1. 1018

@ganeshie8

3. 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!

4. welshfella

|dw:1441468653415:dw|

5. welshfella

|dw:1441468744582:dw|

6. welshfella

- that's Q 23

7. beginnersmind

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

8. welshfella

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

9. welshfella

and r = x^2

10. beginnersmind

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

11. beginnersmind

Oh, right

12. beginnersmind

Area of a circle, lol

13. welshfella

- i dont think so - why the 2 ?

14. welshfella

circle yes

15. welshfella

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

16. sohailiftikhar

as given y=x^2

17. welshfella

|dw:1441469338036:dw|

18. welshfella

we only dealing with x = 0 to x = 1

19. welshfella

|dw:1441469432075:dw|

20. welshfella

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

21. beginnersmind

Anyway, for 24 and 25 use the definition of the total differential: for z = f(x,y) $dz = \frac{\partial f }{\partial x} dx + \frac{\partial f }{\partial y} dy$

22. welshfella

the cross section is a circle

23. sohailiftikhar

yes

24. welshfella

yes

25. welshfella

partial differentials

26. 1018

hey guys! sorry i lost my connection

27. 1018

28. beginnersmind

@welshfella got the answer for 23 in his first two posts. You can ignore the discussion under that. For 24 and 25 use $dz = \frac{\partial f }{\partial x} dx + \frac{\partial f }{\partial y} dy$

29. 1018

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

30. 1018

ok ill try and see if i can get the answer

31. 1018

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

32. 1018

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

33. beginnersmind

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

34. 1018

apply

35. beginnersmind

ok, for 24 |dw:1441470707423:dw|

36. beginnersmind

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

37. 1018

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

38. 1018

it's still part of it?

39. beginnersmind

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

40. 1018

ok thanks! Ill try the other one thanks again!

41. beginnersmind

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

42. 1018

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

43. beginnersmind

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

44. beginnersmind

Ok, it's actually simpler than 24

45. beginnersmind

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

46. beginnersmind

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

47. beginnersmind

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

48. 1018

im still solving haha

49. 1018

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

50. 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.

51. beginnersmind

52. 1018

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

53. beginnersmind

The only thing I can think of is that they actually meant to ask for dz/dt and dx/dt is a typo.

54. 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.

55. 1018

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

56. beginnersmind

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