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Current time:0:00Total duration:23:13

Optimization problem: extreme normaline to y=x²

Video transcript

I just got sent this problem and it's a pretty meaty problem a lot harder than what you normally find in most textbooks so I thought it would help us all to work it out and so one of those problems that when you first read it your eyes kind of glaze over but when you understand what they're talking about it's it's reasonably interesting so they say the curve and the figure above is the parabola y is equal to x squared so this curve right there's y is equal to x squared let us define a normal line as a line whose first quadrant intersection with the parabola is perpendicular to the parabola so this is the first quadrant right here and they're saying that a normal line is something when the first quadrant intersection with the parabola is normal to the parabola so if I were to draw a tangent line that right there this line is normal to that tangent line that's all that thinks this is a normal line right there normal normal line fair enough five normal lines are shown in the figure one two three four five but enough and these all look perpendicular or normal to the parabola in the first quadrant intersection so that makes sense for while the x coordinate of the second quadrant intersection of the normal line with the parabola gets smaller as the x coordinate of the first quadrant intersection gets smaller so let's see what happens is the X quadrant of the first intersection gets smaller so this is where I left off in that dense text so if I start at this point right here so my x quad my x coordinate right there would look something like this let me go down my x coordinate is right around there and then as I move to a smaller x coordinate to say this one right here what happened to the normal line or even more important what happened to the intersection of the normal line in the second quadrant this is the second quadrant right here so when I when I had a larger x value here my normal line intersected here in the second quadrant then when I brought my x value in when I lowered my x value my x value here because this is the next point right here my x value or the intersection here went actually there wording is bad they're saying that the second quadrant intersection gets smaller but it's actually it's not really getting smaller it's getting less negative if you had if you actually had to be if I get smaller could be just absolute value or magnitude but it's just getting less negative it's moving in it's moving in there but it's actually becoming a larger number right it's becoming a less negative but a larger number but if we think an absolute value I guess it's getting smaller right as we went from that point to that point as we move the X in for the intersection in the first quadrant the second quadrant intersection also moved in a bit from that line to that line fair enough but eventually a normal line second quadrant intersection gets as small as it can get so if we keep lowering our x value in the first quadrant so we keep on pulling in in the first quadrant as we get to this point and then this point intersects in the second quadrant right there and then if you go even smaller X values in the first quadrant then your normal line starts intersecting in the second quadrant further and further negative numbers so you can kind of view this as the highest the highest value or the smallest absolute value at which the normal line can intersect in the second quadrant let me make clear that up here you are intersecting when you had a large X on the first quadrant you had a large negative X in the second quadrant intersection and then as you lowered your x value here you had a smaller negative value up until you got to this point right here you got this which is kind of you can kind of view it as the smallest negative value you could get and then when you pulled in your X even more these normal lines started to push out again out in the second quadrant that's I think what they're talking about the extreme normal line is shown as a thick line and the figure right this is the extreme normal line right there so this is the extreme one that that deep bold one extreme normal line after this point when you pull in your X values even more the intersection in your second quadrant starts to push out some and if you know you could extreme think of the extreme case if you draw the normal line down here your intersection in the with the second quarter is going to be way out here someplace although it seems like it's kind of asymptoting a little bit but I don't know let's you read the rest of the problem once the normal line passes the extreme normal line the x-coordinates of their second quadrant intersections with the parabola start to increase and they're really when they say they start to increase they're actually just becoming more negative that wording is bad I should change this to more more negative or they're becoming larger negative numbers because once you get below this then all of a sudden the X intersections start to push out more in the second quadrant fair enough the figures show two pairs of normal lines fair enough the two normal lines of a pair have the same second quadrant intersection with the parabola but one is above the extreme normal line in the first quadrant the other is below it right fair enough for example this guy right here this is when we had a large x value he intersects in the second quadrant there then if you lower and lower the x value if you lowered enough you pass the extreme normal line and then you get to this point and then this point he intersects or actually you go to this point so if you pull in your x value enough you once again extreme it you an intersect at that same point in the second quadrant so hopefully I'm making some sense to you as I try to make some sense of this problem okay now what do they want to know I think I only have time for the first part of this maybe I'll do the second part in the another video find the equation of the extreme normal line well it seems very daunting at first but I think our toolkit of derivatives and what we know about how equations of a line should should be able to get us there so what's the quest the what's the slope of the tangent line at any point on this curve well we just take the derivative of y equals x squared and y prime is just equal to 2x this is the slope of the tangent slope of tangent at any point X at X so if I want to know the slope of the tangent at X naught at some particular