Descartes' method of tangents
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That was arduous. Old RenÃĐ sure loved his quartics ð
Seems like Descartes was sooo close to discovering differential calculus!
So did Fermat, who was arguably even closer! In fact, the two of them fought over who discovered how to find tangents to curves 50 years before Newton and Leibniz did. Another roof has a great video on the topic. th-cam.com/video/xKfEmbWBgvM/w-d-xo.html. Fun fact: the FTC was essentially known by geometric methods before Newton and Leibniz âdiscoveredâ it as well.
Yeah, everyone in that era was clamouring to find out how to do what we call calculus efficiently. Newton and Leibnitz just so happened to have been the ones to come out with the most efficient thorough system so that it became mainstream, including Leibnitz's notation that is now standard (before then, people used all kinds of notations, or even just long verbal constructions. Newton's notation is... interesting as well)
This method is so beautiful and I also appreciate the historical aspect of it. Thank you for sharing it!
Great video. Loved it. Descartes also modified or refined his method to by-pass the circle. It's all in the double root. Late 1630's maybe? For your example:
Center is still C(a,0) with P(3,9)
slope_R = 9/(3-a) ====> slope_T = (a-3)/9
Equation Of Tangent: y - 9 = (a-3)/9 * (x - 3) ===> y = (a-3)/9 *x + (30-a)/3
Intersection of Curve with Tangent: x^2 = (a-3)/9*x + (30-a)/3 ====> x^2 + (3-a)/9 * x + (a-30)/3 must equal x^2 - 6x +9
for a double root at x=3.
Equate co-efficients of x: (3-a)/9 = -6 =====> a = 57
m_t = (57-3)/9 = 6
You can verify the constant term if you want: (57-30)/3 = 9
Easier to force the Tangent to have a double root than the Circle.
=======================================================
Here's a cubic with no circle in sight :Find the slope of the tangent to the curve y = 2x^3 + 4x^2 - 5x + 3 at x = 1 ===> y = 4.
1) The Equation of the tangent: y - 4 = m(x - 1)
2) Intersect with the curve: 2x^3 + 4x^2 - 5x + 3 = mx - m + 4
2x^3 + 4x^2 +x[ -5 -m] + [m-1] = 0 and FORCE a double root at x = 1 (necessary condition for tangency)
3) 2x^3 + 4x^2 +x[ -5 -m] + [m-1] = (x-1)^2*(2x - t)
4)Expand the RHS : 2x^3 + x^2 [-t - 4] + x[2t + 2] - [ t ] and equate coeffiicients.
5) (a) 4 = -t - 4 ===> t = -8
(b) -5 - m = 2t + 2 ====> m = 9
( c) -t = m - 1 not needed, but -8 and 9 verify.
You can also adapt this method for finding Stationary Points. Requires polynomial long division though - but no Calculus.
Thank you for this. Of course it's much easier (and more intuitive) to find the tangent directly (forget the circles) using repeated roots.
correct me if i am wrong but why do we need a double root condition in case of the second example where we are dealing with a cubic curve,i mean the only thing required is that the cubic in step 2 shouldnt have more than 1 distinct real root,that leaves us with just two possibilities,either all of the roots are 1(which in this case is not possible just by looking at the cubic) or it has exacly one real root(which is 1) and the other two being complex conjugate,what am i missing?
@@bhargavsai8014 you need a double root for tangency. A single root would correspond to a secant, i.e. a line simply crossing the cubic on the LHS of equation 2. That would not be a tangent.
You can think of it like this: imagine a line cutting a curve in two points close together. This would be close to being a tangent. Then rotate the line about one of the points of intersection so that the other point of intersection moves towards the fixed point until they coincide. Then we have a tangent and a double point of intersection. This is a double root in the equation that equates the curve to the tangent line.
