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} $