Related Rates problem

[Proximo]

I'm having trouble finding a solution to the following:

If y^2 + x = 3, and dy/dt = 5 when x=2 and y=-1, what is dx/dt?

I've tried multiple times and ways and searched other forums for help and still cannot find the answer. Any help would be greatly appreciated!

 

[JeffM]

I'm having trouble finding a solution to the following:

If y^2 + x = 3, and dy/dt = 5 when x=2 and y=-1, what is dx/dt?

I've tried multiple times and ways and searched other forums for help and still cannot find the answer. Any help would be greatly appreciated!

Please read Read Before Posting.

What are you currently studying? Implicit differentiation by any chance?

Please show what you have tried. Have you tried implicit differentiation?

 

[Proximo]

Please read Read Before Posting.

What are you currently studying? Implicit differentiation by any chance?

Please show what you have tried. Have you tried implicit differentiation?

Yes we are studying implicit differentiation and I did try using it but since we covered "simpler" examples in classes and we are assigned tougher problems to practice at home I don't even know how to set this problem properly from the start.
I tried starting it like a somewhat similar problem I found on another site but the answers I get keep coming up wrong. This is one way another site had as a possible solution:

d(y^2)/dt + d(x)/dt = 0
y^2(dx/dt) + x(dy/dt) + dx/dt = 0

Then I tried to solve it from there. This probably doesn't make any sense but it was the only other help I could find online to a problem similar to mine.

 

[JeffM]

Yes we are studying implicit differentiation and I did try using it but since we covered "simpler" examples in classes and we are assigned tougher problems to practice at home I don't even know how to set this problem properly from the start.
I tried starting it like a somewhat similar problem I found on another site but the answers I get keep coming up wrong. This is one way another site had as a possible solution:

d(y^2)/dt + d(x)/dt = 0
y^2(dx/dt) + x(dy/dt) + dx/dt = 0

Then I tried to solve it from there. This probably doesn't make any sense but it was the only other help I could find online to a problem similar to mine.

The underlying thought here is that both y and x are functions of t. Of course that means that y^2 is also a function of t, as is y^2 + x. And we apply the chain rule.

I am going to go step by step so that you see what is going on.

$\displaystyle y^2 + x = 3 \implies \dfrac{d}{dt}\left(y^2 + x\right) = \dfrac{d}{dt}(3) = 0.$ Make sense?

$\displaystyle But\ \dfrac{d}{dt}\left(y^2 + x\right) = \dfrac{d}{dt}(y^2) + \dfrac{d}{dt}(x) = \dfrac{d}{dt}(y^2) + \dfrac{dx}{dt}.$ Addition rule, right?

$\displaystyle Now \dfrac{d}{dt}(y^2) = \dfrac{d}{dy}(y^2) * \dfrac{d}{dt}(y).$ Chain rule.

$\displaystyle But\ \dfrac{d}{dy}(y^2) = 2y.$ Power rule.

$\displaystyle And\ \dfrac{d}{dt}(y) = \dfrac{dy}{dt}.$

$\displaystyle So\ \dfrac{d}{dy}(y^2) = 2y * \dfrac{dy}{dt}.$ Substituting.

$\displaystyle Giving\ \dfrac{d}{dt}(y^2) + \dfrac{d}{dt}(x) = 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt}.$ More substitution.

$\displaystyle Thus,\ 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt} = 0.$ And more substitution.

Implicit differentiation just skips a bunch of steps and really is a result of the chain rule.

$\displaystyle y^2 + x = 3 \implies 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt} = 0.$ Proceed.

 

[Proximo]

The underlying thought here is that both y and x are functions of t. Of course that means that y^2 is also a function of t, as is y^2 + x. And we apply the chain rule.

I am going to go step by step so that you see what is going on.

$\displaystyle y^2 + x = 0 \implies \dfrac{d}{dt}\left(y^2 + x\right) = \dfrac{d}{dt}(0) = 0.$ Make sense?

$\displaystyle But\ \dfrac{d}{dt}\left(y^2 + x\right) = \dfrac{d}{dt}(y^2) + \dfrac{d}{dt}(x) = \dfrac{d}{dt}(y^2) + \dfrac{dx}{dt}.$ Addition rule, right?

$\displaystyle Now \dfrac{d}{dt}(y^2) = \dfrac{d}{dy}(y^2) * \dfrac{d}{dt}(y).$ Chain rule.

$\displaystyle But\ \dfrac{d}{dy}(y^2) = 2y.$ Power rule.

$\displaystyle And\ \dfrac{d}{dt}(y) = \dfrac{dy}{dt}.$

$\displaystyle So\ \dfrac{d}{dy}(y^2) = 2y * \dfrac{dy}{dt}.$ Substituting.

$\displaystyle Giving\ \dfrac{d}{dt}(y^2) + \dfrac{d}{dt}(x) = 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt}.$ More substitution.

$\displaystyle Thus,\ 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt} = 0.$ And more substitution.

Implicit differentiation just skips a bunch of steps and really is a result of the chain rule.

$\displaystyle y^2 + x = 0 \implies 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt} = 0.$ Proceed.

Yes that makes alot of sense! I did something similar in one attempt and I believe I left out the multiplying of 2y and dy/dt. So plugging in from the original I got 2(-1) * 5 + dx/dt = 0 which gave me the answer dx/dt = 10! Thanks so much for explaining it step by step, you explained it much better than my professor explains the work. Now to try the remaining practice problems... :???: