A student wrote us with this question in regards to the related rates problem " Lamp post casts a shadow of a man walking":
That’s a wonderful explanation, but I’m having a bit of a problem understanding the 3d step. As an eastern European – we use the f’(x) notation more often, so I blatantly just don’t understand the example :D.
Could u give a solution based on v(t) =s’(t) and a(t) =v’(t) ?
I also don’t really get the “in respect to time” part…
Hello there! Welcome to our learning community, and thanks for your questions.
In case it helps, lots (and lots) of students have to put some work in to shift from being focused on how an object’s kinematics quantities are related (x(t), v(t) = x'(t), and a(t) = v'(t)) to how other quantities are related. Related rates problems like this one are all about finding some other relation, like here, for instance, the position of the shadow’s head \ell as a function of the man’s horizontal position x (which we write using the notation of the solution as \ell = \ell(x)). It’s that functional relationship that we develop in Step 2 of our process, and that’s almost always the toughest part of related rates problems for students just learning this stuff.
Let me start with the last bit you wrote:
I also don’t really get the “in respect to time” part…
since the answer highlights why we use Leibnitz (\dfrac{d}{dx}, or \dfrac{d}{dt}, or \dfrac{d}{d\theta}) notation for these types of problems, instead of something like f'(x).
The key thing is that when you take a derivative, you’re finding the rate at which one quantity (like f(x)) changes as another quantity changes (like x, or time t, or angle \theta). When we say “with respect to,” we’re specifying which quantity we’re considering as that second quantity. So the derivative tells you “how sensitive” your function is to changes in input of that second quantity: when you increase (or decrease) that second quantity a tiny bit, how much does the function’s output change?
For instance, if you’re walking up a hill, your vertical position y changes as a function of your horizontal position x, which we typically write as y = y(x). Then if we want to specify how much our vertical position changes with respect to our horizontal position, we want \dfrac{dy}{dx}, which could equal something \dfrac{dy}{dx} = 0.6 \, \tfrac{\text{vertical cm}}{\text{horizontal cm}}. That is, for every 1 cm you move forward, you gain 0.6 cm in height. (You can see more about this very scenario here.)
But maybe you’re not interested in that; instead you want to know how fast, as a function of time, you’re gaining vertical distance as you walk, and you don’t care anything about your horizontal position x. That is, you’re thinking about y = y(t). Then maybe you’d discover that your rate of vertical climb with respect to time t is something like \dfrac{dy}{dt} = 3 \, \tfrac{\text{m}}{\text{min}}: for every minute that passes, you climb 3 meters.
While we’re at it, we could let p equal the price of gold at this moment. Since this doesn’t affect your rate of climb at all, we could write (just to make a point) \dfrac{dy}{dp} = 0: if the price of gold changes a bit, your rate of climb doesn’t change. The derivative can tell you the rate with respect to whatever-you-want, not just always x, or t, or …
Most students prefer prime-notation, and we often do too! But it has the limitation that it doesn’t (easily) tell you what you’re taking the derivative with respect to. There are times its important to specify so you don’t mix up your rates, and this is one of them. In particular, our Step 3 is, “Take the derivative with respect to time of both sides of the equation,” so we mean apply \dfrac{d}{dt} to both sides of the equation (as opposed to \dfrac{d}{dx}, or \dfrac{d}{dp} or something).
For instance, at the end of Step 2 where we had developed the functional relationship between \ell and x that I wrote about above, 0.70\ell = x, or:
Though it’s not written explicitly there, we have \ell as a function of the man’s position x: \ell(x) = \dfrac{1}{0.70}x. So if I find \ell\,'(x), I have
This tells me that if the man moves forward in position say 1 cm, then the shadow’s head moves forward \dfrac{1}{0.70} cm. But this isn’t what the problem is asking for; instead it’s asking how fast his shadow head moves with respect to time. So that’s the derivative I need to take.
It’s important to note that when we use the prime notation, unless we explicitly say otherwise we mean “with respect to the function’s independent variable.” So if I write \ell\,' it looks like I’m finding \dfrac{d\ell}{dx}, which as we saw we don’t want since it doesn’t help us.
Enough of that. We can certainly recast the solution using prime notation, as long as we make perfectly clear that both \ell and x are functions of time, and we now have an equation that relates them. Again picking up from the end of Step 2, now making the function dependence of t clear:
And let’s be perfectly clear that when we write the prime we mean “the derivative with respect to time t.” I can even write \dfrac{d\ell}{dt} = \ell\,' and \dfrac{dx}{dt} = x'; just different ways of writing the same thing now that we’ve agreed everything is a function of time. Then
The problem statement told us that \dfrac{dx}{dt} = x' = 1.5 \, \tfrac{\text{m}}{\text{s}}. I’m guessing you might prefer to recognize this as the man’s horizontal velocity, \dfrac{dx}{dt} = x' = v_{\text{man}}(t) = 1.5 \, \tfrac{\text{m}}{\text{s}}. You’re then finding the quantity \dfrac{d\ell}{dt} = \ell\,' = v_{\text{shadow's head}}=??
I hope this helps. We know (for sure!) that all of this notation can be confusing when you’re first learning it – after all, you’re learning a new language (Calculus notation) at the same time you’re learning these new and sometimes confusing concepts. If we can help with more clarification please ask … and if this is too much and you’d like fewer words and something more direct, please let us know that too. : )
For now, thanks again for asking your questions, and mostly for trying to make sense of these new ideas and new language for yourself! 