X I would just say I would say well let me just say slope it would be 2x not or let me just say f of X naught is equal to 2x not this is the slope at any particular X naught of the tangent line now the normal line slope is perpendicular to this so the perpendicular line and reviewed here but the perpendicular line has a negative inverse slope so the slope of slope of normal normal line at X naught will be the negative inverse of this because this is the slope of the tangent line at X naught so it'll be equal to minus 1 over 2x not fair enough now what is the equation of the of the normal line at X naught let's say that this is my X naught in question what is the equation of the normal line there well we can just use the point-slope form of our equation so this point right here will be on the normal line and that's the point X naught X naught squared because this this is the graph of y equals x naught so x squared so on this the this normal line will also have this point so we could say that the equation of the normal line let me write it down equation of the normal line normal line would be equal to this is just a point-slope definition of a line you say Y minus the Y point which is just X naught squared that's that right there is equal to the slope of the normal line minus 1 over 2 X naught times X minus the X point that we're at minus X minus X naught this is the equation of the normal line so let's see and we what we care about is when X naught is greater than 0 right we care about the normal line when we're in the first quadrant where in all of these values right there so that's me my equation of the normal line and let's get it let's solve it explicitly in terms of X so Y is a function of X well if I add X naught squared to both sides I get Y is equal to actually let me multiply this this guy out I get minus 1/2 X naught times X and then I have plus plus because I have a minus times a minus plus 1/2 the X not in the over the X naught they cancel out and then I have to add this X naught to both sides so what I all I did so far this is just this part right there that's this right there and then I have to add this to both sides of the equation so then I have plus X naught squared so this is the equation of the normal line in kind of MX plus B form this is its slope this is the M and then this is its y-intercept right here that's kind of the the B now what do we care about we care about where this thing intersects we care about where it intersects the parabola and the parabola that's pretty straightforward that's just y equals y is equal to x squared so to figure out where they intersect we just have to set the two y's to be equal to each other so they intersect the x-values where they intersect x squared this Y would have to be equal to that Y so we could just substitute this in for that Y so you get x squared is equal to minus 1 over 2 X naught times X plus 1/2 plus X naught X naught squared fair enough and let's put this in kind of a quadratic equation or quadratic try to try to solve this so we can apply the quadratic equation so let's put all of this stuff on the left hand side so you get x squared plus 1 over 2 X naught times X minus all of this 1/2 plus X naught squared is equal to 0 all I did is I took all of this stuff and I put it on the left hand side of the equation now this is just a standard quadratic equation so we can figure out now where the the X values that satisfy this quadratic equation will tell us where our normal line where our normal line and our parabola intersect so let's just apply the quadratic equation here so the potential x-values where they intersect X is equal to my this be I'm just applying the quadratic equation so minus B is minus 1 over 2 X not plus or minus the square root of B squared so that's that squared so it's 1 over 4x not squared minus 4ac so minus 4 times 1 times this minus thing so I'm going to have a minus times a minus is a plus so it's just 4 times this because there was a 1 there so plus 4 times this right here 4 times this is just 2 plus 4 X naught squared all I did is this is for AC right here this is well minus 4ac the minus and the minus canceled out so you have a plus there's a 1 so 4 times C is just 2 plus let me make that clear 2 plus 4x squared I just multiply this by 2 and of course all of this should be over 2 times a a is just 2 there so let's see if I can simplify this remember what we're doing we're just figuring out where we where the normal line and the parabola intersect now what do we get here this this looks like a little hairy beast here let me see if I can simplify this a little bit so let us factor out let's factor out let me rewrite this so this is equal to well I could just divide everything by 1/2 so this is minus 1 minus 1 over 4 X naught I just divided this by 2 plus or minus 1/2 that's just this 1/2 right there times the square root let me see what I can simplify out of here so if i factor out if i factor out a 4 over X naught squared then what is my expression become this term right here this term right here will become an X to the fourth X naught to the fourth plus now what is this term become this term becomes a 1/2 1/2 X naught squared and just to verify this multiply 4 times 1/2 you get 2 and then the X naught square is cancelled out so write this term x that will equal 2 and then you have plus now we factored out a 4 out of this and the X naught squared so plus 1 over 16 let me scroll over a little bit let me scroll this over a bit and you can verify that this works out if you were to multiply this out you should get this business right here if you just multiply this thing out of that I see the the homestretch here because this should actually factor out quite neatly so what is this equal so the intersection of our normal line and our parabola is equal to this minus 1 over 4 X not plus or minus 1/2 times the square root of this business and the square root this thing right here is 4 over X naught squared now what's this this is actually lucky for us a perfect square and I won't go into the details because then the video will get too long but I think you can recognize that this is X naught squared plus 1/4 and if you don't believe me square this thing right here and you'll get this expression right there and luckily enough this is a perfect square so we can actually take the square root of it and so we get the points at which they intersect our normal line and our parabola and this is quite a hairy problem the points where the intersect is minus 1 over 4 X not plus or minus 1/2 times the square root of this the square root of