Page:Calculus Made Easy.pdf/32

positive increment, the increment which results to $y$ is negative.

Yes, but how much? Suppose the ladder was so long that when the bottom end $A$ was $19$ inches from the wall the top end $B$ reached just $15$ feet from the ground. Now, if you were to pull the bottom end out $1$ inch more, how much would the top end come down? Put it all into inches: $x=19$ inches, $y=180$ inches. Now the increment of $x$ which we call $dx$, is $1$ inch: or $x+dx=20$ inches.

How much will $y$ be diminished? The new height will be $y-dy$. If we work out the height by Euclid I. 47, then we shall be able to find how much $dy$ will be. The length of the ladder is

${\sqrt {(180)^{2}+(19)^{2}}}=181$ inches.

Clearly then, the new height, which is $y-dy$, will be such that

$(y-dy)^{2}=$	$(181)^{2}-(20)^{2}=32761-400=32361$,
$y-dy=$	${\sqrt {32361}}=179.89$ inches.

Now $y$ is $180$, so that $dy$ is $180-179.89=0.11$ inch.

So we see that making $dx$ an increase of $1$ inch has resulted in making $dy$ a decrease of $0.11$ inch.

And the ratio of $dy$ to $dx$ may be stated thus:

${\frac {dy}{dx}}=-{\frac {0.11}{1}}$.

It is also easy to see that (except in one particular position) $dy$ will be of a different size from $dx$.

Now right through the differential calculus we are hunting, hunting, hunting for a curious thing,