How to find the minimum distance from y=sqrt(x) to (3,0). Calculus 1 optimization

[@tradtrad1000]

As we are adults here we don't need to rationalise the denominator

[@scarbotheblacksheep9520]

That amused me to hear him say that.

[@munendersingh5631]

Fre

[@chessandmathguy]

Doesn't exactly serve any purpose.

[@hectornonayurbusiness2631]

I....... am a man now

[@Rsharlan3]

35 after algebra I, I still have a one word answer to sine π/4: "thuhsquairootuhtuovertoo". I wish you could get teachers like BPRP in HS.

[@agfd5659]

You can do this without the derivative just with a simple thought. At 3:40 you can complete the square with the term underneath the square root and notice that the lowest the whole expression can be is when x=5/2, this operation corresponds to finding the vertex of the parabola of the expression underneath the square root.

[@angelmendez-rivera351]

Agfd That still involves the derivative, since the formula for the minimum of a parabola comes from the derivative.

[@agfd5659]

@@angelmendez-rivera351 no, I just completed the square, that's all you need to do, then you look at the expression and notice that x=5/2 is the lowest value of the expression.

[@Alysio]

Alternative to taking the derivative leveraging properties of quadratics : Since the inside of the sqrt function is a 2nd degree polynomial, we know where it's roots are due to the quadratic formula. Knowing that a parabola has symmetry, you can also always know that the extremum of a y = ax^2 + bx + c parabola is at x = -b/2a. Here a = 1 > 0 and b = -5 immediately tells us that x = 5/2 is the minimum of the inside, and since sqrt is an increasing function, it follows that it's also the point where D is minimal :)

[@Alysio]

And I just saw another comment mentioning that exact method, woops ! ^^'

[@ClimateAdam]

Two suggestions: Personally would suggest finding derivative of D^2 as much simpler and minimum will be same! Or alternatively find the gradient of line from (3,0) to any point in the curve = root(x) /(x-3). The closest approach is when the curve's gradient is perpendicular. This boils down to root(x) /(x-3) = -2 root(x), and we get the same answer. Phew!

[@srpenguinbr]

I remember once I found the minimum of D^2 in order to find a tangent circle to a curve. The center was a given point and the radius was the minimum value of D

[@munendersingh5631]

Fer

[@totoxahc]

I couldn't believe he missed this

[@almightyhydra]

Working with D^2 is a standard trick in situations like this, particularly in computing where you don't want to take a square root as it's imprecise and slow.

[@angelmendez-rivera351]

The concept of gradient is only encountered in vector calculus, so for a differential calculus, your approach is unnatural and far from natural. He made the choice to present this approach for a reason. Questioning it merely misses the point. It is not as if he actually made a mistake.

[@bradenscott5126]

No one is acknowledging that this man has the cleanest outfits of any math genius ever .

[@finlayhutchinson7370]

A much easier way to do this is to get (3-x)^2-sqrt(x)^2 = distance^2 and then complete the square (x-2.5)^2 +2.75 and because where the distance squared is greatest is also where the distance is going to be greatest we can just find the x value of the maximum value of the graph which is obviously 2.5 and then sqrt it for the y value

[@harshitbhalla7416]

I'd say it would be better if we consider another function as f(x)=[D(x)]^2, and perform the second derivative test on f(x) so that the calculation becomes easy. This technique is correct as at the point (x,y) where the distance D will be minimum, the square of this distance will too be minimum! And yes we'll get to the answer real quick. Cheers🍻

[@ezequielcentofanti5856]

Another way of solving is to find a point on the curve where the tangent line at that point is perpendicular to the segment passing through (3;0) and that point. Those lines must be perpendicilar because the tangent to the curve is tangent to the circle centered at (3;0) and radius D (minimum distance), and the segment is a radius of the circle.

[@TheKannanmnj]

I am trying to solve a 3d sphere prob same way, but am not getting

[@axelgilbert7220]

Well if you want to take the derivative of D and not D^2, you get a fraction and you may not say "we just need to worry about the top" You always have to worry about the bottom to be non-zero and what's under the square root to be positive. For instance, if one tries to find the point from y=sqrt(x) that is closest to (1/4;0) using the same technique, one will have a surprise (students, try it )

[@pituitlechat3807]

the min of D is the same as the min D² And is more easy to differentiate D² ! It's directly 2x-5!

[@TJStellmach]

Once you have it in the form D²=(x-5/2)²+11/4 you don't even need calculus.

[@brianjohns49]

A great way to optimize the solution. This kind of thinking is so important nowadays for software development.

