## Monthly Archives: July 2018

### Applications of Derivatives: A Quick Review

Section I:

The Derivative as a Rate of Change

In case of a linear function $y=mx+c$, the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x.

As x changes from $x_{0}$ to $x_{1}$, y changes m times as much:

$y_{1}-y_{0}=m(x_{1}-x_{0})$

Thus, the slope $m=(y_{1}-y_{0})(x_{1}-x_{0})$ gives the change in y per unit change in x.

In more general case of differentiable function $y=f(x)$, the difference quotient

$\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}$, where $h \neq 0$

give the average rate of change of y (or f) with respect to x. The limit as h approaches zero is the derivative $dy/dx = f^{'}(x)$, which can be interpreted as the instantaneous rate of change of f with respect to x. Since, the graph is a curve, the rate of change of y can vary from point to point.

Velocity and Acceleration:

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has location/position $x(t)$. Then, at time $t+h$ the position of the object is $x(t+h)$ and $x(t+h)-x(t)$ is the change in position that the object experienced during the time period t to $t+h$. The ratio

$\frac{x(t+h)-x(t)}{t+h-t} = \frac{x(t+h)-x(t)}{h}$

gives the average velocity of the object during this time period. If

$\lim_{h \rightarrow 0} \frac{x(t+h)-x(t)}{h}=x^{'}(t)$

exists, then $x^{'}(t)$ gives the instantaneous rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,

$a(t) = v^{'}(t) = x^{''}(t)$

The speed is by definition the absolute value of the velocity: speed at time t is $|v(t)|$

If the velocity and acceleration have the same sign, then the object is speeding up, but if the velocity and acceleration have opposite signs, then the object is slowing down.

A sudden change in acceleration is called a jerk. Jerk is the derivative of acceleration. If a body’s position at the time t is $x(t)$, the body’s jerk at time t is

$j = \frac{da}{dt} = \frac{d^{3}x}{dt^{3}}$

Differentials

Let $y = f(x)$ be a differentiable function. Let $h \neq 0$. The difference $f(x+h) - f(x)$ is called the increment of f from x to $x+h$, and is denoted by $\Delta f$.

$\Delta f = f(x+h) - f(x)$

The product $f^{'}(x)h$ is called the differential of f at x with increment h, and is denoted by $df$

$df = f^{'}(x)h$

The change in f from x to $x+h$ can be approximated by $f^{'}(x)h$:

$f(x+h) - f(x) = f^{'}(x)h$

Tangent and Normal

Let $y = f(x)$ be the equation of a curve, and let $P(x_{0}, y_{0})$ be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis.

The slope of the tangent to the curve $y = f(x)$ at P is given by $(\frac{dy}{dx})_{(x_{0}, y_{0})}$

Thus, the equation of the tangent to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is $y - y_{0} = (\frac{dy}{dx})_{(x_{0}, y_{0})}(x-x_{0})$

Since PM is perpendicular to PT, it follows that if $(\frac{dy}{dx})_{(x_{0}, y_{0})} \neq 0$, the slope of PN is

$- \frac{1}{(\frac{dy}{dx})_{(x_{0}, y_{0})}} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}$

Hence, the equation of the normal to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is

$y - y_{0} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}(x-x_{0})$

The equation of the normal parallel to the x-axis is $y = y_{0}$, that is, when $(\frac{dy}{dx})_{(x_{0}, y_{0})} = 0$. The length of the tangent at $(x_{0}, y_{0})$ is PT, and it is equal to

$y_{0}\csc{\theta} = y_{0}\sqrt{1+\cot^{2}{\theta}} = y_{0}\sqrt{1+[(\frac{dx}{dy})_{(x_{0}, y_{0})}]^{2}}$

The length of the normal is PN and it is equal to $y_{0}\sec {\theta} = y_{0}\sqrt{1 + [(\frac{dy}{dx})_{(x_{0}, y_{0})}]^{2}}$

If the curve is represented by $x = f(t)$ and $y = g(t)$, that is, parametric equations in t, then

$\frac{dy}{dx} = \frac{g^{'}(t)}{f^{'}(t)}$ where $g^{'}(t)= \frac{dy}{dt}$ and $f^{'}(t) = \frac{dx}{dt}$. In this case, the equations of the tangent and the normal are given by

$y - g(t) = \frac{g^{'}(t)}{f^{'}(t)}[x - f(t)]$ and $[y-g(t)] g^{'}(t) + [x-f(t)]f^{'}(t) = 0$ respectively.

