Applications of the derivative to Geometry

We consider a theorem which is fundamental in all Differential Calculus to Geometry.

Figure 4.1: The geometry of derivatives.

Let

$$y = f(x)$$ (4.5)

be the equation of a curve $AB$.

Now differentiate (4.5) by the General Rule and interpret each step geometrically.

• FIRST STEP. $y + \Delta y = f(x + \Delta x) = NQ$

• SECOND STEP.
  $$\begin{array}{ll} y + \Delta y & = f(x + \Delta x) = NQ\\ ... ...= NR\\ \Delta y &= f(x + \Delta x) − f(x) = RQ. \end{array}$$

• THIRD STEP.
  $$\begin{array}{ll} \frac{\Delta y}{\Delta x} & = \frac{f(x + ... ...an \phi \\ & = {\rm slope\ of\ secant\ line\ } PQ. \end{array}$$

• FOURTH STEP.
  $$\begin{array}{ll} \lim_{\Delta x \to 0} \frac{\Delta y}{\De... ...dy}{dx} = {\rm value\ of\ the\ derivative\ at\ } P. \end{array}$$

But when we let $\Delta x \to 0$, the point $Q$ will move along the curve and approach nearer and nearer to $P$, the secant will turn about $P$ and approach the tangent as a limiting position, and we have also

$$\begin{array}{ll} \lim_{\Delta x \to 0} \frac{\Delta y}{\De... ... \tau\\ & = {\rm slope\ of\ the\ tangent\ at\ } P. \end{array}$$

Hence , $\frac{dy}{dx}$ = slope of the tangent line $PT$. Therefore

Theorem 4.9.1   The value of the derivative at any point of a curve is equal to the slope of the line drawn tangent to the curve at that point.

It was this tangent problem that led Leibnitz^4.6 to the discovery of the Differential Calculus.

Example 4.9.1   Find the slopes of the tangents to the parabola $y=x^2$ at the vertex, and at the point where $x = \frac{1}{2}$.

Solution. Differentiating by General Rule, (§4.7), we get

$$y' = \frac{dy}{dx} = 2x = {\rm slope\ of\ tangent\ line\ at\ any\ point\ on\ curve}.$$

Figure 4.2: The geometry of the derivative of $y=x^2$.

To find slope of tangent at vertex, substitute $x = 0$ in $y'=2x$, giving

$$\frac{dy}{dx} = 0.$$

Therefore the tangent at vertex has the slope zero; that is, it is parallel to the axis of $x$ and in this case coincides with it.

To find slope of tangent at the point $P$, where $x = \frac{1}{2}$, substitute in $y'=2x$, giving

$$\frac{dy}{dx} = 1;$$

that is, the tangent at the point $P$ makes an angle of $45^o$ with the axis of $x$.