## anonymous 2 years ago hello, help with related rates problem. ._. (attached below)

1. anonymous

??

2. anonymous

3. anonymous

4. anonymous

am i able to use the chain rule for this problem?

5. anonymous

Ok so take the derivative with respect of the equation using implicit differentiation. Since both x and y are variables that depend on t you will have to use the chain rule.

6. anonymous

okay so i began doing this: DQ/DT= (2X)(5Y)

7. anonymous

$\frac{dQ}{dt} = -2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}$

8. anonymous

got it.

9. anonymous

im getting there.

10. anonymous

no you can't do the derivative of x and y and the same time. $\frac{ dQ }{ dt }= \frac{ d(x^2y^5) }{ dt }$First take the derivative of the right side with respect to t of the x factor and treat y factor as a constant + the derivative of the right side with respect to t of the y factor treating the x factor as a constant.

11. anonymous

we know that $\frac{dx}{dt} = -2$ so when now find what $\frac{dy}{dt} , at, x = 3, y = 1$

12. anonymous

So solve this $-2 = 2(3)(-2)5(1)^{4}\frac{dy}{dt}$

13. anonymous

@Paynesdad I think this is implicit because they wanted it in terms of Q.

14. anonymous

Seems like you answered it mebs...not helped with it.

15. anonymous

@Paynesdad I think your right....my apologies I get exited when I do math.

16. anonymous

LOL I know the feeling! I am glad I am not the only one.

17. anonymous

Well @mathcalculus does this make sense to you.. I just subbed values that were given to me already from the question into the derivative of Q. and x and y were already defined.

18. anonymous

sorry just doing the math

19. anonymous

yes i see what you mean.

20. anonymous

im not up to the answer yet but close, i just need to check

21. anonymous

i got 48.96.... :? but it's not right..

22. anonymous

would using the chain rule be wrong?

23. anonymous

wait, which one's?

24. anonymous

All of them

25. anonymous

26. anonymous

You can't do what you just did in your 2nd and third step

27. anonymous

oh no, :? why not?

28. anonymous

Whoa whoa whoa.. Hang on here.

29. anonymous

?

30. anonymous

To use the product rule you do one at a time.$\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}$

31. anonymous

Right ^

32. anonymous

oh okay, i understand that.

33. anonymous

i used it twice before.

34. anonymous

Remember d(xy)/dt = y (dx/dt) + x (dy/dt). Also mathcalculus when you took the derivative you left off the dx/dt and dy/dt out of your equations.

35. anonymous

You already know that $\frac{dQ}{dt}= -2$ You know that $\frac{dx}{dt} = -2$ and you know that $x = 3 , y = 1$ just substitute it into the derivative of dQ/dt and solve.

36. anonymous

yes

37. anonymous

So using mebs equation you should get...a fractional answer.

38. anonymous

What did you get?

39. anonymous

@mathcalculus

40. anonymous

gotta start over, i'll keep you posted.

41. anonymous

Ok... let us know.

42. anonymous

okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?

43. anonymous

damn it. wrong again. i got -2/33 this time

44. anonymous

@mebs

45. anonymous

46. anonymous

Show us what you did again if you can.

47. anonymous

ok. 1 sec

48. anonymous

49. anonymous

Got it...Order of operations!

50. anonymous

?

51. anonymous

where did i miss?

52. anonymous

Let me show you what must be done. $-2 = 2\times 3 \times (-2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}$

53. anonymous

yes that i got.

54. anonymous

I used PEMDAS

55. anonymous

You can't add them they both don't have dy/dx dude.

56. anonymous

you have -12 + 45*dy/dt=-2 ...You can't add -12 to 45 until you multiply 45 times dy/dt...It's like 4=-3+7x....you don't simplify that as 4 = 4x ... right

57. anonymous

you get $-2 = -12 + 45\frac{dy}{dx}$

58. anonymous

ohhhhh 45 is part of dy/dx

59. anonymous

yea bro, step It up man dude.

60. anonymous

oops.

61. anonymous

lol yep... it is a coefficient...come on bro!

62. anonymous

i wasn't looking closely.

63. anonymous

p.s not a bro lol but thanks! ^_^

64. anonymous

Its fine I was jking haha.hhaa

65. anonymous

I can tell that by your handwriting haha.haha

66. anonymous

kay, well thanks guys!!! appreciate it tons tons.

67. anonymous

Good question @mathcalculus keep up the hard work.

68. anonymous

@mebs you crack me up

69. anonymous

Keep it up @mathcalculus