Question

Related rates

Answer 1

When a certain commodity is \(p\) dollars per unit, the manufacturer is willing to supply \(x\) thousand units, where \(x^2-2x \sqrt{p}-p^2=31\). How fast is the supply changing when the price is $9 per unit and is increasing at the rate of 20 cents per week?

Answer 2

You are given the equation on how \(x\) and \(p\) are related. So.
Equation given: \(x^2-2x \sqrt{p}-p^2=31\)
You are given rate: \[\frac{dp}{dt}=0.2\]
And you need to find:
\[\frac{dx}{dt}, \text{when }p=9\]

Answer 3

You just differentiate your equation with respect to \(t\) and solve for \({dx}/{dt}\).

Answer 4

Sorry I was AFK, but just differentiate it and plug in values?

Answer 5

\[\frac{d}{dt}(x^2-2x \sqrt{p}-p^2)=\frac{d}{dt}31\]

Answer 6

Yes, essentially. You have to realize that both \(x\) and \(p\) are implicitly functions of \(t\), so you must use the chain rule on the derivatives. That's how you'll get the terms \(dx/dt\) and \(dp/dt\), and yes plug in the value of \(p=9\) and \(dp/dt=0.2\)

Answer 7

So end up with.. \(\frac{d}{dt}(x^2-2x \sqrt{p}-p^2)=\frac{d}{dt}(31)\)
\[\LARGE 2x\frac{dx}{dt}-\frac{1}{\sqrt{p}} \frac{dp}{dt}-2p \frac{dp}{dt}=0\]

Answer 8

I messed up there didn't I? >.<

Answer 9

The middle term. It is like a product rule. Think of \(x\) as "f" and \(\sqrt{p}\) as "g", and use the product rule \(f'g+fg'\)

Answer 10

\[\LARGE 2x\frac{dx}{dt}-(\frac{2x}{\sqrt{p}}+2\sqrt{p}) \frac{dp}{dt}-2p \frac{dp}{dt}=0\]

Answer 11

Um sort of. \[\frac{d}{dt}(-2x \sqrt{p})=-2\left(\frac{d}{dt}x\sqrt{p}\right)=-2\left(\frac{dx}{dt}\sqrt{p}\,+x\frac{1}{2\sqrt{p}}\frac{dp}{dt}\right)\]

Answer 12

That's only for the middle term.