Which point on the curve has the biggest slope?
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- čas přidán 13. 09. 2024
- Which point on the curve has the biggest slope? This is a calculus 1 optimization problem.
This problem is from Single Variable Calculus by James Stewart,
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I appreciate all of these videos, easy and hard. Been long time since I've graduated and it's good to get some knowledge back.
Not sure how some people think, but please keep in mind that he is doing math videos for all, on various levels. He is not doing these videos just to show off his skills with hard problems. I don't think these videos are for fun (well, they are at some point) but majorly educational. This type of problems which some call "boring" are really some critical basics in analysis, and one cannot read or spell words if one cannot sing or know the ABC.
nimmira thank you!!!
I feel math and watching math videos can be both fun and instructional ... I'm a math mouse and likely the oldest dude watching Black Pen, and watching with relish, too ... After a long, grueling career in software development, it's good to get my brain back and return to my first love, math ...
A curve is steeper when the slope is -5 than when the slope is 2, but -5 < 2 ... Should the slope be the absolute value of the gradient?
Using this method, you should be getting the maximum value that can possibly be attained by S(x). So you will get 2 as the biggest slope.
I agree with Neil. The question asks for the largest slope so the answer should be whatever value is largest regardless of sign. The simple solution is to ask which point(s) have the greatest slope. Then the answers are the ones in the video.
Nope, it seems that you two mix up the 'slope' in mathematics and that in daily usage.
A slope, in mathematics, at point A is just a number that indicates how steep the tangent at A is.
Larger ( > 0) slope indicates that the tangent is steeper and going up. Smaller negative slope indicates that the tangent is steeper and going down.
As this video is about finding the 'largest slope', we are finding the slope of the tangent (y = mx + c) that m is the largest positive real number, not the 'steepest tangent', thus what you mentioned is meaningless in this question.
In fact, even if this question is modified to find the steepest tangent, -5 is not an answer since as x → ±∞, the slope of tangent → - ∞, which means there are no such thing as the steepest tangent in this question.
@@Silver_G You have explained it well, thanks. However I found that steepness, as defined in mathematics, is not the same as the slope or gradient, m, but is |m|. So, where the slope of a given curve is -5 the graph is steeper than where the slope is 2.
@@neilgerace355 Yes, steepness itself is defined as |m| because the sign of m only represents whether the tangent is going upward or downward.
So your statement "curve with slope -5 is steeper than curve with slope 2" is indeed, true, but it's kind of irrelevant to this question since the question only cares about the largest slope, not the steepest tangent, as I mentioned.
One more example, observe this set of slopes, {-5, 2, 4} we know that the 4 > 2 > -5 so 4 is the "largest slope" but |2| < |4| < |-5| so even the tangent with slope 4 has the *largest slope* , this doesn't mean that it is the *steepest tangent* .
Solve all the values of x and y such that 16^(x^2+y)+16^(y^2+x)=1
There are none?
-1/2,-1/2 is one of the solution
Technically, you can choose any value for y or x, and then just solve for the other value because you didn't specify that they had to be real solutions.
I've never taken calculus but I know a bit about derivatives. I knew that I had to differentiate y and I did. But then all I did was write
- 15x^4 + 120x^2 = -15(x^4 - 8x^2)
And then complete the square
-15(x^4 - 8x^2 + 16 - 16) = -15((x^2 - 4)^2 - 16)
For maximum -t we need minimum t, which in this case was equivalent to minimum (x^2 - 4)^2, which is 0. Solving (x^2 - 4)^2 = 0 obviously gives x = +-2, and the slope was also 240. Here you saw a solution from a soon-to-be 10th grader
Wow! Are you in 10th grade now?
From burger King to a mathematician .I salute Ur spirit (wish I knew ur name)
his name is steve (chow?)
Mathedidasko oh come on, you should call me by 1/(1-x)
@@blackpenredpen haha ok
Jacket so hard thank you for the help :)
What if the question was asking for the tangent line with the steepest slope to an ellipse?
😇😇😇 I am blessed now
thank you!
Well this is quite easy. I mean after the headache that were those crazy IIT JEE integrals.
: ))))
I had this example on my exams, and I'm not quite sure if i did it well:
What are
monotonicity of functions:
f(x)= x+3/x+9/x^3+27/x^5...
I figure out, that this function if sum of: (3^n)*(x^(1-2n) but wasn't sure what to do next.
Nicely done! Thank you!
*scratches head* But points don't have a slope.
Please make more videos with that Helium voice😂 I am replaying it over and over again🤣
Abha Vishwakarma lol I am glad that you like it!
And I was going to comment same lol!
Shoot a video about what is t: a ^ b = b ^ a * t
Just to be clear, a positive slope is by definition bigger than a negative slope. Negative numbers are by definition smaller than positive numbers. I fail to see why people think a negative slope can be considered larger than a positive slope here.
Larger in magnitude, not in terms of a number line. Larger can be interpreted as "more massive", which would mean the abs(s(x)). But you are correct. Positive will always be greater than negative.
They mean the absolute value
Thank you so much for the explanation. What is your native language? Curious.
You should check the limits at infinity too, right?
The slope tends to -inf at both ends of the number line thus technically the slope is smallest (most negative) there ...?
S(x)=120x^2-15x^4=-15(x^2-4)^2+240 max S(x)=240 at x=±2
It is interesting that "the largest slope" or "the biggest slope" does not mean the steepest slope, which might also be the most negative. By the way, the plural of "point of inflection/inflexion" is "points of inflection/inflexion".
Julian Locke Not particularly. It makes sense, and it just follows directly from the definition of "greatest". I don't care if |-5| > |2|, because regardless of this, -5 will forever always be less than 2.
