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

??

2. mathcalculus Group Title

3. mathcalculus Group Title

4. mathcalculus Group Title

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 Group Title

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

7. mebs Group Title

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

8. mathcalculus Group Title

got it.

9. mathcalculus Group Title

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 Group Title

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

12. mebs Group Title

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

13. mebs Group Title

@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 Group Title

@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 Group Title

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.

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sorry just doing the math

19. mathcalculus Group Title

yes i see what you mean.

20. mathcalculus Group Title

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

21. mathcalculus Group Title

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

22. mathcalculus Group Title

would using the chain rule be wrong?

23. mathcalculus Group Title

wait, which one's?

24. mebs Group Title

All of them

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26. mebs Group Title

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

27. mathcalculus Group Title

oh no, :? why not?

Whoa whoa whoa.. Hang on here.

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?

30. mebs Group Title

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 Group Title

oh okay, i understand that.

33. mathcalculus Group Title

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 Group Title

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 Group Title

yes

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

What did you get?

@mathcalculus

40. mathcalculus Group Title

gotta start over, i'll keep you posted.

Ok... let us know.

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

43. mathcalculus Group Title

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

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@mebs

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Show us what you did again if you can.

47. mathcalculus Group Title

ok. 1 sec

48. mathcalculus Group Title

Got it...Order of operations!

50. mathcalculus Group Title

?

51. mathcalculus Group Title

where did i miss?

52. mebs Group Title

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 Group Title

yes that i got.

54. mathcalculus Group Title

I used PEMDAS

55. mebs Group Title

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 Group Title

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

58. mathcalculus Group Title

ohhhhh 45 is part of dy/dx

59. mebs Group Title

yea bro, step It up man dude.

60. mathcalculus Group Title

oops.

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

62. mathcalculus Group Title

i wasn't looking closely.

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

64. mebs Group Title

Its fine I was jking haha.hhaa

65. mebs Group Title

I can tell that by your handwriting haha.haha

66. mathcalculus Group Title

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

67. mebs Group Title

Good question @mathcalculus keep up the hard work.

@mebs you crack me up

Keep it up @mathcalculus