I am having a particular problem understanding related rates problems. I think this is a good example to show my two issues:
Two planes are flying towards the same point. The first is going to
the east, and the other to the north. The first goes at a speed of
$450 km/h$ and the other at $600km/h$.When the two planes are at $150km$ and $200km$ respectively from the
collision point, at what rate is the distance between them changing?
Well. Some data:
The distance between the first plane and the collision point is $x$, so:
$$\frac{dx}{dy} = 450km/h$$
Whereas the distance between the second plane and the collision point is $y$:
$$\frac{dy}{dt}=600km/h$$
The distance between the two planes is $z$, and it can be calculated as
$$z^2=x^2+y^2$$
We want to find the rate of change of $z$, that is
$$\frac{dz}{dt} = \ ?$$
The previous formula is the distance between planes, but since we want the rate of change, we need its derivative…
$$2z\cdot\frac{dz}{dt} = 2x\cdot\frac{dx}{dt} + 2y\cdot\frac{dy}{dt}$$
Now, this particular step is what confuses me. I understand that we need the derivative of the distance function, but what I don't grasp is: the derivative with respect to what? To $x$? To $y$? To $z$?
Suppose that the distance formula is actually $z^2 = x^2\cdot y^2$ (I made this up), we would have to calculate the derivative, yes? For $z^2$ it would be $2z$, but what would happen with $x^2\cdot y^2$? If I were doing this with respect to $x$, the answer would be $2x\cdot y^2$, but if I were doing it with respect to $y$ it would be $2y\cdot x^2$.
The second thing that confuses me is, why are we multiplying $2z$ by $\frac{dz}{dt}$, $2x$ by $\frac{dx}{dt}$ and $2y$ by $\frac{dy}{dt}$? The derivative of $z^2$ is exactly $2z$ (by power rule), so why are we multiplying by $\frac{dz}{dt}$ too?
Anyway, if I ignore my two issues, we can go on I suppose:
We need the current value of $z$, which is calculated using the distance formula:
$$z^2 = 150^2 + 200^2$$
$$z = 250$$
Now we just plug into the derivative of the distance formula:
$$2(250) \cdot \frac{dz}{dt} = 2(150)\cdot450 + 2(200)\cdot600$$
Which yields
$$\frac{dz}{dt} = -750km/h$$
Best Answer
$x$ and $y$ are functions of $t$, so it would be more appropriate to write $z$ as $$z=\sqrt{[x(t)]^2+[y(t)]^2}$$ Using the chain rule for $t$, $$\frac{\mathrm{d}z}{\mathrm{d}t}=\frac{\partial z}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial z}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}$$ $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $\frac{\mathrm{d}y}{\mathrm{d}t}$ are already known, so all you have to do is find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.
In the end, you're taking the derivative with respect to time, $t$. This becomes clearer if you simply write $$z=f(x,y)$$ instead of $$z^2=g(x,y)$$ as you have done.
I would avoid doing it like you did, to avoid implicit differentiation.