## mathcalculus Group Title hello, help with related rates problem. ._. (attached below) one year ago one year ago

??

2. mathcalculus

3. mathcalculus

4. mathcalculus

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

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. mathcalculus

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

7. mebs

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

8. mathcalculus

got it.

9. mathcalculus

im getting there.

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. mebs

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

12. mebs

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

13. mebs

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

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

15. mebs

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

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

17. mebs

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. mathcalculus

sorry just doing the math

19. mathcalculus

yes i see what you mean.

20. mathcalculus

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

21. mathcalculus

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

22. mathcalculus

would using the chain rule be wrong?

23. mathcalculus

wait, which one's?

24. mebs

All of them

25. mathcalculus

26. mebs

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

27. mathcalculus

oh no, :? why not?

Whoa whoa whoa.. Hang on here.

29. mathcalculus

?

30. mebs

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

Right ^

32. mathcalculus

oh okay, i understand that.

33. mathcalculus

i used it twice before.

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. mebs

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. mathcalculus

yes

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

What did you get?

@mathcalculus

40. mathcalculus

gotta start over, i'll keep you posted.

Ok... let us know.

42. mathcalculus

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

43. mathcalculus

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

44. mathcalculus

@mebs

45. mathcalculus

Show us what you did again if you can.

47. mathcalculus

ok. 1 sec

48. mathcalculus

Got it...Order of operations!

50. mathcalculus

?

51. mathcalculus

where did i miss?

52. mebs

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. mathcalculus

yes that i got.

54. mathcalculus

I used PEMDAS

55. mebs

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

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. mebs

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

58. mathcalculus

ohhhhh 45 is part of dy/dx

59. mebs

yea bro, step It up man dude.

60. mathcalculus

oops.

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

62. mathcalculus

i wasn't looking closely.

63. mathcalculus

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

64. mebs

Its fine I was jking haha.hhaa

65. mebs

I can tell that by your handwriting haha.haha

66. mathcalculus

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

67. mebs

Good question @mathcalculus keep up the hard work.