I'm reading through the book "Calculus Made Easy" by Silvanus P. Thompson
In the page 12, there's an example of a ladder of 181 inches resting against a vertical wall and calculating how much the top end of the ladder goes down as we separate the bottom end of the ladder from the wall, i.e. the ratio where dy decrease while we increase dx
At the end of the example. Silvanus says:
And the ratio of dy to dx may be stated thus:
$ \frac{dy}{dx} = - \frac{0.11}{1} $
Its also easy to see that (except in one particular position) $ dy $ will be of a different size from $ dx $
I understand that if $ dy = dx $ then:
$ 1 = - \frac{0.11}{1} $
And this doesn't make sense. What particular position is the author referring to?
Let the vertical height of the ladder be $y$ and the horizontal length be $x$
Now, whatever position the ladder may be in, the following is always true by pythagoreas theorem. $$x^2+y^2=181^2$$ Now when you differentiate the equation with respect to $x$, you get $$2x+2y{dy\over dx}=0$$ $${dy\over dx}=-{x\over y}=-{\cot}{\theta}$$ Now, this rate of change is equal to the instantaneous slope of the ladder. When the ladder is steep (${\tan{\theta}}$ is large), a small increase in $x$ will result in a large decrease in $y$. You can experiment this at home by taking a slab of cardboard and sliding it along the wall and floor of a room. Clearly, the rate of change is equal (${dy\over dx}=1$) only when $x$ is equal to $y$.