like this one. Probably missed it on some tests before, and actually got it on an exam. The completing the square step, and getting this thing setup to be factored is all pretty tricky. It's never the Calculus, and always the Algebra.

[@abhinavnreddy6073]

I think an interesting note would be that I dont think you need to compute x^2 at all. Since all you need is the slope value which is 2x, so now you have two different slopes and the point (1,-1) which we know also passes through the tangent lines. So that would make setting up the equation even easier. (y+1) = m(x-1) and you can change m to the two roots of the quadratic and you get two lines, because ttwo slopes

[@KazACWizard]

i did it this way. notice first and foremost, the only possible intercepts of your general tangent lines are negative. this is CRUCIAL to the method. use the gernel fact of the line equation but here ts y=mx-c. then plug in y=x^2 use the discriminant property with the negative c to get m^2=4c. rearrange to get c in terms of m. plug back into your line equation since this restriction means the form of the line WILL be tangent to the graph only. now restrict the equation further to only pass through 1,-1. by plugging in the values of course. find your m values. then plug back into the equation and find your c values. resulting equations are the same as in the solution in the video.

[@bogdand6129]

I may be making a huge mistake but instead of using that cool method of making the 2 slope expressions equal, a simpler way to get that polynomial expression for x is to use point-slope form and use the values 1 and -1 for x1 and y1. Otherwise, deceptively simple problem. Thank you for the awesome video Math Sorcerer!

[@dirk_math6794]

Just intersect the pencil of lines through (1,-1) i.e. m(x-1)+n(y+1)=0, with m and n arbitrary real numbers, with the curve y=x² and put the discriminant equal to 0 (to force a double intersection point).

[@LinusE]

fell asleep after a slow work day to this video, woke up to "a basic calculus problem" and was startled by the math on it at first 😂

[@rogerlie4176]

You actually don't need to calculate the y-values. After you find that x = 1 ± √2 you use that k = 2x. Since the lines go through ( 1, -1) you get the two equations 1) y + 1 = 2(1 + √2)(x - 1) 2) y + 1 = 2(1 - √2)(x -1)

[@danielstuart3270]

That was a very cool problem, but I was never exposed to that kind of problem in my Calculus classes, but now that I see it, the solution seemed simple. I might have to pick up that book. 🤓

[@bhamidipativasu2397]

We can use coordinate geometry S1^2=S S11 to get pair of tangents from a point

[@lastchance8142]

Tough problem, but simple Calculus here. This requires imagination, which only develops after many, many practice problems!

[@TheMathSorcerer]

yup!

[@danielmrosser]

No new concepts beyond calculus 1 are involved in the problem shown in this video - it is simply using (applying) the concept of a derivative- that’s it. Please note the Silverman book is titled “Calculus with Applications”. The emphasis on applications in the Silverman book and many other great math texts sets it apart from the standard calculus textbooks. Applications of calculus unfortunately are not emphasized enough in math classes and oftentimes relegated to extra credit assignments.

[@lorenpearson1230]

Interesting review of a book I actually used. The irony of the modern books is that they traded space for examples and descriptions for images, and although they say a picture is worth a thousand words, the reality is the books my kids used were poorer for it. Math is ultimately more cerebral than most would like to admit, and the sad part about images is that they lock you into a slice of the whole graph, a plane projection, as opposed to understanding the full surface that a curve describes. Images are thus constraining, while a book like this forced you to think of these things first and appreciate a modern graph and 3D representation on video later. I feel that going the other way is much much harder. As for the problem: it is easy to do. The real issue not addressed was that the students were clearly not aware of how to decode a math statement, and thus understand what tools they needed and thence the approach. Perhaps tangents confused them, or perhaps the various cases. I frankly did not like your solution path: I would have gone through one line at a time fully. Finish the full solution from start to finish, then repeat. Doing the conceptual leap to generalized answers is the final step where you compare those answers and show how a general solution gives both.

