### Applications of Differentiation Part III

Stationary points
Stationary points occur when $\frac{dy}{dx} = 0$

A stationary point can be a:
• maximum point: $\frac{d^2y}{dx^2} \quad \text{lesser than 0}$
• minimum point: $\frac{d^2y}{dx^2} > 0$
• point of inflexion: $\frac{d^2y}{dx^2} = 0$ Second derivative is inconclusive
Example 1
Given $y =13-20x + 4x^2 +3x^3$ Find the maximum and minimum values of y and state where those values occur.

\begin{aligned} \frac{ dy }{ dx } &= -20 +8x +9x^2 \\ \text{To find stationary values of x} \\ \frac{ dy }{ dx } &= 0 \\ -20 +8x + 9x^2 &=0 \\ ( 9x - 10) ( x + 2) &= 0 \\ x &= \frac{10}{9} \qquad x = -2 \\ \text{next we determine the second derivative} \\ \frac{d^2y}{dx^2} &= 18x + 8 \\ \text{When} \: x = \frac{10}{9} \qquad \frac{d^2y}{dx^2} = 18 \times \frac{10}{9} + 8 = 28 \\ \text{hence value of y is minimum when} \: x = \frac{10}{9} \\ \text{When} \: x = \frac{10}{9} \qquad y = 13-20(\frac{10}{9}) + 4(\frac{10}{9})^2 +3(\frac{10}{9})^3 \\ \end{aligned}

Example 2
A curve has the equation $y = \frac{2x-4}{x+3}$ Obtain an expression for $\frac{dy}{dx}$ and hence explain why the curve has no turning points.

To find $\frac{dy}{dx}$ we make use of quotient rule
\begin{aligned} \frac{dy}{dx} &= \frac{(x+3)(2) - (2x-4) }{(x+3)^2} \\ &= \frac{2x + 6 -2x +4 }{(x+3)^2} \\ &= \frac{10}{(x+3)^2} \\ \end{aligned}

$\frac{dy}{dx} =0$ has no solutions since $10 \neq 0$ and denominator cannot equal zero. Hence curve has no turning points.

Example 3
The diagram shows a cross section of a cone with height 15 cm and radius 6cm and a cylinder with radius R cm and height H cm. The upper edge of the cylinder is in contact with the cone

a) Making use of similar triangles, express H in terms of R

Solution
\begin{aligned} \frac{R}{6} &= \frac{15-H}{15} \\ 6H &= 90 -15R \\ H &= 15 - \frac{5}{2} R \\ \end{aligned}

b)Hence show that the volume of the cylinder is given by $\frac{1}{3} \pi (15R^2 - \frac{5}{2} R^3)$

\begin{aligned} V &= \frac{1}{3} \pi R^2 H \\ &=\frac{1}{3} \pi (R^2)(15 - \frac{5}{2} R) \\ &=\frac{1}{3} \pi (15R^2 - \frac{5}{2} R^3) \\ \end{aligned}

c) Given that R can vary, find the volume of the largest cylinder in terms of $\pi$ that can fit inside the cone

From (b): $\ V = \frac{1}{3} \pi (15R^2 - \frac{5}{2} R^3)$

To find max volume,
• Find $\frac{dV}{dR}$
• Set $\frac{dV}{dR} = 0$
• Find value of R
• Find volume

Find $\frac{dV}{dR}$
$\frac{dV}{dR} = \frac{1}{3} \pi (30R - \frac{5}{2} (3R^2))$

Set $\frac{dV}{dR} = 0$
\begin{aligned} \frac{dV}{dR} &= 0 \\ \frac{1}{3} \pi (30R - \frac{5}{2} (3R^2)) &= 0 \\ 30R - \frac{5}{2} (3R^2) &= 0 \\ 30 - \frac{5}{2} (3R) &= 0 \\ \frac{5}{2} (3R) &= 30 \\ R &= 4 \\ \end{aligned}

\begin{aligned} V &= \frac{1}{3} \pi (15R^2 - \frac{5}{2} R^3) \\ &= \frac{\pi}{3}( 15 \times 4^2 - \frac{5}{2} \times 4^3)\\ &= \frac{\pi}{3}(80) \\ &= \frac{80\pi}{3}\\ \end{aligned}