1018
  • 1018
please help. numbers 24 & 25 (picture)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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1018
  • 1018
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madhu.mukherjee.946
  • madhu.mukherjee.946
@ganeshie8
1018
  • 1018
oh and i tried to answer #23 but my answer didnt show up among the choices. if anyone can answer that would be helpful too. thanks!

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More answers

welshfella
  • welshfella
|dw:1441468653415:dw|
welshfella
  • welshfella
|dw:1441468744582:dw|
welshfella
  • welshfella
- that's Q 23
beginnersmind
  • beginnersmind
Shouldn't the integrand be \(2\pi\ x^2 dx\) ?
welshfella
  • welshfella
no the area of a 'slice' is pi r^2
welshfella
  • welshfella
and r = x^2
beginnersmind
  • beginnersmind
Sorry meant \(2\pi\ (x^2)^2 dx \)
beginnersmind
  • beginnersmind
Oh, right
beginnersmind
  • beginnersmind
Area of a circle, lol
welshfella
  • welshfella
- i dont think so - why the 2 ?
welshfella
  • welshfella
circle yes
welshfella
  • welshfella
no that is not the curve the curve is y - x^2
sohailiftikhar
  • sohailiftikhar
as given y=x^2
welshfella
  • welshfella
|dw:1441469338036:dw|
welshfella
  • welshfella
we only dealing with x = 0 to x = 1
welshfella
  • welshfella
|dw:1441469432075:dw|
welshfella
  • welshfella
it is the sum of all the discs with height dx and radius x^2
beginnersmind
  • beginnersmind
Anyway, for 24 and 25 use the definition of the total differential: for z = f(x,y) \[dz = \frac{\partial f }{\partial x} dx + \frac{\partial f }{\partial y} dy\]
welshfella
  • welshfella
the cross section is a circle
sohailiftikhar
  • sohailiftikhar
yes
welshfella
  • welshfella
yes
welshfella
  • welshfella
partial differentials
1018
  • 1018
hey guys! sorry i lost my connection
1018
  • 1018
what are your answers?
beginnersmind
  • beginnersmind
@welshfella got the answer for 23 in his first two posts. You can ignore the discussion under that. For 24 and 25 use \[dz = \frac{\partial f }{\partial x} dx + \frac{\partial f }{\partial y} dy\]
1018
  • 1018
oh yes! squared. i forgot x^2 still needs ^2. thanks @welshfella thanks @beginnersmind
1018
  • 1018
ok ill try and see if i can get the answer
1018
  • 1018
@beginnersmind hey can you show me your work, i think im close to it but i dont fully understand. thanks
1018
  • 1018
i mean when im deriving already. i think im missing something
beginnersmind
  • beginnersmind
You mean show why my formula is true? Or apply it?
1018
  • 1018
apply
beginnersmind
  • beginnersmind
ok, for 24 |dw:1441470707423:dw|
beginnersmind
  • beginnersmind
Remember to take the partial derivative with respect to x you just treat y as a constant.
1018
  • 1018
oh i should still write the dx and dy in my answer?
1018
  • 1018
it's still part of it?
beginnersmind
  • beginnersmind
Sure, they are asking for dz. The answer will be in a form: (something) * dx + (something else) * dy
1018
  • 1018
ok thanks! Ill try the other one thanks again!
beginnersmind
  • beginnersmind
No problem. Let me know if you want me to check your solution.
1018
  • 1018
oh no wait can you help me out on 25, hahaha. that is a hard one
beginnersmind
  • beginnersmind
Oh, 25 is different because x and y are a function of t.
beginnersmind
  • beginnersmind
Ok, it's actually simpler than 24
beginnersmind
  • beginnersmind
They are only asking for dx/dt. You can ignore everything except the definition of x(t).
beginnersmind
  • beginnersmind
It's weird though, I'm getting \( \pi/2\), for 25. Paging @welshfella
beginnersmind
  • beginnersmind
-pi/2 that is. But it's not one of the options.
1018
  • 1018
im still solving haha
1018
  • 1018
do i just ignore the e? i will just derive the exponent then sub to the orig value?
beginnersmind
  • beginnersmind
IDK, I checked my work and I'm still getting -pi/2 Here it is in full if someone wants to go over it. x(t) = t*cost dx/dt = cost - t*sint evaluate at t=pi/2 dx/dt = 0 - (pi/2)*1 = -pi/2.
beginnersmind
  • beginnersmind
@ganeshie8 can you check please?
1018
  • 1018
that's my sol'n for x. haha at least i'm getting somewhere
beginnersmind
  • beginnersmind
The only thing I can think of is that they actually meant to ask for dz/dt and dx/dt is a typo.
beginnersmind
  • beginnersmind
I tried calculating dz/dt for t = pi/2 and got -[(pi)^3]/8. I think I'll just have to give up here. Something looks fishy, especially giving z(x(t),y(t)) but then only asking for dx/dt.
1018
  • 1018
maybe there really is an error, hey thanks anyway. big help!
beginnersmind
  • beginnersmind
No problem. Just for completeness's sake this is the formula for dz/dt|dw:1441473257074:dw|

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