Calculus - Jacobians

[weak turtle]

in b), i'm supposed to show that x = x(z) and y = y(z) exists in a limited section, which i did by using jacobians

in c), i'm supposed to calculate (x_z)'(2) and (y_z)'(2) either by using the implicit function theroem or switching x and y to x(z) and y(z) in the equation system, and thereafter plugging in z = -2

i tried the second method, but i must have done it wrong because i can't seem to get anywhere

help is appreciated

[timid tiger]

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[rugged trail]

Do you know the values of x(-2) and y(-2) in that region?

Judging from the first question it seems like you do

Because (4, -2, -2) is in your set of interest

So x(-2) = 4 and y(-2) = -2

Then finding the derivatives is just solving an affine system

[weak turtle]

???

[rugged trail]

I mean think about it

You show that within the region, which is a neighborhood of (4, -2, -2), x and y are uniquely determined given z

[weak turtle]

yes

how the heck did you arrive at x(-2) = 4 and y(-2) = -2

[rugged trail]

Now I take the point (4, -2, -2) itself

The point fits the bill

And by unicity it is the only one

[weak turtle]

y(-2) = -2, what happened here

[rugged trail]

Uniqueness sorry

[weak turtle]

you are inserting -2 as z

[rugged trail]

(x=4, y=-2, z=-2)

Which is obviously in its own neighborhood

Let me try to reformulate it in other words

(4, -2, -2) is in your set of interest S

So you know that (4, -2) is at least a candidate for (x(-2), y(-2))

Ok so far?

[weak turtle]

they are output values, possible ones

for those input values of z

[rugged trail]

Yes

But now you use question 1

This shows that a candidate is unique

So it is one, and the only candidate

[weak turtle]

don't know what you mean here, what are you referring to

"use question 1"

[rugged trail]

Your answer to question 1

[weak turtle]

question 1 i don't need broski

i proved question b) by using jacobian

i need help with c)

[rugged trail]

Within a neighborhood of (4, -2,-2) x and y are uniquely determined with z

Yeah sorry question b

My bad

[weak turtle]

no need for b)

c) is where i need help

[rugged trail]

Well you need to reuse the result from question b

[weak turtle]

b) is only about proving that x(z) and y(z) exists

[rugged trail]

Yes, so x and y are unique wrt z

In a region R

[weak turtle]

"show that [...]"

[rugged trail]

Which includes (4, -2, -2) itself

[weak turtle]

i didn't get anything from b)

[rugged trail]

You did though

[weak turtle]

x(-2) = 4

[rugged trail]

I see an answer to question b already

[weak turtle]

y(-2) = -2

[rugged trail]

Yeah exactly

Because if you set z, you only have one solution for x and y

And now, we found a solution (4, -2), which means it is the only one

Therefore you must have (x(-2), y(-2)) = (4, -2)

It cannot be anything else, otherwise it would not be unique

Is it more or less clear? Or should I try explaining in another way

[weak turtle]

i understand so far

c) is not answered though

[rugged trail]

That is good

All you gotta do is replace the values of x(-2) and y(-2) in your system

Then it is an affine system with 2 unknowns and 2 equations

You can solve it

[weak turtle]

how would there be two unknowns if we plug those in

[rugged trail]

Because your unknowns are x'(-2) and y'(-2)

You don't know the derivatives yet

[weak turtle]

which system are you talking about

[rugged trail]

And you want to find that out

[weak turtle]

the last one that i arrived at?

[rugged trail]

Yep

I hope that my explanation for why this approach is valid makes sense to you

Since you need to justify why you know x(-2) and y(-2)

[weak turtle]

i ended up with the same answer as i got with another method

all i did was plug in z = -2 everywhere

in z and also in x(z) and y(z)

[rugged trail]

Well that is a good sign

Isn't it?

[weak turtle]

should be

[rugged trail]

Cool

[weak turtle]

thanks man

[rugged trail]

Np man

[weak turtle]

i got related questions - if i struggle, i'll send here if you don't mind @rugged trail

same question is a to e i believe

[rugged trail]

Np just ping me. I will answer whenever available

As in ping me here

[weak turtle]

👌 will do

[weak turtle]

d) determine the tangent line @rugged trail

"hint: above means that we can use z = t to parameterize gamma close to (4, -2, -2). velocity vector at t = -2 for this parameterization gives the direction of the tangent line"

[rugged trail]

Ah ok

Can't you reuse the previous result here?

