How to find the maximum curvature of y=e^x

Рет қаралды 13,312

Ай бұрын

Calculus tutorial on finding the maximum curvature of y=e^x.
Here's how to get the curvature formula: • A faster way to derive...
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Пікірлер: 37

@user-fp5qd7fi6w Ай бұрын

Bro's life flashed before his eyes at 3:47 😂

@daniel_77. Ай бұрын

U had a existential crisis at 3:47

@shruggzdastr8-facedclown Ай бұрын

I thought my computer had frozen or my WiFi connection had crashed for a moment there

@niom9446 Ай бұрын

3:48 I thought my computer froze 😂

@Titurel Ай бұрын

Me too

@mikehood6745 Ай бұрын

Made me smile at 9:18. Generally BPRP steps through the algebra, writing it out line by line. Then BLAM! "We have to get rid of the 2, then the e, then the 2, so x is equal to that," in one fell swoop 😲.

@cdkw2 Ай бұрын

What about the oscillating trig function? They might be fun to get curvature of!

@Kettwiesel25 Ай бұрын

Pretty boring obviously infinite

@zaeemameer8701 Ай бұрын

Why not take it a step further and evaluate the curvature ofthe weierstrass function

@jesuspinales8738 Ай бұрын

Just restrict the domain to [0,2pi]

@michaelz2270 Ай бұрын

If you want to simplify the algebra... it suffices to find the minimum of κ^2, given by e^(2x)/(1 + e^(2x))^3. Letting u = e^(2x) you are minimizing u/(1 + u)^3 = 1/(1 + u)^3 - 1/(1 + u)^2. The derivative of this is -3/(1 + u)^4 + 2/(1 + u)^3, which equals zero when -3 + 2(1 + u) = 0 or u = 1/2. Then κ^2 = (1/2) / (1 + 1/2)^3 = (1/2)*(8/27) = 4/27, so that κ is the square root of this or 2/(3 sqrt(3)).

@ianfowler9340 Ай бұрын

Try my differential equation which is a necessary condition. See previous video. y ' ' ' [1 + (y ')^2 ] - 3 y' (y' ' )^2 = 0 . Once you have this general condition for any y = f(x) , it is very powerful !! In our case: where y = y' = y ' ' = y ' ' ' = e^x Substituting gives: e^x[ 1 + e^(2x) ] -3 [ e^x][e^(2x) ] = 0 and divide through by e^x 0 1 + e^(2x) - 3e^(2x) = 0 1 - 2e^(2x) = 0 etc . . . . Note that -ln(2)/2 = ln[1/sqrt(2)] You can also find the y-value at that point to get y = 1/sqrt(2). Point is: (ln(1/sqrt(2) , 1/sqrt(2) ) which makes perfect sense for y = e^x. Who would have guessed sqrt(2)? Didn't see that coming when I started. One last thing: If i were to make an educated guess at the start I might have chosen (0.1) - but no cigar. Close though.

@romanvolotov Ай бұрын

the reasoning at 7:55 is slightly wrong, should've said that both are greater than zero for every x. based on what you've said one one them can be >0 and the other one

@vinuthomas7193 Ай бұрын

e^(-2x/3) is always > 0 and so is e^(4x/3)

@romanvolotov Ай бұрын

@@vinuthomas7193 ikr. but bprp didn't say it

@vinuthomas7193 Ай бұрын

Oh I see what you mean

@fzlagges5849 Ай бұрын

I was able to solve it by using v=e^x in the original K, and it simplified both the derivative and the solving parting.

@joe4582 Ай бұрын

We did have a similar exercise in with we needed to find the smallest line form a point on a e Funktion and its solution of the tangensequtaion

@MrKoteha Ай бұрын

I've tried doing it with other functions and it's interesting that for the natural logarithm the max curvature turns out to be the same as for e^x (Although with a minus if you don't take the absolute value) Also I noticed that the integral from -∞ to +∞ of the curvature function for y = e^x is equal to 1

@gerryiles3925 Ай бұрын

Not that interesting considering they are inverse functions, so are reflections in the line y = x. They are the "same" curve so must have the same, but strictly negative, max curvature...

