# More Applications of Differentiation - Examples

On this page we display several examples to illustrate the concepts and theorems discussed in the More Applications of Differentiation theme.

Problem
A ball of ice with radius $r=6 \ \mathrm{cm}$ is melting. How fast is the volume decreasing if the radius is decreasing with $0.5 \ \mathrm{cm/h}$?

Solution

We have that $V(t) = \frac{4}{3}\pi r^3(t)$ This implies that $V'(t) = 4\pi r^2(t)r'(t) =4\pi(6 \ \mathrm{cm})^2 \cdot 0.5 \ \mathrm{cm/h} =72\pi \ \mathrm{cm^3/h},$ so the volume of the ball is decreasing at a rate of $72\pi \approx 226.2 \ \mathrm{cm^3/h}$.

## Newton's Method

Problem
a) Explain why the equation $x\ln x =1$ only has one solution.

b) Use Newton's Method to approximate this solution.

Solution
a) Let $f(x) = x\ln x -1$. We see that this function is defined on $(0, \infty$), and $f'(x) = 1+\ln x$. Furthermore, $f'(x)< 0$ on $(0, \mathrm{e^{-1}})$, and $f'(x)> 0$ on $( \mathrm{e^{-1}}, \infty )$. Since $\lim_{x \to 0+}f(x) = -1$ we must have that $f(x)\leftarrow1$ for $x \in (0, \mathrm{e^{-1}})$. Because $f$ is increasing for all $x \in ( \mathrm{e^{-1}}, \infty )$, it can be zero at most once in this interval. Since $f(\mathrm{e^{-1}}) <0$ and $f(\mathrm{e}) >0$, the Intermediate value theorem gives that $f(c)=0$ for some $c \in ( \mathrm{e^{-1}}, \mathrm{e})$. This therefore has to be the only solution.

b) We let $x_0=1$, and define $x_{n+1} =x_n -\frac{x_n\ln x_n-1}{1+\ln x_n}$. Using a calculator or Maple, this gives $x_1 = 2.0, \ x_2 = 1.772, \ x_3 = 1.7632, \ x_4 = 1.7632,$ and so a good approximation to the solution is $1.7632$.

## Using l'Hôpital's rule

Problem

Calculate $\lim_{x \to 0}\frac{\tanh x}{x}$.

Solution

We see that as $x \to 0$, $(\tanh x)/x$ approaches $0/0$, so we can use L'Hôpital's Rule. This gives $\lim_{x \to 0}\frac{\tanh x}{x} =\lim_{x \to 0}\frac{(\tanh x)'}{(x)'} = \lim_{x \to 0}\frac{1}{\cosh^2 x} =1.$

## Finding extreme values

Problem
Let $f(x) = \frac{x^2}{\sqrt{4-x^2}}$. Determine if the function has absolute maximum or absolute minimum values, and find them if you can.

Solution
The domain of $f$ is given by those $x$ such that $4-x^2 \ge 0$, i.e. $x \in (-2,2)$. Since $\lim_{x \to 2-} f(x) = \infty$, it's clear that $f$ doesn't have an absolute maximum value. To determine if $f$ has an absolute minimum value we use Theorem 6: a local extreme value can only occur in a point $x$ where $f'(x)$ is zero (critical point) or undefined (singular point), or in an endpoint. Since $(-2,2)$ doesn't include any endpoints, and $f'$ is defined everywhere in $(-2,2)$, it's enough to examine where the derivative is zero. We have that $f'(x) = \frac{2x(4-x^2)-x^2(-2x)}{4-x^2} = \frac{8x}{4-x^2},$ so the only point of interest is $x=0$. Since $f'(x)$ is negative to the left of $x=0$ and positive to the right, we conclude that $f(0) = 0$ is a local minimum. Since $f(x) > 0$ for $x \neq 0$, we find that $0$ is the global minimum (and it only occurs for $x=0$).

## Concavity

Problem
Let $f(x) = \ln(x)^2$. Determine the concavity of $f$.

Solution
We have that $f'(x) = 2\frac{\ln(x)}{x}, \qquad f''(x) = 2\frac{1-\ln(x)}{x^2}.$ Considering the sign of $f''(x)$, we see that it is convex for $x \in (0,e)$, and concave for $x \in (e,\infty)$.

## Approximating a function using polynomials

Problem
Use the second order Taylor polynomial for $f(x) = \ln(x)$ about $1$ to approximate $\ln(1.3)$, and give a bound on the error of the approximation.

Solution
We have that $f'(x) = \frac{1}{x}$, $f''(x) = \frac{-1}{x^2}$. The Taylor polynomial about $1$ is $P(x) = f(1) + f'(1)(x-1) + \frac12 f''(1) (x-1)^2 = (x-1) - \frac12 (x-1)^2.$ The approximation is thus $P(1.3) = 0.255$. We can give an estimate of the error of this approximation using Taylor's Theorem, which tells us that $f(1.3)-P(1.3) = \frac{f'''(s)(s-1)^3}{3!},$ where $s$ is some number between $1$ and $1.3$. Since $f'''(x) = \frac{2}{x^3}$, we get $|f'''(s)| < 2$ and so $|f(1.3)-P(1.3)| < \left|\frac{2(1.3-1)^3}{3!} \right| = 0.009.$ We conclude that $\ln(1.3) \in (0.246,0.264)$. Actually, since we must have $f'''(s)>0$, we can even say that $\ln(1.3) \in (0.255,0.264)$.