Or I guess not so much

X(t) = x(Z(t))
Y(t) = y(Z(t))
Z(t) = t

So with P = (X, Y, Z) the position vector, we want dP/dt the velocity vector

We could try the derivative of each component

dX/dt(t) = dZ/dt(t) * dx/dz(Z(t))
dY/dt(t) = dZ/dt(t) * dy/dz(Z(t))
dZ/dt(t) = 1

but if Z(t) = -2 this computation is nearly trivial

[weak turtle]

my first idea was the tangent line equation?

[rugged trail]

What does that give?

[weak turtle]

not sure, i'm trying right now

not sure what it is that's supposed to be parametricized

[weak turtle]

@rugged trailgamma is the intersection between the two equations given initially

the intersection obviously yields a circle

it should have the radius 576 and it is a bit tilted

due to the plane being tilted

[rugged trail]

Yep, I can visualize that

the intersection of a sphere and a hyperplane yields a circle

[weak turtle]

yeah

but that tangent line at that point

is the mystery right now

[rugged trail]

why is it?

you can use the circle equation

this hyperplane is but a 2D plane, so you could attempt to change basis

and take an orthonormal basis where 2 axes are on the hyperplane

then the circle equation is essentially just u² + v² = R²

and you've just got to express the tangent point in terms of u, v, w instead of x, y, z

it's an idea

[weak turtle]

but surely, they want it expressed in the reference cartesian system

the one we are using

x, y, z

[rugged trail]

yeah, just change base back once you're done with computations

it's just a linear, and invertible transform

but this approach sounds super tedious

[weak turtle]

it is

don't think this is the way

[rugged trail]

I think my approach with P(t) was a bit better

because dP/dt(t) is colinear with the tangent vector

[weak turtle]

familiar with

l(t) = r(t_0) + r'(t_0) * (t - t_0)

?

[rugged trail]

you can think of P(t) as the position of a ball that runs at constant speed in said circle

[rugged trail]

yeah but r'(t_0) is just my dP/dt(t_0)

isn't it

or well, more generally, what you call r(t) is what I call P(t)

so yes that approach is equivalent

[weak turtle]

if that's true then z'(t) = 0, no

[rugged trail]

to avoid confusion, I denoted them X, Y, Z instead of x, y, z

[weak turtle]

yes

[rugged trail]

where x is a function of z

and X is a function of t

[weak turtle]

and P is just position vector

first term

r(t_0)

[rugged trail]

Then we don't really define z'(t), but Z'(t) = 1

because Z(t) = t

(using the hint)

[weak turtle]

i mean

[weak turtle]

what is missing then

[rugged trail]

Well, with my formulation, the problem is pretty trivial

[weak turtle]

we have P, dP/dt

[rugged trail]

Here

[rugged trail]

Yeah and that's enough

you can compute the tangent line

[weak turtle]

how

[rugged trail]

dX/dt(t) = dZ/dt(t) * dx/dz(Z(t))
dY/dt(t) = dZ/dt(t) * dy/dz(Z(t))
dZ/dt(t) = 1

X(t) = x(Z(t))
Y(t) = y(Z(t))
Z(t) = t

Very important: X is a function of t (time), but x is a function of z (spatial location), the same as you computed in the previous questions

So here we only ask you to compute it assuming that Z(t) = -2

It is clear that P(-2) = (4, -2, -2), since Z(t) = -2

but now dP/dt(-2) is also computable

since we know dx/dz(-2) and dy/dz(-2) from question c

[weak turtle]

wdym P(-2), P is just the point

why the -2

[rugged trail]

P at time t = -2

[weak turtle]

since z = t

i assume

and z = -2

[rugged trail]

It'd be the proper way of writing it yes

I'm just writing for clarity, uppercase letters for functions of time

and lowercase letters for functions of z

it's important not to confuse the two for differentiation

[weak turtle]

thanks, gonna go to sleep and read this tomorrow

hopefully i understand and can get the tangent line

[weak turtle]

how is this computable while we have those other things you mentioned right after, i don't understand or follow where they are related

dP/dt itself is what we want, isn't it

[rugged trail]

Well just try to plug t = -2

[weak turtle]

we want the derivative at that point

[rugged trail]

It holds that Z(t) = -2 as well

Then dx/dz(Z(t)) is x'(-2) which you computed previously

And dZ/dt is always 1

[weak turtle]

you mean 0

[rugged trail]

So now you know dX/dt(-2)

[weak turtle]

Z(t) = -2
then dZ/dt = 0

[rugged trail]

No, Z(t) = t

We are just evaluating at a particular time t=-2

[weak turtle]

ok, then what

i don't know how this is related to dP/dt

which is what we want i believe

[rugged trail]

^

[weak turtle]

is it possible to get r'(t_0) @rugged trail