hello, help with related rates problem.
._. (attached below)

- anonymous

hello, help with related rates problem.
._. (attached below)

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- schrodinger

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- anonymous

??

- anonymous

##### 1 Attachment

- anonymous

@Paynesdad It's posted ^_^

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## More answers

- anonymous

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

- 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

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

- anonymous

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

- anonymous

got it.

- anonymous

im getting there.

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

sorry just doing the math

- anonymous

yes i see what you mean.

- anonymous

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

- anonymous

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

- anonymous

would using the chain rule be wrong?

- anonymous

wait, which one's?

- anonymous

All of them

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

oh no, :? why not?

- anonymous

Whoa whoa whoa.. Hang on here.

- 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

Right ^

- anonymous

oh okay, i understand that.

- anonymous

i used it twice before.

- 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

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

yes

- anonymous

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

- anonymous

What did you get?

- anonymous

@mathcalculus

- anonymous

gotta start over, i'll keep you posted.

- anonymous

Ok... let us know.

- anonymous

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

- anonymous

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

- anonymous

@mebs

- anonymous

@Paynesdad

- anonymous

Show us what you did again if you can.

- anonymous

ok. 1 sec

- anonymous

##### 1 Attachment

- anonymous

Got it...Order of operations!

- anonymous

?

- anonymous

where did i miss?

- 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

yes that i got.

- anonymous

I used PEMDAS

- anonymous

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

- 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

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

- anonymous

ohhhhh 45 is part of dy/dx

- anonymous

yea bro, step It up man dude.

- anonymous

oops.

- anonymous

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

- anonymous

i wasn't looking closely.

- anonymous

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

- anonymous

Its fine I was jking haha.hhaa

- anonymous

I can tell that by your handwriting haha.haha

- anonymous

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

- anonymous

Good question @mathcalculus keep up the hard work.

- anonymous

@mebs you crack me up

- anonymous

Keep it up @mathcalculus

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