please help. numbers 24 & 25 (picture)

- 1018

please help. numbers 24 & 25 (picture)

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- madhu.mukherjee.946

@ganeshie8

- 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

|dw:1441468653415:dw|

- welshfella

|dw:1441468744582:dw|

- welshfella

- that's Q 23

- beginnersmind

Shouldn't the integrand be \(2\pi\ x^2 dx\) ?

- welshfella

no the area of a 'slice' is pi r^2

- welshfella

and r = x^2

- beginnersmind

Sorry meant \(2\pi\ (x^2)^2 dx \)

- beginnersmind

Oh, right

- beginnersmind

Area of a circle, lol

- welshfella

- i dont think so - why the 2 ?

- welshfella

circle yes

- welshfella

no that is not the curve
the curve is y - x^2

- sohailiftikhar

as given y=x^2

- welshfella

|dw:1441469338036:dw|

- welshfella

we only dealing with x = 0 to x = 1

- welshfella

|dw:1441469432075:dw|

- welshfella

it is the sum of all the discs with height dx and radius x^2

- 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

the cross section is a circle

- sohailiftikhar

yes

- welshfella

yes

- welshfella

partial differentials

- 1018

hey guys! sorry i lost my connection

- 1018

what are your answers?

- 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

oh yes! squared. i forgot x^2 still needs ^2. thanks @welshfella thanks @beginnersmind

- 1018

ok ill try and see if i can get the answer

- 1018

@beginnersmind hey can you show me your work, i think im close to it but i dont fully understand. thanks

- 1018

i mean when im deriving already. i think im missing something

- beginnersmind

You mean show why my formula is true? Or apply it?

- 1018

apply

- beginnersmind

ok, for 24
|dw:1441470707423:dw|

- beginnersmind

Remember to take the partial derivative with respect to x you just treat y as a constant.

- 1018

oh i should still write the dx and dy in my answer?

- 1018

it's still part of it?

- beginnersmind

Sure, they are asking for dz. The answer will be in a form:
(something) * dx + (something else) * dy

- 1018

ok thanks! Ill try the other one thanks again!

- beginnersmind

No problem. Let me know if you want me to check your solution.

- 1018

oh no wait can you help me out on 25, hahaha. that is a hard one

- beginnersmind

Oh, 25 is different because x and y are a function of t.

- beginnersmind

Ok, it's actually simpler than 24

- beginnersmind

They are only asking for dx/dt. You can ignore everything except the definition of x(t).

- beginnersmind

It's weird though, I'm getting \( \pi/2\), for 25. Paging @welshfella

- beginnersmind

-pi/2 that is.
But it's not one of the options.

- 1018

im still solving haha

- 1018

do i just ignore the e? i will just derive the exponent then sub to the orig value?

- 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

@ganeshie8 can you check please?

- 1018

that's my sol'n for x. haha at least i'm getting somewhere

- 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

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

maybe there really is an error, hey thanks anyway. big help!

- beginnersmind

No problem. Just for completeness's sake this is the formula for dz/dt|dw:1441473257074:dw|

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