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mathcalculus
Group Title
hello, help with related rates problem.
._. (attached below)
 11 months ago
 11 months ago
mathcalculus Group Title
hello, help with related rates problem. ._. (attached below)
 11 months ago
 11 months ago

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mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@Paynesdad It's posted ^_^
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
am i able to use the chain rule for this problem?
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay so i began doing this: DQ/DT= (2X)(5Y)
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{dQ}{dt} = 2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
got it.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im getting there.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
we know that \[\frac{dx}{dt} = 2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
So solve this \[2 = 2(3)(2)5(1)^{4}\frac{dy}{dt}\]
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
@Paynesdad I think this is implicit because they wanted it in terms of Q.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Seems like you answered it mebs...not helped with it.
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
@Paynesdad I think your right....my apologies I get exited when I do math.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
LOL I know the feeling! I am glad I am not the only one.
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
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.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
sorry just doing the math
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes i see what you mean.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im not up to the answer yet but close, i just need to check
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got 48.96.... :? but it's not right..
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
would using the chain rule be wrong?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
wait, which one's?
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
You can't do what you just did in your 2nd and third step
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh no, :? why not?
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Whoa whoa whoa.. Hang on here.
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Right ^
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh okay, i understand that.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i used it twice before.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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.
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
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.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
So using mebs equation you should get...a fractional answer.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
What did you get?
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
@mathcalculus
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
gotta start over, i'll keep you posted.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Ok... let us know.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
damn it. wrong again. i got 2/33 this time
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@mebs
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@Paynesdad
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Show us what you did again if you can.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ok. 1 sec
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Got it...Order of operations!
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
where did i miss?
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
Let me show you what must be done. \[2 = 2\times 3 \times (2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes that i got.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
I used PEMDAS
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
You can't add them they both don't have dy/dx dude.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
you get \[2 = 12 + 45\frac{dy}{dx}\]
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ohhhhh 45 is part of dy/dx
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
yea bro, step It up man dude.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oops.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
lol yep... it is a coefficient...come on bro!
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i wasn't looking closely.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
p.s not a bro lol but thanks! ^_^
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
Its fine I was jking haha.hhaa
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
I can tell that by your handwriting haha.haha
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
kay, well thanks guys!!! appreciate it tons tons.
 11 months ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
Good question @mathcalculus keep up the hard work.
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
@mebs you crack me up
 11 months ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Keep it up @mathcalculus
 11 months ago
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