[@soldenstoll8495]

Another avenue of completion you could take is just writing out the formula for a tangent line at t. It’s y=f’(t)(x-t)+f(t). Plugging in f(x)=x^2 into our equation let’s us easily solve for the two possible values of t and leads to the same result.

[@scotthennebeul1705]

That book helped me ten years ago in my calc class!!!

[@AnkhArcRod]

This is a 3rd year high school problem, right? If so, I can understand why some calc students may not have managed to solve it.

[@ValenceFlux]

There's this point in understanding it where you can visualize the graph points out of the numbers. It's a good feeling when it finally makes sense to you.

[@GhostyOcean]

So we need the equations of the lines that are both tangent to f(x)=x² and pass through (1,-1). Setting up the equation of the tangent line at x=a we get y = f'(a)(x-a)+f(a) y = 2ax-a² But since we want the lines that pass through (1,-1), plug those values in to x and y respectively and solve for a. So -1= 2a(1) - a² a² -2a-1=0 => a = 1 ± √2. Plugging these values of a into the tangent line equation gives us y = (2+2√2)x -3-2√2 y = (2-2√2)x -3+2√2. These are valid solutions since they both pass through (1,-1) and are tangent to f(x)=x² at x=1±√2.

[@nightowl32]

getting my feet wet with calculus.....see immediately how algebra is applied....gotta really think outside of the box

[@captainsnake8515]

I'm not sure I see what's so challenging about this problem. The solution seemed like a very direct aplication of the techniques taught in a standard calculus class, with no fancy tricks required.

[@johnstfleur3987]

CONSCIOUSNESS IS ETERNAL FOR EVERYONE INCLUDING THE GOOD ONE NOW.

[@cameronhunt5967]

I solved this by making a general formula for the tangent line and lines that pass though the point (1,-1), setting them equal to each other with the same slope and solving for the x values that made it true

[@feraudyh]

If I remember right Silverman translated a lot of mathematics books into English before he started to write his own, based on the best ideas he'd got from the previous books.

[@Tabu11211]

I picked this up after you suggested the publisher! Turns out I should probably grab another book to supplement it haha.

[@Dover0486]

That Dover book is in very good condition to have been owned for years.

[@VeteranVandal]

There are 3 regions in a parabola in this situation. The interior, where no tangent crosses a point inside it, the parabolic line region per se, in which the tangent is one function which is the line the derivative evaluated at said point is and external points, which will have 2 solutions. Working out the above is fairly easy, so I won't do it, but regarding this problem in particular, I'll claim there are 2 solutions since f(1)>yo=-1 with yo denoting the y value of the point which our tangent must cross, so our point coordinates are denoted by (xo,yo), and the resolution does bear that out - even if Wolfram Alpha doesn't at first, you can make it show you by force both solutions. Your solution and mine followed the basic same principles, except I did an even slower analysis based on discovering the coordinate where we cross Y in the tangent lines, the intercepts. Why people are making mistakes here I don't know. I found plenty of exercises online that are analogous to this one, and there are faster techniques for this problem (based on the point slope expression of the straight line, you can abreviate the math slightly for a more elegant solution using slopes and the point (1,-1)) than the one displayed in the video, plenty online as well, and I found them in other books. I cannot believe people got this wrong, tho, in an exam, you have limited time to spend solving a problem you are seeing, likely, for the first time. Remember: those with the answers can do everything a student in a test is doing at least 3 times faster, because we are more used to it and because we already did the exact exercise in the test. I attribute not nailing this largely to the fact that exams don't examine as much as we'd hope.