@@bhargavsai8014 Let's use his simple example : y = x^2 at (3,9). Skipping the details we get m = 6. I don't think he found the equation - just the slope. The equation of the tangent is : y = 6x - 9. Intersect that with y = x^2 and we get : x^2 = 6x-9 ====> x^2 - 6x + 9 = 0 ====> (x-3)^2 = 0, a double root at x = 3. You get away with "set the discriminant = 0" to find m ONLY because the original curve was a quadratic. i.e. x^2 - mx - 3m -9 = 0. B^2-4AC = 0 gives m^2 - 4(3m-9) = 0 and (m-6)^2 = 0.
Think of a cubic. The tangent at some point on the cubic will intersect the cubic at only 2 points, not 3 - yet there are 3 real solutions. Hence the double root. Try this for yourself:
y = x^3 - 5x^2 + 7 @ P(2-5).
Find the equation of the tangent : y = -8x + 11
Intersect the tangent and the cubic: x^3 - 5x^2 + 7 = -8x + 11 and solve this equation. x^3 - 5x^2 + 8x - 4 = 0. You already know that x = 2 MUST be a solution.
Divide by x-2 to get (x-2)(x^2 - 3x +2) = 0 and factor again to get: (x-2)(x -2)(x-1) = 0 and you can see the double root at x = 2.
The tangent only intersects the cubic at (2,-5) and (1,3)
So one last comment on the "circle method". If you try y = x^3 at (2,8) you quickly discover that the y^2 term in the circle equation becomes x^6. That means forcing a double root gets you (x - 2)^2 * [quartic] . No wonder Descartes revised his method. I bet he got a lot of push back on that one.
On the other hand, if you intersect y = x^3 with the tangent you get : x^3 - mx + (2m - 8) = 0 and then force a double root at x = 2: (x - 2)^2*(x - t). You very easily get t = -4 and m = 12.
One of the giant's shoulders that Newton mentions he stood upon was most certainly Descartes ( not to mention Kepler, Galileo, ...). Pretty fine praise from the tallest giant of them all. Thanks very much Michael for such a great classic solution.
Here is another cool method - no calculus. Given f(x) = 5x^4 - 3x^3 + 2x^2 - 4x + 3. Divide by x-2. to get constant remainder R and quotient Q(x). Evaluate Q(2) and note remaiber R. Now use come calculus to find the equation of the tangent at x=2 and voila! Can you prove this always works?
I can see professor's pain when avoiding himself to use derivates (which wasn't invented yet).
Another cool video might be: Fermat's Method for Finding Tangents.
This is just creepy.
I was just thinking about what a version of Taylor series where you approximate the function using circles instead of polynomials, and it reminded me of Descartes's method of finding tangent lines. Now I see a video about it in my feed.
Is Michael stalking me?
I really love this video, keep up the good work sir.
Bruh everything can be bruteforced in coordinates
U could even just use linear equations and some geometry for this
I thought it was some secret method
Turns out it was something SIMILAR to what i did in highschool
I think I imagined that mathematicians before the development of calculus weren't even trying to find tangents to curves, but of course that's too important of a mathematical idea to leave behind. So *of course* they were finding tangents. Never even wondered how they did it until now!
Isn't the hypotenuse the longest side of a right triangle? If so, O-P couldn't be the hypotenuse. Did I hear that wrong?
The hypotenuse is the side opposite the right angle which would be OP if OQ were perpendicular and Q were not P. However due to catchy Schwarz and the parallel postulate the hypotenuse must be sqrt(a^2+b^2) which is greater than a and b but he was using that to show that OP and l are perpendicular.
Question: What is the symbol Michael uses for "at" at 7:49. It looks like a 3 with a dot above it. I've been doing mathematical proofs for over 40 years, and I am not familiar with it.
The use of a symbol there except @ struck me as well, I'm not familiar with it, I thought he was just drawing a semicolon (;) but maybe the intent was something different. I'd love to hear more insights.
I've never seen it before either.
I think it is another form of 'and' &, .. ampersand ... crossed epsilon Google is a friend and so is wiki
en.wikipedia.org/wiki/Ampersand scroll down to Histort section/parahraph
@@Alan-zf2tt I thought of that as a possibility, but he says "at". But even the ampersand would have dots above and below. But so far, that's the best explanation (the only one too).