The Angle between Two Curves

The angle of intersection of two curves is defined as the angle between the two tangents at the point of intersection. Let $y = f(x)$ and $y=g(x)$ be two curves, and let $P(x_{0}, y_{0})$ be their point of intersection. Also, let $\psi$ and $\phi$ be the angles of inclination of the two tangents with the x-axis, and let $\theta$ be the angle between the two tangents. Then,

$\tan {\theta} = \frac{\tan{\phi}-\tan{\psi}}{1+\tan{\phi}\tan{\psi}} = \frac{g^{'}(x) - f^{'}(x)}{1+f^{'}(x)g^{'}(x)}$

Example 1:

Write down the equations of the tangent and the normal to the curve $y = x^{3} - 3x + 2$ at the point $(2,4)$.

Solution 1:

$\frac{dy}{dx} = 3x^{2}-3 \Longrightarrow \frac{dy}{dx}_{(2,4)} = 3.4 - 3 = 9$.

Hence, the equation of the tangent at $(2,4)$ is given by $y-4 = 9(x-2) \Longrightarrow 9x-y-14=0$ and the equation of the normal is $y - 4 = (-1/9)(x-2) \Longrightarrow x+9y -38=0$.

Rolle’s Theorem and Lagrange’s Theorem:

Rolle’s Theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) f(x) is continuous on $[a,b]$, (ii) f(x) is derivable on $(a,b)$, and (iii) f(a) = f(b). Then, there exists a $c \in (a,b)$ such that $f^{'}(x)=0$.

For details, the very beautiful, lucid, accessible explanation in Wikipedia:

https://en.wikipedia.org/wiki/Rolle%27s_theorem

Lagrange’s theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) $f(x)$ is continuous on $[a,b]$, and (ii) $f(x)$ is derivable on $(a,b)$. Then, there exists a $c \in [a,b]$ such that

$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$

Example 2:

The function $f(x) = \log {\sin(x)}$ satisfies the conditions of Rolle’s theorem on the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$, as the logarithmic function and $\sin (x)$ are continuous and differentiable functions and $\log {\sin (\frac{5\pi}{6})} = \log {\sin (\pi - \frac{\pi}{6})} = \log{\sin{(\frac{\pi}{6})}}$.

The conclusion of Rolle’s theorem is given at $c=\frac{\pi}{2}$, for which $f^{'}(c) = \cot (c) = \cot (\pi/2) =0$.

Rolle’s theorem for polynomials:

If $\phi(x)$ is any polynomial, then between any pair of roots of $\phi(x)=0$ lies a root of $\phi^{'}(x)=0$.

Monotonicity:

A function $f(x)$ defined on a set D is said to be non-decreasing, increasing, non-increasing and decreasing respectively, if for any $x_{1}, x_{2} \in D$ and $x_{1} < x_{2}$, we have $f(x_{1}) \leq f(x_{2})$, $f(x_{1}) < f(x_{2})$, $f(x_{1}) \geq f(x_{2})$ and $f(x_{1}) > f(x_{2})$ respectively. The function $f(x)$ is said to be monotonic if it possesses any of these properties.

For example, $f(x) = e^{x}$ is an increasing function, and $f(x)=\frac{1}{x}$ is a decreasing function.

Testing monotonicity:

Let $f(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then,

(i) for $f(x)$ to be non-decreasing (non-increasing) on $[a,b]$ it is necessary and sufficient that $f^{'}(x) \geq 0$ ($f^{'}(x) \leq 0$) for all $x \in (a,b)$.

(ii) for $f(x)$ to be increasing (decreasing) on $[a,b]$ it is sufficient that $f^{'}(x)>0$ ($f^{'}(x)<0$) for all $x \in (a,b)$.

(iii) If $f^{'}(x)=0$ for all x in $(a,b)$, then f is constant on $[a,b]$.

Example 3:

Determine the intervals of increase and decrease for the function $f(x)=x^{3}+2x-5$.

Solution 3:

We have $f^{'}(x) = 3x^{2}+2$, and for any value of x, $3x^{2}+2>0$. Hence, f is increasing on $(-\infty, -\infty)$. QED.