@@angelmendez-rivera351 Really I was commenting on the use of language. If I asked you "Which is greater: my $5 debt or your $2 credit?", you might have to think about what I mean by "greater". I suspect that, in a normal conversation, you would probably ask "What do you mean by 'greater'?"
Out of curiosity, what is the percentage of women watching this stuff? I say 75 % men 25 % women
There is no way to know.
0% men
0% women
100% maths lovers
Integrate::((sec^2x)-7)/(sin^7x)
Dont we have to check the x => infinity?
You do, but in this case it goes to negative infinity, which means that it cannot have a greater slope.
@@salwanshathar6753 Yes, and the same as x -> -inf
ignore me, im an idiot, its -3x^5, so it goes form +infinity to -inf, we usualy had the stuff the other way around (with highest exponent first) so i thought it goes from - to + making + infinity the highest slope
@@CubemasterXD yes, that is totally correct because it's a polynomial
This is the biggest slope going up from the left to the right, but not going down.
The 'slope' (a number) here is not the daily 'slope' (a noun, though a number is also a noun lol) we heard.
The slope of a point A is a number that indicates how steep the tangent at A is. Larger slope indicates steeper tangent going up, negative slope indicates steeper tangent going down.
In this video, he just wants to find the points which have a tangent with largest slope.
Therefore, the original statement does not need to mention "going up from left to right" or "going dowb from right to left".
As long as he is finding the points that attain the largest slope, we do not care about whether the tangent is increasing or decreasing.
"biggest slope going up from left to right, but not going down"
you must not understand the definition of biggest. A positive slope is BY DEFINITION bigger than a negative slope.
So I understand why we did the second derivative and found local max and min, but just by looking at the function, it would seem that the slope gets even larger at both negative and positive infinity. This graph is like the function x^3, it gets larger at both ends. Which would make the slope (tangent line) steeper.
Are those just local max values?
when you put infinity into he cubic ,the graph is really negative aka the gradient is concaving down ie the gradient is steep in the negative so is the least slope. however plugging in neg infinity you get a really positive number aka the gradient is concaving up. Now.... this could mean many things, it could mean that the gradient of the original function is positive and is increasing( indicating a max slope), or is really negative and is increasing (not what were looking for) . plug in neg inf into the first derivative and it will be negative indicating it is a really neg gradient getting more positive( not what were looking for)
summary - plug inf and neg inf into 1st derivative of original function. they are both negative so cannot be largest slope
Does this not just solve for the greatest positive slope? Not the greatest absolute value slope?
stevemonkey6666 The problem is not asking for the slope with the greatest absolute value, it is asking for the greatest slope. Big diffrence.
Let's do it with e _ funktion & use of Assymptopics. Highest degree potenz in a Function is =5 & event. 3x^5 . Costant = 1 therefor ..... 3x^5 @ 1( hire i use symbole ".. @..." to show Assymptopic ------> x=0.803
Now y= e^x & vlue 0.803 for y ------> x= inv ln o.803 -----> X = 2.232 . Now dig this this is the meaning of coordinate ( 2, 2) & also 4 more x values( because of x^ 5 ). Therefor the most probeble POINTS are :
P1=(2,2)..p2= ( 2, 23)....p3=( 2, 232) ....p4= (2, 2322).... p5= (2, 23222).
Now which POINT is the answer. The one in the mittel because A = 1/2 ( a+ b ). Therefor the POINT 3( 2 , 232 ).
What is i! and !i
You know what, I hav no clue
Black Pen Red Pen BayBay
godsend
do u tutor tho. my exam is in two days.
Great!
You forgot to check whether the slope gets bigger than the local maxima at infinity or negative infinity. It was kind of obvious, because the left and right critical points are both maxima, but you didn't mention it.
No, look at a graph. Necessarily the slope tends to +/- infinity as you approach those points, and what would you call them? How would you find those points? Any point you choose always has another further away with a greater slope.
Why is it when I graph this I see no slope at x=0 ?
In the video we see that 2, -2 are the inputs for maximun slope. However, 0 is minimun slope. You can check with S(x) that S(0) = 0. So, yeah, you should not see slope at x=0.
jojojorisjhjosef that's because that is a particular case in which something else than normal happens: you can see in your graph that left to 0, the slope is increasing (a positive slope). To the right of 0, you can see that the slope is also positive. Now, what you mean is that the value of the first derivative, which represents the slope, is 0. But that also means that the second derivative will be 0 at x=0. This is because the *rate at which the slope changes* at x=0 is also 0.
Great
Get this guy a proper *Microphone* ...
Can you post the graph of the equation with both points marked. Just curious to see
I came to know your name is 1/1-x
I thought that the second derivative gave the turning points where the slope = 0, eh? I need to re-read my notes before I go back lol
If im not mistaken, second derivative gives concavity and for f"(x)=0 is inflection point if there is change of concavity
Jeff if you want to know where the slope = 0, you just set the value of the first derivative to 0. If you want to know how fast the slope changes, you use the second derivative. The "turning points" about which you're talking are the points at which the change of the slope, computed by the second derivative, go from negative to positive or reversed. That's when you set the second derivative of the slope to 0.
That's true for the original function but he was trying to find the max of the slope so all the work got shifted up one derivative if it asked for the points of inflection of the slope he would need to go to the third derivative
"Faaassitive" 😀😀
Uhhh, the 5th degree polynomial is NOT in standard form, reverse the order of the terms please.
Yes second
Please step up the difficulty. These problems are so boring.
Jan Gjerdum Then don't watch. We don't need ungrateful watchers like you. The videos are aimed for his students above all else. And they are meant to educational rather than entertaining. If you want to have fun, go gambling or something.
2=1+1
Dr.Peyam and bprp ftw!!😂😂😂
5th comment
Sorry but Too easy
Get some hard problems for us please
BTW thanks for all these videos
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