[@shivamprakash6398]

For minimum distance (3,0) passes through the normal. We can write equation of normal through curve in polar coordinates. Then we can find the coordinates of the point on curve. This one is easier method

[@10GeorgeJJJ]

Couldn't you just find the minimum point of x^2 - 5x +9 by using the formula xmin =-b/2a? You know the trinominal ia always positive so you know that youll get the right answer

[@lukasmbrito]

Yes that's right, because the square root function is strictly increasing, so if you know the minimum of a function you know the mininum of the square root of this function. In general it can be made for any increasing function and decreasing, althought in the latter the minimum becomes the maximum anf vice-versa.

[@Actanonverba01]

This guy is awesome.

[@blackpenredpen]

Matt Curcio Thanks!

[@CoasterMagicX2]

@ 3:34 you could have set distance equal to a variable (let's call it A) and then square both sides so A^2=x^2 - 5x + 9 And then take the derivative. So 2a (da/dx) = 2x-5 and now plug in da/dx =0. This simplifies the calculations a lot because you don't have to worry about the Chen Lu. The down side to this is you have to implicit.

[@MusicalInquisit]

I don't find either method that difficult.

[@sardarbekomurbekov1030]

Thank you. Chan Lu always saves a day.

[@Reliquancy]

Optimizations involving distances come up so much trying to do practical things but it’s awful the limitations. For example finding the way to put n points on the surface of a sphere so the sum of all the distances between them is maximized is a famous unsolved problem in math and that’s just about the simplest thing you would want to do... I found a continued fraction approximation for distance that you can truncate and it helps but there are caveats to when you can use it. idk

[@seanfraser3125]

It would have been easier to minimize x^2 - 5x + 9 using either calculus or the vertex formula. The x-value that minimizes D^2 is necessarily the same one that minimizes D, since we’re only looking at x positive.

[@coconutflour9868]

Great solution, but I found another neat one! At the point closest to the given point, the line connecting the given point to the point on the curve has to be normal to the curve! So we can find the slope k of the tangent at x0 (the point we're looking for), and -1/k is the slope of the normal, then using the given point find the y-intercept of the normal and you have the equation of the normal! Then just equate that with the equation of the curve and solve for x0 :D

[@wotchadave]

There's actually a second solution given by y=sqrt(5/2), namely y=-sqrt(5/2). Logically this makes sense, since the point is at y=0 and the curve continues into the negative y quadrant.

[@wotchadave]

Unless it's defined purely as the positive square root... which I guess it is, given how the function is drawn

[@anuradhasharma6308]

You are simply the best! Love from India! 😊😍🤗😇

[@flypelican1555]

First!!!! ( Love from India 🤩 )

[@Jozin2626]

This helped me understand optimization better! Thanks

[@emmeeemm]

Like other commenters, I minimized D^2, rather than D itself. f(x)=x^2 is a strictly increasing function when x > 0, so if (D(a))^2 < (D(b))^2, then D(a) < D(b) (distances are only positive, so squares of positives are all we care about). And because I didn't feel like picking arbitrary test values at 3 AM local time, I did a second derivative test instead on D^2. d^2/dx^2(D^2)=2, which is positive, so D^2 is concave up, making x=5/2 a local minimum for D^2 and therefore for D. Why do actual work when the math does itself?

[@angelmendez-rivera351]

Because it is an objectively good habit to be able to do work like that. In the real world, strategies like that will seldom be available.

[@slickomodee]

Since sqrt(x) is strictly increasing it's enough to maximize/minimize what's inside, isn't it. Take derivative of x^2-5x+9 and set zero to obtain 2x-5=0 which yields x=5/2 without using chen lu

[@98danielray]

you dont even need that property. the derivative of (g(f))' will be a quotient multiplied by f', so f' =0 is enough

[@trevorteske9104]

Youre the best! Thanks

[@kinyutaka]

I really need to read up on derivatives. It's the one part of these videos that truly gets me lost.

[@yordyne.menendez7744]

dang homie you wearing some supreme there

[@buraningen9521]

A sum of square is positive, so we know x^2-5X+9 is positive or equal to 0 We have just to find the minimum is (beta,alpha)

[@markcentral]

Please Generalize to finding minimal distance from y=f(x) to point (a,b)

[@SoyFerchow]

Ey, I love you. Saludos desde México

[@husklyman]

What is the closest point to (x,y) from y=f(x)?

