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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

$$\frac{dz}{dt} = -750km/h$$

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$x$ 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 $\frac{\partial z}{\partial y}$.

I understand that we need the derivative of the distance function, but what I don't grasp is: the derivative with respect to what?

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.

The derivative of $z^2$ is exactly $2z$ (by power rule), so why are we multiplying by $\frac{dz}{dt}$ too?

I would avoid doing it like you did, to avoid implicit differentiation.

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  • $\begingroup$ So let's see if I get this straight: $z$, $x$, $y$ are all functions of $t$ (and they are actually $z(t)$, $x(t)$ and $y(t)$), so $$[z(t)]^2$$ would have as derivative $$2\cdot z(t)$$ and by the chain rule, I have to multiply this by the derivative of $z(t)$ (which is $\frac{dz}{dt}$) and therefore the full derivative of $[z(t)]^2$ is $$2\cdot z(t) \cdot \frac{dz}{dt}$$? $\endgroup$ – Zol Tun Kul Nov 9 '15 at 0:56
  • $\begingroup$ Yes, that's it. $\endgroup$ – HDE 226868 Nov 9 '15 at 0:57
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Ya, you have to derive with respect to time. and do take note that the derivative should be negative since your distance towards the collision point is decreasing. so if you make a sketch of the problem, you get a right triangle. Hence, you can use the pythagorean theorem to describe the distances. $x^2 +y^2 = z^2$ then we differentiate in terms of time $t$. Basically, you assume here that $x,y,z$ are functions defined on $t$ implicitly. Hence, we we also differentiate with respect to $t$ implicitly. That is, $2x \frac{dx}{dt} + 2y\frac{dy}{dt}=2z\frac{dz}{dt}$

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