Recent content by Poetria

"The trajectory of a particle is a semi-circle contained in the x≤0 half-plane." - this one is wrong. There were several options in this problem. The right one: The trajectory of a particle is a semi-circle contained in the x≥0 half-plane. Thank you very much. I know this is easy but I was...

Great, I understand everything now. Many thanks. 😍

Ok, I got it. Many thanks. But why is my answer wrong then? :rolleyes:

The trajectory of a particle is a semi-circle contained in the x≤0 half-plane. Well, this is somewhat weird. I have come across examples with x(t)=cos(t), y(t)=sin(t) and not the other way round. By the way, my answer is wrong but I don't know why. This is probably silly. :(

You may want to benefit from licentia poetica. :) A long time ago I have written a post about Émilie du Châtelet: https://blogs.otago.ac.nz/emxphi/emilie-du-chatelet-and-experimental-philosophy-i/ Elinor is my avatar's name in Second Life.

Perhaps a gradient would be also a good idea: Gradient ##\vec (0.21, 0.105171)## A tangent plane: 0.21*x + 0.105171*y-0.32 Slope of the gradient: 0.500814

I got: -10.3022 + 13.3022*x = y x=1.1 Approximation of the change: 4.33022-3=1.33022 A beautiful metaphor: ##t\longmapsto f(t)## is the path, and you should plant the flowers at ##t=1## and ##t=1.1.##

Many thanks. But I can't read this: "The linear approximation is the secant through these points: You can't use 'macro parameter character #' in math mode ? A nice quote from Leibniz. :) Émilie du Châtelet (my avatar) was Leibnizian. :)

##f_x=3*x^2+y## ##f_y=2*y+x## ##(3*(t^2)^2+e^{t-1})*2*t+(2*e^{t-1}+t^2)*e^{t-1}## Well, I am not sure how to evaluate it. I got a wrong result by multiplying by 0.1, i.e. ##((3*(t^2)^2+e^{t-1})*2*t+(2*e^{t-1}+t^2)*e^{t-1})*0.1## I guess it is trivial but I am lost. :(

Anyway I got it right. :)

Or is my question silly? :(

Wow. :) I got it. :) Wonderful. Many thanks :)

Well, I have studied the definition of the orthogonal matrix: https://mathworld.wolfram.com/OrthogonalMatrix.html "The rows of an orthogonal matrix are an orthonormal basis. That is, each row has length one, and are mutually perpendicular." There is a constraint: -1/sqrt(2) in the left top...

I think so. I will get to the bottom of it sooner or later. :)

Sorry, I should have sent an image.