[@JCCyC]

Oooooookay. I paused the video at 4:04 and went on to solve the problem, knew exactly how to... and made a couple of stupid pass-to-the-other-side-and-sum errors. Here's the same reasoning with the correct values: f'(x) = 2x Tangent at point (k, k²): y = d + 2kx Finding d, substituting (k, k²): k² = d + 2k² => d = -k² Generic tangent: y = 2kx - k² Finding k, substituting (1, -1): -1 = 2k - k² k² - 2k - 1 = 0 => Quadratic formula => k = (2 +- sqr(4 + 4))/2 => k = 1 + sqr(2) or k = 1 - sqr(2) And k² = 3 + 2sqr(2) or 3 - 2sqr(2), respectively. Tangents: y = (2+2sqr(2))x - (3+2sqr(2)) y = (2-2sqr(2))x - (3-2sqr(2))

[@an-bk5vx]

u can also jus consider all straight lines passing thorough (1,-1) which are of the form y=m(x-1)-1 then notice all the tangent lines must be included in these lines. equate this to x^2 then notice gradient must be the same therefore m=2x so yh same thing from there

[@dalitlegreenfuzzyman]

No calculus needed. We know our two lines are of the form y=mx-m-1 for some slope m. AND we know that we want them to cross y=x^2 at one distinct point due to tangency. Hence this simply amounts to letting the discriminant of the equation x^2=mx-m-1 be equal to 0 and solving for m. This gives us m=2+/-2root2.

[@mrhk57]

Can't believe I came up with exactly the same solution as yours.

[@Etothe2iPi]

You don't need calculus to solve this problem. Just solve the intersection problem by setting the discriminant to zero. I find it easiest if you shift the point to the origin. The lines become y=mx and f(x) becomes (x+1)^2+1.

[@malignusvonbottershnike563]

I realised after solving it using calculus that you actually don't even need to use calculus to solve this problem! The general equation of a line passing through (1,-1) is y = mx - (m+1), and we want that line to intersect y = x^2 exactly once for it to be a tangent. So, set x^2 = mx - (m+1), which leads to x^2 - mx + (m+1) = 0, and because that equation only has one solution, the discriminant must be equal to 0. That gives us (-m)^2 - 4(1)(m+1) = m^2 - 4m - 4 = 0, which gives us the correct solutions for the gradient for us to just plug back in! I dunno, that made me happy when I realised you could do it that way haha

[@jamesmelendez9971]

This is a great alternate way to solve .vs. The analytic geometry way. Always loved Dover books

[@jamesmelendez9971]

I'm toying with the idea of writing a book that will explain how to solve all kinds of math problems without first having to memorize all that rigmerole/drudgery!

[@sylowlover]

A simpler way to write the solution is y=ax-a-1 for a=2±2√2, looks a little cleaner.

[@malawigw]

This is a quite simple problem, I give it on regular exams for my 17y old students.

[@rnistuk]

Cool! Bought the book! Sadly, I got the electronic version, so it doesn't smell quite so nice.

[@playerguy2]

While I know very little about the US education system, I figured this shouldn't have been *too* difficult, granted I'm at a bit higher level than this was meant for. my method was to 'simply' write down the functions I know at the outset, then substitute and simplify my way to the solution: parabola defined by `f` where f(x) = x^2 tangent `t` being t(x) = kx+m and just started plugging in values conventions used were x and y denote any pair of values forming a point (x,y) on the graph and (x_t, y_t) being the point where the tangent t meets the parabola. so we know the tangent passes through (1,-1), meaning t(1) = -1, or k*1 + m = -1 we also know the tangent meets the parabola at (x_t, y_t) t(x_t) = f(x_t) k*x_t + m = x_t^2 we also know the tangent's slope is the same as the parabola at (x_t, y_t) k = f'(x_t) k = 2x_t everything is set up, now we start substitution and simplification k*x_t + m = x_t^2 k = 2x_t thus 2x_t*x_t + m = x_t^2 m = -x_t^2 we have the equations for k and m, so our tangent is defined by t(x) = kx+m k = 2x_t m = -x_t^2 t(x) = (2x_t)x + (-x_t^2) the tangent passes through (1, -1), so t(1) = -1 k*1 + m = -1 (2x_t)1 + (-x_t^2) = -1 0 = -x_t^2 + 2x_t + 1 solving the quadratic results in x_t = 1 + sqrt(2) or 1 - sqrt(2) FINALLY plug this into t(x) = (2x_t)x + (-x_t^2) we get t(x) = (2 + 2*sqrt(2))x - (3 + 2*sqrt(2)) or (2 - 2*sqrt(2))x - (3 - 2*sqrt(2)) While no one should trust any link on the internet, here's how that looks on a graph: www.desmos.com/calculator/ompox1udzz

