anonymous
  • anonymous
hello, help with related rates problem. ._. (attached below)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
??
anonymous
  • anonymous
anonymous
  • anonymous
@Paynesdad It's posted ^_^

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anonymous
  • anonymous
am i able to use the chain rule for this problem?
anonymous
  • 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.
anonymous
  • anonymous
okay so i began doing this: DQ/DT= (2X)(5Y)
anonymous
  • anonymous
\[\frac{dQ}{dt} = -2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]
anonymous
  • anonymous
got it.
anonymous
  • anonymous
im getting there.
anonymous
  • 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.
anonymous
  • anonymous
we know that \[\frac{dx}{dt} = -2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]
anonymous
  • anonymous
So solve this \[-2 = 2(3)(-2)5(1)^{4}\frac{dy}{dt}\]
anonymous
  • anonymous
@Paynesdad I think this is implicit because they wanted it in terms of Q.
anonymous
  • anonymous
Seems like you answered it mebs...not helped with it.
anonymous
  • anonymous
@Paynesdad I think your right....my apologies I get exited when I do math.
anonymous
  • anonymous
LOL I know the feeling! I am glad I am not the only one.
anonymous
  • 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.
anonymous
  • anonymous
sorry just doing the math
anonymous
  • anonymous
yes i see what you mean.
anonymous
  • anonymous
im not up to the answer yet but close, i just need to check
anonymous
  • anonymous
i got 48.96.... :? but it's not right..
anonymous
  • anonymous
would using the chain rule be wrong?
anonymous
  • anonymous
wait, which one's?
anonymous
  • anonymous
All of them
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
You can't do what you just did in your 2nd and third step
anonymous
  • anonymous
oh no, :? why not?
anonymous
  • anonymous
Whoa whoa whoa.. Hang on here.
anonymous
  • anonymous
?
anonymous
  • anonymous
To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]
anonymous
  • anonymous
Right ^
anonymous
  • anonymous
oh okay, i understand that.
anonymous
  • anonymous
i used it twice before.
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
yes
anonymous
  • anonymous
So using mebs equation you should get...a fractional answer.
anonymous
  • anonymous
What did you get?
anonymous
  • anonymous
@mathcalculus
anonymous
  • anonymous
gotta start over, i'll keep you posted.
anonymous
  • anonymous
Ok... let us know.
anonymous
  • anonymous
okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?
anonymous
  • anonymous
damn it. wrong again. i got -2/33 this time
anonymous
  • anonymous
@mebs
anonymous
  • anonymous
@Paynesdad
anonymous
  • anonymous
Show us what you did again if you can.
anonymous
  • anonymous
ok. 1 sec
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Got it...Order of operations!
anonymous
  • anonymous
?
anonymous
  • anonymous
where did i miss?
anonymous
  • anonymous
Let me show you what must be done. \[-2 = 2\times 3 \times (-2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]
anonymous
  • anonymous
yes that i got.
anonymous
  • anonymous
I used PEMDAS
anonymous
  • anonymous
You can't add them they both don't have dy/dx dude.
anonymous
  • 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
anonymous
  • anonymous
you get \[-2 = -12 + 45\frac{dy}{dx}\]
anonymous
  • anonymous
ohhhhh 45 is part of dy/dx
anonymous
  • anonymous
yea bro, step It up man dude.
anonymous
  • anonymous
oops.
anonymous
  • anonymous
lol yep... it is a coefficient...come on bro!
anonymous
  • anonymous
i wasn't looking closely.
anonymous
  • anonymous
p.s not a bro lol but thanks! ^_^
anonymous
  • anonymous
Its fine I was jking haha.hhaa
anonymous
  • anonymous
I can tell that by your handwriting haha.haha
anonymous
  • anonymous
kay, well thanks guys!!! appreciate it tons tons.
anonymous
  • anonymous
Good question @mathcalculus keep up the hard work.
anonymous
  • anonymous
@mebs you crack me up
anonymous
  • anonymous
Keep it up @mathcalculus

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