The quadratic does not need to be irreducible. Try y = x^3-9x^2+24x-3 at x=3. So cool. What is happening at x = 3??
You can tell this is a historical method of solving a problem because it requires knowing half a dozen highly specific Polynomial Facts
I like this.
You sound liked "Dr Peyam" ! ðĪĢ
Descartes should have discovered the derivatives instead ðĪ
Here are my thoughts on why I think the quadratic factor, (x^2+bx+c), in (x-3)^ (x^2+bx+c) is automatically irreducible.
In general a parabola and a circle can have as many as 4 intersection points. But here you are forcing the circle to have its center on x-axis and the parabola to open up or down. Such a circle and such a parabola can intersect in at most 2 points. Also, you are forcing the circle to be tangent to the parabola - hence the double root. So if the quadratic
(x ^2 + bx + c) on the RHS : (x-3)^2*(x^2 + bx + c) was reducible then it would produce 2 more real roots ====> 3 intersection points between the circle (center on the x - axis) and the parabola (opening up). This cannot happen. Therefore x^2 + bx + c = 0 will have no real roots. If the parabola opens right or left then just place the center on the y-axis.
Thoughts?
OP2 + PQ2 = OQ2 !!!!
can you explain how thats the irreducible form of a quadratic?
Because (x-h)^2 is always non-negative and k^2 is always positive (given that kâ 0). Therefore their sum can never be 0, i.e irreducible (by the factor theorem of polynomials)
@@015Fede Had to work that out in my head when he said that, but that is what I came up with also.
You could also use (x-3)^2(x^2+bx+c)
@@ianfowler9340 yes, but once you get solutions you will have to show it is irreducible.
@@ingiford175 I don't think it does needs to be irreducible. You could, for example, find the equation of the tangent at an inflection point - say a cubic at the IP. In such a case you will end up getting a triple root, (x-a)^3 on the RHS. And even if it is irreducible, x^2+bx+c will automatically be irreducible.
Try y = x^3-9x^2+24x-3 at x=3
Why don't you use the Ruffini theorem?
It was discovered in 1799. Descartes died in 1650. He wouldn't have used a technque that was only invented a century and a half after his death.
3:00 shouldn't it be OP^2 + PQ^2 = OQ^2? OP is not the hypotenuse!
Bro both works, did u listen
If OQ were perpendicular to l and P,Q distinct it would be.
@3:38 why does this imply P=Q?
Think of an isosceles triangle ABC where BC lay on the same line and AB=AC yet B does not equal to C
OQ is perpendicular to L by definition of Q, which means that OQ is the smallest distance from any point on L to O. I believe the existence and uniqueness of such a point is either an axiom of Euclidean geometry or a definition or a trivial result. In any case, this means that if another point P is on L such that OP=OQ then P=Q. I don't think this justification makes the argument circular or contradictory.
@@aniruddhvasishta8334 gotcha thanks
But how we know such a Q exists?
This just shows if thereâs such a Q then P=Q. But we donât a priori know that a tangent hits the circle at a perpendicular point, just that any such point on said line would be on the circle but how do we know there are any such Qâs?
Not from this argument it seems
You're right. Here's a fix: OQ^2 + PQ^2 âĪ OQ^2 as in the video, which implies PQ^2 âĪ 0 -> PQ = 0 -> P = Q
@@aadfg0
Perfect thank you. I think this should have been mentioned.
In any case thereâs still the issue of showing such a Q exists. He showed uniqueness, but not existence.
@@Happy_Abe Q is defined as the point such that OQ is perpendicular to l. Do you wish for a proof that perpendiculars exist? There many at various levels depending on what you're willing to accept.
What!
