Direction of a curve

It was shown in §4.9, that if

$$y = f(x)$$

is the equation of a curve (see Figure 6.1),

Figure 6.1: The derivative = slope of line tangent to the curve.

then

$$\frac{dy}{dx}\ =\ \tan\ \tau\ =\ {\rm slope\ of\ line\ tangent\ to\ the\ curve\ at\ any\ point\ } P.$$

The direction of a curve at any point is defined to be the same as the direction of the line tangent to the curve at that point. From this it follows at once that

$$\frac{dy}{dx}\ =\ \tan\ \tau\ = \ {\rm slope\ of\ the\ curve\ at\ any\ point\ } P.$$

At a particular point whose coordinates are known we write

$$\left [ \frac{dy}{dx} \right ]_{x=x_1, y=y_1}\ = \ {\rm slope\ of\ the\ curve\ (or\ tangent)\ at\ point\ } (x_1,y_1).$$

At points such as D, F, H, where the curve (or tangent) is parallel to the axis of $x$, $\tau = 0^o$, therefore $\frac{dy}{dx} = 0$.

At points such as A, B, G, where the curve (or tangent) is perpendicular to the axis of $x$, $\tau = 90^o$, therefore $\frac{dy}{dx} = \infty$.

At points such as E, where the curve is rising^6.1,

$$\tau \ = \ {\rm an\ acute\ angle;\ therefore\ } \frac{dy}{dx}\ =\ {\rm a\ positive\ number}.$$

The curve (or tangent) has a positive slope to the left of B, between D and F, and to the right of G in Figure 6.1. At points such as C, where the curve is falling,

$$\tau \ = \ {\rm an\ obtuse\ angle;\ therefore\ } \frac{dy}{dx}\ =\ {\rm a\ negative\ number}.$$

The curve (or tangent) has a negative slope between B and D, and between F and G.

Example 6.1.1   Given the curve $y = \frac {x^3}{3} - x^2 + 2$ (see Figure 6.2).

(a) Find $\tau$ when $x=1$.

(b) Find $\tau$ when $x=3$.

(c) Find the points where the curve is parallel to the $x$-axis.

(d) Find the points where $\tau = 45^o$.

(e) Find the points where the curve is parallel to the line $2x - 3y = 6$.

Figure 6.2: The graph of $y = \frac {x^3}{3} - x^2 + 2$.

Differentiating, $\frac{dy}{dx} = x^2 - 2x$ = slope at any point.

(a) $\tan \tau = \left [ \frac{dy}{dx} \right ]_{x=1} = 1 - 2 = -1$; therefore $\tau = 135^o =3\pi/4$.

(b) $\tan \tau = \left [ \frac{dy}{dx} \right ]_{x=3} = 9 - 6 = 3$; therefore $\tau = \arctan\, 3 = 1.249...$.

(c) $\tau = 0^o$, $\tan \tau = \frac{dy}{dx} = 0$; therefore $x^2 - 2x = 0$. Solving this equation, we find that $x = 0$ or $2$, giving points C and D where the curve (or tangent) is parallel to the $x$-axis.

(d) $\tau = 45^o$, $\tan \tau = \frac{dy}{dx} = 1$; therefore $x^2 - 2x = 1$. Solving, we get $x = 1 \pm \sqrt{2}$, giving two points where the slope of the curve (or tangent) is unity.

(e) Slope of line = $\frac{2}{3}$; therefore $x^2 - 2x = \frac{2}{3}$. Solving, we get $x = 1 \pm \sqrt{\frac{5}{3}}$, giving points E and F where curve (or tangent) is parallel to $2x - 3y = 6$.

Since a curve at any point has the same direction as its tangent at that point, the angle between two curves at a common point will be the angle between their tangents at that point.

Example 6.1.2   Find the angle of intersection of the circles

(A) $x^2 + y^2 - 4x = 1$,

(B) $x^2 + y^2 - 2y = 9$.

Solution. Solving simultaneously, we find the points of intersection to be $(3, 2)$ and $(1, -2)$. This can be verified ``by hand'' or using the Sage solve command:

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: x = var("x")
sage: y = var("y")
sage: F = x^2 + y^2 - 4*x - 1
sage: G = x^2 + y^2 - 2*y - 9
sage: solve([F == 0,G == 0],x,y)
[[x == 1, y == -2], [x == 3, y == 2]]

Figure 6.3: The graphs of $x^2 + y^2 - 4x = 1$, $x^2 + y^2 - 2y = 9$.

Using (A), formulas in §5.35 give $\frac{dy}{dx} = \frac{2 - x}{y}$. Using (B), formulas in §5.35 give $\frac{dy}{dx} = \frac{x}{1 - y}$. Therefore,

$$\left [ \frac{2 - x}{y} \right ]_{x = 3, y = 2} = -\frac{1}{2} =\ {\rm slope\ of\ tangent\ to\ (A)\ at\ } (3, 2).$$

$$\left [ \frac{x}{1 - y} \right ]_{x = 3, y = 2} = - 3 =\ {\rm slope\ of\ tangent\ to\ (B)\ at\ }(3, 2).$$

We can check this using the commands

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: x = var("x")
sage: y = function("y",x)
sage: F = x^2 + y^2 - 4*x - 1
sage: F.diff(x)
2*y(x)*diff(y(x), x, 1) + 2*x - 4
sage: solve(F.diff(x) == 0, diff(y(x), x, 1))
[diff(y(x), x, 1) == (2 - x)/y(x)]
sage: G = x^2 + y^2 - 2*y - 9
sage: G.diff(x)
2*y(x)*diff(y(x), x, 1) - 2*diff(y(x), x, 1) + 2*x
sage: solve(G.diff(x) == 0, diff(y(x), x, 1))
[diff(y(x), x, 1) == -x/(y(x) - 1)]

The formula for finding the angle between two lines whose slopes are $m_1$ and $m_2$ is

$$\tan\theta = \frac{m_1 - m_2}{1 + m_1 m_2},$$

by item 55, §1.1. Substituting, $\tan \theta = \frac{-\frac{1}{2} + 3}{1 + \frac{3}{2}} = 1$; therefore $\theta = \pi/4= 45^o$. This is also the angle of intersection at the point $(1, -2)$.