The following is a simple criterion for determining the sign of $f^{'}(x)$:

If $a,b \geq 0$, then $(x-a)(x-b)>0$ iff $x > \max (a,b)$ or $x < \min(a,b)$;

$(x-a)(x-b)<0$ if and only if $\min(a,b) < x < \max(a,b)$

Maxima and Minima:

A function has a local maximum at the point $x_{0}$ if the value of the function $f(x)$ at that point is greater than its values at all points other than $x_{0}$ of a certain interval containing the point $x_{0}$. In other words, a function $f(x)$ has a maximum at $x_{0}$ if it is possible to find an interval $(\alpha, \beta)$ containing $x_{0}$, that is, with $\alpha < x_{0} < \beta$, such that for all points different from $x_{0}$ in $(\alpha, \beta)$, we have $f(x) < f(x_{0})$.

A function $f(x)$ has a local minimum at $x_{0}$ if there exists an interval $(\alpha, \beta)$ containing $x_{0}$ such that $f(x) > f(x_{0})$ for $x \in (\alpha, \beta)$ and $x \neq x_{0}$.

One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at all points sufficiently close to the point of local maximum. Similarly, the local minimum is the smallest value only in comparison to the values of the function at all points sufficiently close to the local minimum point.

The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point $x_{0}$ of the function $f(x)$ is that $f^{'}(x_{0})=0$, or $f^{'}(x_{0})$ does not exist. The points at which $f^{'}(x)=0$ or $f^{'}(x)$ does not exist, are called critical points.

First Derivative Test:

(i) If $f^{'}(x)$ changes sign from positive to negative at $x_{0}$, that is, $f^{'}(x)>0$ for $x < x_{0}$ and $f^{'}(x)<0$ for $x > x_{0}$, then the function attains a local maximum at $x_{0}$.

(ii) If $f^{'}(x)$ changes sign from negative to positive at $x_{0}$, that is, $f^{'}(x)<0$ for $x, and $f^{'}(x)>0$ for $x > x_{0}$, then the function attains a local minimum at $x_{0}$.

(iii) If the derivative does not change sign in moving through the point $x_{0}$, there is no extremum at that point.

Second Derivative Test:

Let f be twice differentiable, and let c be a root of the equation $f^{'}(x)=0$. Then,

(i) c is a local maximum point if $f^{''}(c)<0$.

(ii) c is a local minimum point if $f^{''}(c)>0$.

However, if $f^{''}(c)=0$, then the following result is applicable. Let $f^{'}(c) = f^{''}(c) = \ldots = f^{n-1}(c)=0$ (where f^{r} denotes the rth derivative), but $f^{(n)}(c) \neq 0$.

(i) If n is even and $f^{(n)}(c)<0$, there is a local maximum at c, while if $f^{(n)}(c)>0$, there is a local minimum at c.

(ii) If n is odd, there is no extremum at the point c.

Greatest/Least Value (Absolute Maximum/Absolute Minimum):

Let f be a function with domain D. Then, f has a greatest value (or absolute maximum) at a point $c \in D$ if $f^(x) \leq f(c)$ for all x in D and a least value (or absolute minimum) at c, if $f(x) \geq f(c)$ for all x in D.

If f is continuous at every point of D, and $D=[a,b]$, a closed interval, the f assumes both a greatest value M and a least value m, that is, there are $x_{1}, x_{2} \in [a,b]$ such that $f(x_{1})=M$ and $f(x_{2})=m$, and $m \leq f(x) \leq M$ for every $x \in [a,b]$.

Example 4:

a) $y=x^{2}$, with domain $(-\infty, \infty)$. This has no greatest value; least value at $x=0$

b) $y=x^{2}$ with domain $[0,2]$. This has greatest value at $x=2$ and least value at $x=0$.

c) $y=x^{2}$ with domain $(0,2]$. This has greatest value at $x=2$ and no least value.

d) $y=x^{2}$ with domain $(0,2)$. This has no greatest value and no least value.

Some other remarks:

The greatest (least) value of continuous function $f(x)$ on the interval $[a,b]$ is attained either at the critical points or at the end points of the interval. To find the greatest (least) value of the function, we have to compute its values at all the critical points on the interval $(a,b)$, and the values $f(a), f(b)$ of the function at the end-points of the interval, and choose the greatest (least) out of the values so obtained.

We will continue with problems on applications of derivatives later,

Nalin Pithwa.