1,4,9 = 3,5,4, while 19 = (a,0 & 3,9) , 1/6= (0,54) gremlins
Noice
The proof that the radius is perpendicular to the tangent is flawed. In particular, OP = OQ does not (by itself) imply P = Q. (They could be sides of an Isosceles triangle, so you have to bring in the fact that the base angles would then both be 90°, meaning that the triangle was degenerate.) A simpler argument is to go back to the inequality OQ^2 + PQ^2 âĪ OQ^2, hence PQ^2 âĪ 0, from which it follows that P = Q since PQ^2 âĨ 0.
Going through all that for such a simple problem makes me really appreciate differential calculus. Imagine if we could only calculate slopes at points, and every time we had to do all of that!
For the radius being perpendicular to the circle I think a proof by contradiction would have been easier to follow.
Suppose OP is not perpendicular to the tangent. Let Q â P be such that OQ is perpendicular to the tangent. That makes OQ the shortest distance from O to the tangent and therefore OQ < OP. But OP is the hypotenuse of the right triangle OPQ and therefor OP > OQ. By contradiction then the assumption that OP was not perpendicular to the tangent must have been wrong.
The fact the y-co-ord of the tangent circle is zero comes from a symmetry argument?
Na, there are infinitely many tangent circles. He just chose one centered at y=0 for simplicity
@@015Fede (duh) Of course! Any circle centred on the normal to the tangent would do; Penn just makes use of the fact that the normal is not parallel to y = 0; i.e., that the tangent is not vertical
Can you explain why a symmetry argument is insufficient?
I think he made a mistake. OP^2+PQ^2=OQ^2.
He made no mistake.
Now generalise it .... see you next year.
OQ^2+QP^2 = OP^2 REALLY? A small mishap. You are actually very good and relaxing.
@@JustNow42 this is in fact correct, he's assuming that the angle in Q is right to prove that Q=P
I had the same thought initially
Were you listening? Or you just saw the equation and jumped to comment?
hes setting up a contradiction
assuming OQ perpendicular to l and Q not P its correct. Which is the point as since Q is outside the circle OQ is bigger than OP a contradiction with pythagoras. we can go one of four ways deny Pythagoras and thus parallel postulate ,two deny P as a point of tangency deny OQ perpendicular to l or deny Q and P distinct. All three alternatives are less attractive then P=Q so we choose P=Q
Euclid, Descartes, Newton and Einstein:
0D (indivisible, subatomic, non-natural): not fundamental
1D, 2D, 3D and 4D (divisible, atomic, natural): fundamental
Leibniz:
0D (indivisible, subatomic, non-natural): fundamental
1D, 2D, 3D and 4D (divisible, atomic, natural): not fundamental
Can someone explain to me how stuff that can be divided further is fundamental and stuff that cannot be divided further is not fundamental?
In geometry any new dimension has to contain within it all previous dimensions.
Kinda seems like Euclid, Descartes, Newton and Einstein sucked at geometry.
The notion of things being "fundamental" is an artifact of the axiomatic method. You need to pick somewhere to start in math, but often you could have started somewhere else and gotten to the same place.
It's like a choosing an origin and basis. You could have picked another one and gotten the same result. None of them is the "true" one, they all work.
2:16 ??? The right angle is at the point P, because the line l is a tangent to the radius OP => OQ must be the hypontenuse !???
Nice try, LLM
â@@wolliwolfsen291 He picked a point Q on the line such that there is a right angle between the segment OQ and line L. The proof is trying to show that doing so leads to the conclusion that point Q must be point P, showing that the radius at that point is perpendicular to the tangent line.
@strangeWaters
Yeah fundamental and not fundamental aren't the current scientific lingo, it's locally real and not locally real.
They mean the same thing though. Leibniz said necessary and contingent. Different ways of talking about the same thing.
It's all in reference to 0D and non-zero dimensions. Non-natural and natural, quantum and classical.
We already proved the observable natural dimensions (1D-4D) are not locally real so it's contradictory that two years after that Nobel Prize we're still also calling the observable universe fundamental.
Kinda crazy it took a Nobel Prize to prove stuff that can be divided further (natural, divisible stuff) is not fundamental. We really aren't that advanced. It's unfortunate how we learn about the fundamental subatomic stuff last.