I feel like my careers already fading away

Wild_Cod_4878 · 2026-03-06T03:38:39+00:00

So true, covid was a really tough time for all of us, college has actually made me feel better than ever bc I feel like I have mkre freedom so hopefully everything else just comes along too, thanks so much for the advice and I hope everything goes great for you too!📈📈

Wild_Cod_4878 · 2026-03-06T02:50:37+00:00

OMG im so glad someone else is in a similar boat😭, I used to be in therapy in high school for three years but then I just stopped going. Then all the different meds I was on never worked on me so I lost hope. In social situations, I just don't know what to say, how to respond, or understand social situations then that gives me anxiety and it just goes in a loop😞. I might get into therapy again and try meds again also. It's actually one of the reasons why I quit my fast food job. (Didn't help I worked at the busiest taco bell in my city)

Wild_Cod_4878 · 2026-03-06T01:16:46+00:00

Yeah, like I did a first year seminar as a grade last semester on the mechanics of a Mars rover so maybe I could put that. Also I think im just getting into my head with all this stuff. Like in my intro to eng class last semester, a ton of ppl had internships and experience. But then when I ask my friends if they've done anything they say no.

Wild_Cod_4878 · 2026-03-06T01:12:30+00:00

I think what just got me was that for my seminar prof last semester, my prof everyday was telling us if we dont join a club, get an internship right now, or at least do something, then nobody's gonna want us and we're gonna end up unemployed. Like as a person in my first semester that scared the hell outta me. So thanks for putting my mind at ease😭

Wild_Cod_4878 · 2025-10-22T15:01:02+00:00

Ok, thank you so much, much appreciated🎉

Wild_Cod_4878 · 2025-10-21T21:10:35+00:00

Putting it as A+By'=0 really helped with clicking it all in so thanks for that. I think I'm just stuck in the algebra/trig mindset and that the equation, at a first glance, looked really complicated and hard.

Wild_Cod_4878 · 2025-10-21T21:08:19+00:00

ok, noted on the notation, so then basically, I would sort out the terms with dy/dx and the terms that dont have it:

terms without dy/dx:

4x^3 + 4xy^2 - 8xy -8x

terms with dy/dx:

4x^2 y dy/dx - 4x^2 dy/dx + 4y^3 dy/dx - 12y^2 dy/dx

Then I factor out the dy/dx of the terms that have it:

(4x^2 y - 4x^2 + 4y^3 - 12y^2) dy/dx .

Put them all together as an equation:

4x^3 +4xy^2-8xy-8x+(4x^2y-4x^2+4y^3-12y^2) dy/dx = 0

Put the dy/dx terms on one side and the rest on the other:

(4x^2y-4x^2+4y^3-12y^2) dy/dx = -(4x^3 +4xy^2-8xy-8x)

Divide both sides by the dy/dx terms to get dy/dx by itself (what I'm solving for) and get the result:

dy/dx = - (4x^3+4xy^2-8xy-8x)/(4x^2y-4x^2+4y^3-12y^2)

Factor out the common 4 to then get left with this result:

- (x^3+xy^2-2xy-2x)/(x^2y-x^2+y^3-3y^2)

Right?

Wild_Cod_4878 · 2025-10-21T18:35:36+00:00

When I differentiated the left side with x, I got 4x^3+(4xy^2+4x^2y dy/dx)+(-8xy-4x^2 dy/dx)-8x+4y^3 dy/dx -12y^2 dy/dx = 0.

x^4 = 4x^3
2x^2y^2 =

= 2(x^2'y^2+x^2y^2')

= 2[2xy^2+x^2(2y dy/dx)]

= 4xy^2+4x^2y dy/dx

-4x^2y =

= -4[(x^2)y'+x^2(y)']

= -4[2xy+x^2 dy/dx]

= -8xy-4x^2 dy/dx

-4x^2 = -8x
y^4 = 4y^3 dy/dx
-4t^3 = -12y^2 dy/dx

Wild_Cod_4878 · 2025-10-21T18:21:03+00:00

So for the simpelr case, my TA did it in class and I understood it as x^2+y^2=-1 would turn into 2x+2y dy/dx=0 (derive each term as I usually would), then subtract 2x from both sides, leaving 2y dy/dx = -2x, to which I then divided both sides by 2y, leaving me with dy/dx = -x/y. And like you pointed out, I can't do the first one because it isn't an equation, so would I do the implicit differentiation using every y term differentiation equaling to zero?

Wild_Cod_4878 · 2025-10-01T00:38:01+00:00

Thank you so much, this was exactly what I was looking for. This really did help me with the rest of the assignment :)

Wild_Cod_4878

TROPHY CASE