# 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.

## Related Rates

**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 \(\ln(1.3) \in (0.255,0.264)\).