[@mismis3153]

In my school system, we learnt that the tangent equation at a is : y = f'(a)(x-a) + f(a) Just this makes all these kinds of problems trivial to solve IMO

[@JNCressey]

I did it a different way: Say a tangent touches y=f(x) at x=t. That tangent will cross the line x=1 at some y-coordinate =g(t). For such a tangent to pass through the point (1,-1), it would mean t is a solution to g(t)=-1. So we aim to find an expression for g(t) and solve g(t)=-1. We get a function for g(t) like this: g(t) = (the height of y=f(x) at x=t) + (the "rise" along the tangent for the "run" from x=t to x=1) = f(t) + (1-t)*m(t), where m(t) is the gradient of the tangent. The gradient of the tangent is the derivative of f(x) at x=t, so m(t) = f'(t) = 2t. Subbing in f(t) and m(t), we now get g(t) in terms of only t: g(t) = t²+(1-t)*2t. Solve g(t)=-1 t²+(1-t)*2t=-1 ... t = 1 ± sqrt(2) We can use our solutions for t to get values for the gradients: m(t) = 2*(1 ± sqrt(2)). We can use the point-gradient form of the line equation with the m(t) solutions and the point (1,-1) to get two lines: [ y+1 = 2*(1 + sqrt(2))*(x-1) ] and [ y+1 = 2*(1 - sqrt(2))*(x-1) ]

[@koenth2359]

Let's find the equation for the tangent line that touches the parabola at point (t, t²). A (non-vertical) line has to be of the form y = ax+b. The derivative of t² is 2t which tells us the slope, so we know: a = 2t. In order for ax+b to pass through (t, t²) we set x=t, y=t² and plug that into y=ax+b: t² = at+b Since a is known in terms of t, we can solve for b: b = t² - at = t² - 2t² = -t² b = -t² Now that a and b are expressed in t, we have our tangent line expressed in terms of t: y = 2tx - t² In order for the line to pass through (1,-1) we set x=1 and y=-1 to get -1 = 2t•1 - t² which is rewritten as t² -2t - 1 = 0 Solving this for t gives t = 1 ± √2. Plug the solutions for t back into the line equation to get y = 2(1 ± √2) x - (1 ± √2)² Simplifying gives : line 1 : y = (2 + 2√2) x - 3 - 2√2 line 2: y = (2 - 2√2) x - 3 + 2√2

[@koenth2359]

If we'd replace the point (1,-1) by an arbitrary point (X, Y), our quadratic has (a=1, b=-2X, c= Y) and solves as t = 2X ± √( X^2-Y)² and the discriminant X^2-Y very neatly shows that any point below the parabola has two tangent lines going through it, points above the parabola have none, and points on the parabola have, surprise, one tangent line going through them.

[@nikos4677]

The difficult part of this problem is the numbers. When i saw these numbers i got scared that I did a mistake

[@stumbling]

Two years without any calc, so I'm pleased with myself that I managed to figure this out. :)

[@SuperMaDBrothers]

Solution: write down the constraints: 1) y = x^2 2) the slopes match, so (the slope of y=x^2) = 2x = (the slope of the line we're making) = (y-(-1))/(x-(1)) Done, problem solved. Constraints aren't used enough to explain math even though they're literally the whole thought process... even in this video you jump into solving before there's a clear picture of what you're doing...

[@seabraalves600]

I recently did a question like this and people are not waiting for this point being the point that the tangent lines cross, they actually confuses each other with the tangency