[@blank0s162]

the point on y = f(x) closest to the point (h, k) is the point (t, u) f'(t) = (h - t)/(f(t) - k) u = f(t)

[@Riiisuu]

Omer create a function of the distance, which is the the Pythagorean sum of the x and y components (like shown in the video), differentiate your distance sum to find the minimum (which would be the shortest point between the curve and (x, y) coords. Hope this helped

[@marks9618]

Omer (a,b) minimizes distance if a is a zero of -(2(x-a)+2f'(x)(f(x)-b))/(2sqrt((x-a)^2+(f(x)-b)^2)), and f'(x) for all values less than a is a different sign than those greater. The formula is too long to memorize, so just do it by hand, use Chen Lu Edit: grammar

[@Green_Eclipse]

Define: f is the function you want to compare. (Xp,Yp) is the point you are comparing distance to. Therefore: Xp and Yp must be constants. Know: D^2= (X-Xp)^2 + (Y-Yp)^2. D= sqrt( (X-Xp)^2 + (f(x)-Yp)^2 ) Now take d/dx: D'=1/(2D) * (2(X-Xp)(1) + 2(f(x)-Yp)(f'(x)) D'=1/D * (X-Xp + (f(x)-Yp)(f'(x)) Some notes:f is not clearly defined and thus we do not know its derivative and cannot simplify further. However we can note that a minimum distance is when D' = 0. Therefore we can further note that: 0= X-Xp + (f(x)-Yp)(f'(x)) When that is true, that is a possible minimum value of distance but could also be a max or bounce back and be neither. So one should use the numberline check for D'(x) to make sure it changes from negative to positive to ensure it is a minimum.

[@marks9618]

densch123 I would imagine that that Omer was asking about generalizing the result in the video to an arbitrary function, restricting one point to the function's path. (x,y) was a bad choice of notation; (a,b) would be better.

[@darwinschuppan8624]

Can you please do a video on Y=Xth root of X and why it peaks at e

[@blackpenredpen]

Darwin Schuppan You can check out my e^pi vs pi^e video

[@olahalyn4139]

The Xth root of X is the same as X^(1/X). If you take the natural log of both side, then use implicit differentiation. You get dy/dx is equal to 1/X(1-lnX). Setting dy/dx equal to 0 - looking for its maximum point, mean either 1/X is equal to 0(which shows the gradient tending to zero as X tends to infinity), or 1-lnX is equal to zero. Therefore 1 is equal to lnX and e^1 is X. So, the peak is at coordinates: (e, e^(1/e)). Does that answer your question?

[@darwinschuppan8624]

@@olahalyn4139 Yes! Thank you!

[@jakemoll]

There’s another way to do it, by minimising the determinant of the vector from (3,0) to (t^2, t), which is on the curve

[@thestrategict.v.499]

I don't know if you know this but this exact same problem was on the JEE MAIN exam held in India about a week back!

[@blackpenredpen]

The Strategic T.V. Wow!! This is just a HW problem that I assigned to my students.

[@郭典昀]

using tangent line vector perpendicular to distance vector .

[@whichwitch_1]

I know we can use Lagrange multipliers too, then how about you to show us using them? I'm college student, highly interested in multivariate functions and calculus.

[@Veggie13]

The other way I did this was to find the perpendicular to the curve that passed through (3,0). I THINK this is valid (it gives the right answer, anyway). The slope at (x, sqrt(x)) is m = 1/(2 sqrt(x)). The perpendicular is m* = -1/m = -2 sqrt(x). D = (3-x, -sqrt(x)), so m* = -sqrt(x)/(3-x) = -2 sqrt(x). Solution is x=5/2.

[@sireevictineerivero342]

The solution provided is more general. I dont think that works for every case

[@Veggie13]

@@sireevictineerivero342 It's a necessary condition. It's probably not sufficient though. Let our function be y = f(x). Let our point be (X, Y). Let D = sqrt((Y-y)^2 + (X-x)^2). Find 0 = D' = (2(Y-y)(-y') + 2(X-x)(-1))/2D = (-(Y-y)y' - (X-x))/D. Thus y' = -(X-x)/(Y-y) is true at the solution. Let the slope be m = y'(x). Let the perpendicular be m* = -1/m = (Y-y)/(X-x). This shows the perpendicular at (x,y) passes through (X,Y).

[@The_Math_Enthusiast]

I did this by completing the square and it worked equally well.

[@hisxmark]

My first impulse was to rotate the graph pi/2 counter-clockwise, Then find the find the intersection of the line through (0, 3) with the slope of =1/2, then rotate the answer back -pi/2. Then you don't have to deal with 1/(x)^1/2. But then, I'm kind of simple minded.