@tobybartels8426 Ай бұрын

Whenever you're dividing by a power or a root, it's simpler to change that to a negative exponent and use the Product Rule, than to use the Quotient Rule.

@DEYGAMEDU Ай бұрын

8:00 due to chain rle there will be an e^x multiplied at last

@yassiryassir-rp4to 25 күн бұрын

what is y = lnx

@vinuthomas7193 Ай бұрын

When would k be used instead of y''? And why is that the formula for k?

@ianfowler9340 Ай бұрын

See previous video for the development of k = y' '/ [ 1 + (y ')^2] ^ (3/2)

@bobh6728 Ай бұрын

K, the curvature, not only depends on how fast the first derivative is changing, which is the second derivative, but also the current value of the first derivative at that point.

@ianfowler9340 Ай бұрын

@@bobh6728 Yes. Because the "per unit arc length" depends on the value of the first derivative.

@kingforgotten9090 Ай бұрын

I might be wrong but I think when you were taking the derivative of the function k you did the derivatives of e to the powers wrong where they should be -2/3e^(-5/3) and 4/3e^(1/3) so you brought the powers down correctly for the power rule just didn't subtract 1 from the exponent.

@wambleeska1035 Ай бұрын

You can’t use the power rule to differentiate exponential functions. For the power rule you have the variable, say x, raised to a power so x^2 or x^(6/7) etc… e^x is an exponential function where the variable is in the exponent. For these functions you can’t use the power rule to differentiate. The derivative of the function e^x is special because it is actually the same as the function itself e^x - in fact this is similar for all exponential functions where the derivative of the function is proportional to the function itself. It just so happens that in the case of e^x, the constant of proportionality is equal to one so the derivative of e^x is e^x. If we look at an example from the video, y = e^(-2/3 x), we can use the chain rule to differentiate it. If we let u = -2/3x, then the derivative du/dx = -2/3. The original function y=e^(-2/3x) can now be written as y=e^u and the derivative of y with respect to u is just e^u (remember the derivative of e^u is just itself). Using the chain rule now we see that the derivative will be equal to dy/du * du/dx = e^u * -2/3. Now we can replace any u with the original substitution -2/3x to find that the derivative is -2/3 e^(-2/3x). A quick way to differentiate e to some power is just to bring the derivative of the exponent in front of the function and then just leave the original unchanged. So for the other example e^(4/3 x) we see the derivative of the exponent is just 4/3 so we can bring this down to the front and then leave the rest of the function unchanged: 4/3 e^(4/3x).

@kingforgotten9090 Ай бұрын

@wambleeska1035 you're right, I was thinking of e being the variable not x. However thank you for explaining my mistake.

@tobybartels8426 Ай бұрын

9:58 : You don't need to do a first-derivative test, since you already verified that the limit in either direction is 0, and the value here is larger than that.

@joeythreeclubs Ай бұрын

Please do tan(x) next please, I couldn't figure it out.

@yassiryassir-rp4to Ай бұрын

im like a VIP audience

@yassiryassir-rp4to Ай бұрын

9 june 2024 ???

@fsisrael9224 Ай бұрын

use the Chen Lu!

@abacaabaca8131 Ай бұрын

How do you prove : taking 1 part from a one-third part is equal to 3 ? i.e 1/(1/3)= 3 And also why (a.b)^n = a^n . b^n ? because (2x3)² != 2²x 3² .

@F_A_F123 Ай бұрын

1) x = 1/(1/3) x * (1/3) = 1 (x * 1/3) * 3 = (1) * 3 x * (1/3 * 3) = 3 x * 1 = 3 x = 3 2) I don't know how to prove that, it's probably hard. But (2 * 3)² DOES equal 2² * 3²: (2 * 3)² = 6² = 36 2² * 3² = 4 * 9