The Calculus Problem Nobody Could Solve

In this video I go over a book and then do a harder calculus problem. The book is called Essential Calculus with Applications and it was written by Richard Silverman. The calculus problem is one which I put on various exams every semester as extra credit. Nobody was ever able to get it completely correct.

(this is my affiliate link)

0:00 Introduction
3:34 The Problem
12:05 Finishing Up

If you enjoyed this video please consider liking, sharing, and subscribing.

There are several ways that you can help support my channel:)

************Udemy Courses(Please Use These Links If You Sign Up!)*************
Abstract Algebra Course

Advanced Calculus Course

Calculus 1 Course

Calculus 2 Course

Calculus 3 Course

Calculus 1 Lectures with Assignments and a Final Exam

Calculus Integration Insanity

Differential Equations Course

Differential Equations Lectures Course (Includes Assignments + Final Exam)

College Algebra Course

How to Write Proofs with Sets Course

How to Write Proofs with Functions Course

Trigonometry 1 Course

Trigonometry 2 Course

Statistics with StatCrunch Course

Math Graduate Programs, Applying, Advice, Motivation

Daily Devotionals for Motivation with The Math Sorcerer

Thank you:)

[nexusnexus]

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.

[koningen]

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.

[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 :)

[justintroyka]

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.

[pacopascal]

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.

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.

[galzajc]

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)

[valoraz]

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

[JR-ucnk]

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

[fuoridisenno]

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.

[bubbotube]

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

[davidellis]

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.

[donovanthompson]

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

[michaelzumpano]

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 :-)

[martinepstein]

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)

[paul]

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.

[perplexedon]

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.

[williamguest]

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.

[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.

[WzardsChss]

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!