Multivariable Calc Help Needed

[distant igloo]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[hollow talon]

@cloud cloud sorry to ping but would u know how to do this?

[mild violet]

Do you know the formula of the gradient ?

@hollow talon

[hollow talon]

I'm hella confused. Am I dumb or is it rlly this easy?

the slope is given by f(x,y) = -x^2 - y^2

fx = -2x
fy = -2y

∇f (a) = (fx(a), fy(a))

here a is the point (1,0)

So the gradient = (-2(1), -2(0)) = (-2, 0) ?

[mild violet]

Well yes I’m guessing it’s just that lol

[hollow talon]

yeah has to be that..

I had another follow up question

I’m confused about the multiplication in the numerator. We r multiplying -g with the gradient of f. But the gradient of f is the collection of partial derivatives, which can be considered as a vector. I’m not sure how we can get a value for the acceleration when multiplying a constant with a vector.

But then I was told that..... The acceleration a is also a vector. Your answer to 1(b) should be a vector of two values.

So... how do we get that lmao

[mild violet]

Well yes here a is a vector so if you multiply g by a vector you’ll get a vector

i don’t understand your question it seems clear a= (-g * grad(f))/(1+N(grad(f))^2)

Here you are multiplying a scalar by grad(f) which gives you a vector

Which will be your expression of a which is also a vector

[hollow talon]

(-2,0) is a vector right?

the thing is so

it's not (grad(f))^2

they said this:

[mild violet]

Yes this is a scalar

[hollow talon]

So is that just (-2,0) dot (-2,0) = 4?

[mild violet]

It’s the norm of grad(f)

[mild violet]

Yes

[hollow talon]

ohh

since m is 1

we can just ignore it?

[mild violet]

Yeah m * a=a

[hollow talon]

I'm so confused

If we have to solve for a

the gradient is supposed to be based off a point

but the question doesn't give us a point

[quartz crow]

just solve for a by scaling both vectors by 1/m no?

$\vec{a}=\frac{-g}{m(1+\norm{\vec{\nabla}f})^2}\vec{\nabla}f$

[pliant epoch]

Omegabet_

[hollow talon]

my final answer has to be a vector tho

like a specific answer like (x,y) ... i think?

[quartz crow]

that's a vector

it's a scalar multiple of the gradient

and gradient is a vector

you can write out the components if you feel the need to

$\vec{a}=\frac{-g}{m(1+f_x^2+f_y^2)}[f_x,f_y]$

[pliant epoch]

Omegabet_

[hollow talon]

hmm

but we gotta plug in values still

[quartz crow]

sure ig

idk, I dont have the question

[hollow talon]

.

[quartz crow]

nowhere does the question you've posted says to

yeah

that's preamble

[hollow talon]

.

I was told to do this:

[quartz crow]

and where in that.. do you see "find the acceleration at this point?"

for future: post the entire question

yeah

you just plug in that m

you get a=(mess)

so... QED

move on

again

nowhere does it say the acceleration at a point

it says find the acceleration

period

if you feel the incessenant need to make the dependence explicit you can

$\vec{\nabla}f\equiv\vec{\nabla}f(x,y)$

[pliant epoch]

Omegabet_

[hollow talon]

yeah but the acceleration should be a vector of two points...

[quartz crow]

AND IT IS

it's a vector

[hollow talon]

oh

[quartz crow]

with 2 entries

[quartz crow]

.

look at those entries

there are 2.

therefore

a is a vector, with 2 entries.

[hollow talon]

ye I understand that but that would be simply restating the formula with m and g substituted. But we need to acc solve it

[quartz crow]

what does "solve it" mean to you?

if not

get a=(stuff)?

[hollow talon]

Solve it mean get something like a = (4,0)

[quartz crow]

mind you, it's only 1 point as well

so... very little has to be done

that requires knowing a point

which you dont

it's quite literally

just asking what is a= as a function

[hollow talon]

ohh

so we r assuming that

it's the same as with a specificed point but just now that we would use an arbitrary point

[quartz crow]

sure

it's just asking you to solve the given equation for a

nothing more

nothing less

there's no point to plug in, so there's no reason to think of plugging a point in

[hollow talon]

so ∇f is the gradient?

So if we were going to solve the acceraltion at a specific point, it would have said solve the acceleration at for example (2,0)?

[hollow talon]

∇f = (-2x(x,y) , -2y(x,y)) right?

[quartz crow]

no

f_x=-2x
f_y=-2y
grad(f)=[-2x,-2y]=-2[x,y]

for that specific f ofc

gradient is just a vector with the partials as the entries.

[hollow talon]

But isn't that how the gradient is? The (x,y) would just be a point to be filled in?

[quartz crow]

not at all

$\vec{\nabla}f(x,y)=[f_x(x,y),f_y(x,y)]$

[pliant epoch]

Omegabet_

[quartz crow]

not f_x times (x,y)

[hollow talon]

But

f_x is -2x

So why is -2x(x,y) wrong notation?

[hollow talon]

We also have

So would that just be equal to -2x^2 = 4x^2 and -2y^2 = 4y^2

So would the final answer then be a = (-10/(1+4x^2+4y^2)) (-2x, -2y) ?

[quartz crow]

yes.

[quartz crow]

cause... that's a vector

it's saying the derivative of a function from R to R is... a vector?

when, from calc1, you know the derivative of such a function is a number/scalar

[hollow talon]

+close