Page:Calculus Made Easy.pdf/207

all alike, clearly ${\dfrac {dy}{dx}}=a$, if we reckon $y$ as the total of all the $dy$’s, and $x$ as the total of all the $dx$’s. But whereabouts are we to put this sloping line? Are we to start at the origin $O$, or higher up? As the only information we have is as to the slope, we are without any instructions as to the particular height above $O$; in fact the initial height is undetermined. The slope will be the same, whatever the initial height. Let us therefore make a shot at what may be wanted, and start the sloping line at a height $C$ above $O$. That is, we have the equation

$y=ax+C$.

It becomes evident now that in this case the added constant means the particular value that $y$ has when $x=0$.

Now let us take a harder case, that of a line, the slope of which is not constant, but turns up more and more. Let us assume that the upward slope gets greater and greater in proportion as $x$ grows. In symbols this is:

${\frac {dy}{dx}}=ax$.

Or, to give a concrete case, take $a={\tfrac {1}{5}}$, so that

${\frac {dy}{dx}}={\tfrac {1}{5}}x$.

Then we had best begin by calculating a few of the values of the slope at different values of $x$, and also draw little diagrams of them.