[@TheKannanmnj]

I didn't get u, 918547238291 can u please watzap dat

[@hisxmark]

@@TheKannanmnj Were you addressing me?

[@TheKannanmnj]

@@hisxmark yes hisxmark

[@hisxmark]

@@TheKannanmnj Let u = (x)^1/2 => v = u^2, then du/dv = -2 and the shortest distance between the point (u =0, v = 3) has the slope -1/2. Solve for the intersection of that line with the parabola, then, simply do the inverse transform. Or, put the problem in the complex plane, multiply by i solve for xi + y find the distance from (0, 3i) and multiply by -i.

[@TheKannanmnj]

@@hisxmark I have a prob can I solve this way

[@emperorpingusmathchannel5365]

Can you do the closest points between 2 curves now?

[@98danielray]

@Nemo Clanc it depends on the second derivative might as well be the furthest or neither

[@angelmendez-rivera351]

Nemo Clanc Not quite. You are supposed to use a distance formula. It is more convoluted than what you stated.

[@fireblastification]

Thank you so much

[@yge1035]

I think you could've also take the drivitive of f(x), and multiply that by the slope of the line that the points create, and that should be equal to -1 solution: [0.5×sqrt(x)]×[(sqrt(x)-0)/(x-3)]=-1 (0.5)/(x-3)=-1 1=-2x+6 x=5/2 :) #yay

[@NonTwinBrothers]

In precal we did this problem using x=-b/2a, hehe

[@Sasquatch4lifeX]

Do more videos with multi variables!!! Calc 3!

[@pachi06]

you don't have to differentiate d(x) because min d comes from min d^2=x^2-5x+9=(x-5/2)^2+r

[@theperpetualprocrastinator9776]

Could you do a video using the calculus of variations? I'm trying to wrap my head around it.

[@johnquest3102]

Me too, I only ever saw one example of it, the tautochrone problem.

[@windwave42]

youre awsome!

[@seangrand3885]

Why did you not find what x can be in the denominator? What if 5/2 makes the square root negative or the inside quadratic function equal to 0?

[@joluju2375]

Of course you can easily verify that 5/2 doesn't lead to problems, but it's useless since 5/2 is the only candidate solution. A problem at 5/2 would mean there is no solution, which is impossible, there has to be a closest point.

[@rogerkearns8094]

Chen we lu it? Yes, we chen!

[@diegobotto6245]

I'm in college i passed calculus 1 a year ago, and i forgot how to do this, i feel so embarassed

[@ulymoni9072]

Great work my guy and Elizabeth doesn’t like you

[@redactdead]

Can't you also take the derivative of D² and find the same minima because √x is an increasing function?

[@cameronkhanpour3002]

Oh lord this is giving me ptsd of related rates and optimization

[@Eternap]

"We are adults now"

[@orlando._.575]

Orthogonal projection 🤔

[@timothyvowell4773]

Got it right 😎

[@KaiKunstmann]

When x -> infinity, then D'(infinity) = 1/(infinity) = (almost) 0; ergo, x=infinity is another solution. right?

[@TJStellmach]

That is essentially a zero derivative _maximum_ value, not a minimum.

[@KaiKunstmann]

I have an error in my limit anyway: forgot the x in the numerator, so when x -> infinity, then D'(infinity) -> 1.

[@angelmendez-rivera351]

x = infinity cannot be a solution to anything because it is not a number.

[@shandyverdyo7688]

*PLEASEEE HELP MEEE I'M DYING* i can't solve integral of 1/(x^8 + 1)

[@angelmendez-rivera351]

Shandy Verdyo You can factor x^8 + 1 using the roots of negative unity. Then multiply together complex conjugate pairs to get 4 irreducible quadratic factors, and then perform a partial fraction decomposition. It becomes awfully easy to integrate each fraction separately, and all you need do is add the integrals.

[@shandyverdyo7688]

@@angelmendez-rivera351 Yeah thanks anyway. But i stuck on PFD cz it has very much decomposition.

[@marco-xe9je]

Wow, this is actually an a clever idea, is it common in calc classes? I'm in high school tho

[@TJStellmach]

Yes, this is a pretty standard calculus approach to finding a minimum value. In this particular case, though, your high school math will do. Since the minimum distance is also the minimum _squared_ distance, you can get this in the form D²=(x-5/2)² + 11/4 and you're home free. The minimum is then obviously where x-5/2=0.