this is the square root of this which is just 2 over X naught times the square root of this which is X naught squared plus 1/4 and if I were to rewrite all of this I get minus 1 over 4 X naught plus let's see this one happened this to cancel out right so these cancel out so plus or minus now I just have a 1 over X naught times X naught squared so I have 1 over X naught oh sorry let me I have we have to be very careful X naught squared divided by X naught is just X naught let me do that in a yellow color because you know what I'm dealing with this term divided more multiplied by this term is just X naught and then you have a plus 1/4 X naught and this is all a parenthesis here so these are the two points at which the normal curve and our parabola intersect let me just be very clear those two points are for if this is my X naught that we're dealing with right there it's this point and this point and we have a plus or minus here so this is going to be the plus version and this is going to be the minus version in fact the plus version should simplify into X naught let's see if that's the case let's see if the plus version actually simplifies to X naught so these are our two points if I take the plus version that should be our first quadrant intersection so X is equal to minus 1/4 X not plus X not plus 1/4 X not and good enough it does actually cancel out that cancels out so X naught is one of the points of intersection which makes complete sense because that's how we even define the problem but so this is the first quadrant intersection so that's first quadrant intersection the second quadrant intersection will be when we take the minus sign right there so X I'll just call it in the second quadrant intersection it'd be equal to minus 1/4 X naught minus this stuff over here - this stuff there so minus X naught minus 1 over 4 minus 1 over 4 X naught now what do we have so let's see we have a minus 1 over 4 X naught minus 1 over 4 X naught so this is equal to minus X naught minus X naught minus 1 over 2 X naught if I take 1 minus 1/4 minus 1/4 I get minus 1/2 and so my second quadrant intersection all this work I did got me this result second quadrant intersection I hope I don't run out of space my second quadrant intersection of the normal line and the parabola is minus X naught minus 1 over 2 X naught now this by itself is a pretty neat result we just got but we're unfortunately not done with the problem because the problem wants us to find that point the maximum point of intersection they call this the extreme normal line the extreme normal line is when our second quadrant intersection essentially achieves a maximum point I know they call it a the smallest point but it's a smallest negative value so it's really a maximum point so how do we figure out that maximum point well we have our second quadrant intersection as a function of our first quadrant X I could rewrite this as my second quadrant intersection as a function of X naught is equal to minus X minus 1 over 2 X naught so this is going to reach a minimum or a maximum point when its derivative is equal to 0 and now this is a very unconventional notation and that's probably the hardest thing about this problem but let's take this derivative with respect to X naught so my second quadrant intersection the derivative of that with respect to X naught is equal to this is pretty straightforward it's equal to minus 1 and then I have a minus 1/2 times this is the same thing as X to the minus 1 so it's minus 1 times X naught to the minus 2 right I could have rewritten this is minus 1/2 times X naught to the minus 1 so you just put its exponent out front and decrement it by 1 and so this is the derivative with respect to my first quadrant intersection so let me simplify this so X my second quadrant intersection the derivative of it with respect to my first quadrant intersection is equal to minus 1 the minus 1/2 and the minus 1 the the minuses can't or they become a positive when you multiply them so plus half taught over x-naught squared now this will reach a maximum minimum when it equals zero so let's set that equal to zero and then solve and then solve this problem right there well we add one to both sides we get 1 over 2 X naught squared is equal to 1 or you could just say that that means that 2 X naught squared must be equal to 1 if we just invert both sides of this equation or we could say that X naught squared X naught squared is equal to 1/2 or if we take the square roots of both sides of that equation we get X naught is equal to 1 over the square root of 2 so we're really really really close now we've just figured out the X naught value that gives us our extreme normal line this value right here let me do it a nice deeper color this value right here that gives us the extreme normal line that over there is X naught is equal to 1 over the square root of 2 now they want us to figure out the equation of the extreme normal line well the equation of the extreme normal line we already figured out right here it's this the equation of the normal line is that thing right there so if we want the equation of the normal line at this extreme point right here the one that creates the extreme normal line I just substitute 1 over square root of 2 and for X naught so what do I get I get and this is the homestretch again this is quite a beast of a problem y minus x naught squared X naught squared is 1/2 right 1 over square root of 2 squared is 1/2 is equal to minus 1 over 2 X naught so let's be careful here so minus 1/2 times one over X not 1 over X naught is the square root of 2 right all of that times X minus X naught so that's 1 over the square root of 2 X naught is 1 over square root of 2 so let's simplify this a little bit so the equation of our normal line assuming I haven't made any careless mistakes is equal to so y minus 1/2 is equal to let's see if we multiply this minus square root of 2 over 2 X and then if I multiply these square root of 2 over this this becomes 1 and then I have a minus and a minus so then I have a plus 1/2 I think that's right yeah plus 1/2 so that's my right this times this times that is equal to plus 1/2 and then we're at the home stretch so we just add 1/2 to both sides of this equation and we get our extreme normal line equation which is y is equal to minus square root of 2 over 2 X if you add 1/2 to both sides of this equation you get +1 and there you go that's the equation of that line they're assuming I haven't made any careless mistakes but even if I have I think you get the idea of hopefully how to do this problem which is quite a beastly one