# Related rates ladder: why not substitute x=8 earlier?

A student wrote us regarding the problem on the main site “At what rate does the angle change as a ladder slides away from a house?”:

I know its a few years later but why do we use x as a variable when developing the cos (theta) equation for developing the equation. Don’twe know x is 8? Or is it because its not a constant?

Thank you for the excellent question!

You are correct there: we use x as a variable, rather than immediately substituting x = 8, because its value varies with time. That is, x is a function of t, an idea that is typically written mathematically as x = x(t). Similarly, \theta is also a function of time: \theta = \theta(t).

If we were to make the substitution x = 8 too early, we would lose the time-dependence in our key equation

\cos \Big(\theta(t)\Big) = \frac{x(t)}{10}

We don’t usually write the time-dependence explicitly in there as \theta(t) and x(t), because it’s annoying to do so, but it’s important to remember that the time-dependence is there. Then in Step 3 of the problem solution, when we take the derivative of that equation, we correctly obtain

-\sin \Big(\theta(t)\Big) \dfrac{d\theta}{dt}= \frac{1}{10} \dfrac{d\Big(x(t)\Big)}{dt}

and can proceed from there.

By contrast, if we incorrectly substitute x = 8 before we take the derivative, then we’ve lost the time dependence and so obtain

\begin{align*} \dfrac{d}{dt}\Big[ \cos \Big(\theta(t)\Big)\Big] &\overbrace{=}^? \dfrac{d}{dt}\left[ \frac{8}{10}\right] \quad \text{[INCORRECT]}\\[8px] -\sin \Big(\theta(t)\Big) \dfrac{d\theta}{dt}&\overbrace{=}^? 0 \quad \text{[since we took the derivative of a constant]} \end{align*}

If you were to obtain that last line in a solution of your own on an exam, say, it’d be a sign that you’ve made a substitution too soon, and need to go back and use the variable instead. (This is actually a pretty common student error so it’s important to be able to recognize it for yourself – and is also why this is such a great question.)

The key take-away here is to leave all quantities that vary with time as variables until after you take the derivative, and at the end (Step 4) are substituting values for the particular instant the problem specified, like “when the base is 8 feet from the house” in this problem.

We hope that all makes sense, and trust you’ll ask for clarification if not. For now thanks again for the great question, and please let us know what others you have!