[@marco-xe9je]

@@TJStellmach From where did this D^2 function and that 11/4 came from? And what do you mean by minimum squared distance? Maybe these questions sound a bit trivial, but my math skills and my english aren't that good dude, if this is the case I'm sorry haha

[@atp429]

@@marco-xe9je He basically means that if we just find the minimum of (x^2 - 5x + 9), we'll get the minimum of it's square root too. That D^2 expression comes from a method called completing the square. Basically, we need to write x^2 - 5x + 9 in the form of a squared expression + a constant. Kinda hard to explain how it's done, but if you work out (x-5/2)^2 + 11/4, you'll get x^2 - 5x + 9. Therefore we know that the expression will be minimum when x = 5/2

[@marco-xe9je]

@@atp429 Oh, now I see, thanks man!

[@marco-xe9je]

@@TJStellmach Thank you too bro!

[@danieldyszkant3245]

Very impressive but can you find the formula of the parabola x^2 rotated θ degrees around the origin

[@Veggie13]

Define your parabola as the parametric function (t, t^2). Then apply the 2D rotation matrix [[cosθ, -sinθ],[sinθ, cosθ]] to that vector. This gives a new parametric function (t cosθ - t^2 sinθ, t sinθ + t^2 cosθ).

[@giovanni1946]

Great ! Actually I had this to do as a math exercise 1 week ago with (2, 0) xd

[@Myquerencia]

Solve all the values of x and y such that 16^(x^2+y)+16^(y^2+x)=1

[@aarondevon8144]

Can you tech some basics for ap calc and high Shcool students

[@JamalAhmadMalik]

#yay!😘 I'm a wannabe mathfreak!👌

[@felipevidotti9245]

Why 2X-5=0 ?

[@blackoutemzz2674]

He found the derivative of D(x). And equalled it to zero. So he can find x and y.

[@alpag123]

We have been used to more difficult...

[@kingarth0r]

I like how you say you don't need to rationalize the denominator because all calc students are adults. Meanwhile I'm taking calculus at 14 lol

[@blackpenredpen]

KingArthur Lol, you are a math adult!!!

[@MathNerd1729]

Same here Lol

[@balazsb2040]

I'm glad I'm not the only 14-year old interested in Calculus.

[@plopap]

I’m 7 lol

[@mrrahal7703]

Pleaseeeee do this integral : Integral of 10/(3sinx + 5cosx)^2 ❤❤❤👊

[@blackpenredpen]

MR RAHAL is this another IIT JEE one again?

[@PackSciences]

So here is my attempt. D(x) = |O(x,y) - M(x,y)|_2 = sqrt((xO-xM)^2 + (yO - yM)^2) = sqrt((X-3)^2 + (Y-0)^2). O lies on Y = sqrt(X) D(x) = sqrt((x-3)^2 + X) = sqrt(x^2 - 6x + 9 + x) = sqrt(x^2 - 5x + 9) = sqrt((x-5/2)^2 + 11/4) (x-5/2)^2 + 11/4 is a second-order real-coefficient polynomial with the dominant coefficient being positive, therefore its extremum is a minimum. 11/4 is positive, so (x-5/2)^2 + 11/4 is always positive. So min D(x) = sqrt(min (x-5/2)^2 + 11/4) = sqrt(11/4), reached when x = 5/2 so when y = sqrt(5/2).

[@Foxxey]

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[@aarondevon8144]

Please be my math tutor

[@CaptainJohnPrice11]

i love you

[@MalekiRe]

In the first day Ya!

[@diogofaustino6893]

I love this channel but I simply can't understand why he chooses not to take the (-0) out from one step to the other and then differentiates like a madman. He's either very cautious and slow on the explanation or skips steps.

[@anand.suralkar]

Hey

[@arghyadeepchatterjee6100]

No disrespect but this is not the quickest and easiest way to do this. Think about it in a geometrical way. The point that is closest to (3,0) would lie on the intersection of curve and the normal to the curve passing through the point (3,0) . So let us parametricize and denote a general point on the curve as (t^2,t) keeping in mind that t must be positive. And then just find the equation of normal at the parametric point and make it pass through (3,0) To find the equation of normal at any point we could use differentiation or , use the concept of finding the equation of tangent to any 2nd degree equation and then just finding the equation of normal. Solve it to find t=5/2 See the solution here :- plus.google.com/photos/photo/100774575622085894963/6649612158703815490?authkey=COXa7dfjsL-6lgE

[@TheKannanmnj]

Can u come watzap I have an interesting prob to solve this way 8547238291

[@vondrez]

lol i just did this type of question in class today :L

[@peanut12345]

sqrt 9 = 3 tada

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[@obiwan8972]

51st!!!!