### Applications of differentiation

#### Find the equations of tangents and normals to a curve

• Tangent to the curve of a function at a point is a line touches that point.
Normal to the curve of a function at a point is a line perpendicular to the tangent to the curve at that point.
Say, f(x) is continuous and differentiable (continuity and differentiability are out of the context of DSE Math) at $$x=x_0$$. Equation of the tangent is given by
$$y=f'(x_0)x-f'(x_0)x_0+f(x_0)$$
where f'(x) denotes the first derivatives. Putting $$x=x_0$$ in f'(x) yield the slope of tangent at $$x=x_0$$
• Find the equations of the tangents to the curve $$y = - {x^2} + 6x + 2$$ where the tangents pass through (-1, 4).
• Let $$({x_1}{\rm{ }},{\rm{ }}{y_1})$$ be the point of contact of a required tangent. Since $$({x_1}{\rm{ }},{\rm{ }}{y_1})$$ lies on the curve,
∴ $${y_1} = - {x_1}^2 + 6{x_1} + 2\,...................\;(1)$$
$$\frac{{dy}}{{dx}} = - 2x + 6$$
Slope of the tangent at $$({x_1},{\rm{ }}{y_1}) = {\left. {\frac{{dy}}{{dx}}{\kern 1pt} } \right|_{{\kern 1pt} ({x_1}{\rm{ }},{\rm{ }}{y_1})}} = - {\rm{ }}2{x_1} + 6$$
Slope of the line joining $$({x_1}{\rm{ }},{\rm{ }}{y_1})$$ and $$( - {\rm{ }}1{\rm{ }},{\rm{ }}4)$$$$= \frac{{{y_1} - 4}}{{{x_1} + 1}}$$
∴ $$\begin{array}{c}\frac{{{y_1} - 4}}{{{x_1} + 1}} = - {\rm{ }}2{x_1} + 6 \\{y_1} - 4 = ( - {\rm{ }}2{x_1} + 6)({x_1} + 1) \\{y_1} = - {\rm{ }}2{x_1}^2 + 4{x_1} + 10\,........\;(2) \end{array}$$
(1) - (2): $$\begin{array}{c}0 = {x_1}^2 + 2{x_1} - 8\\({x_1} + 4)({x_1} - 2) = {0^{}}\end{array}$$
$${x_1} = - {\rm{ }}4$$ or $${x_1} = 2$$

When $${x_1} = - {\rm{ }}4$$,
$$\begin{array}{c}{y_1} = - {\rm{ }}{( - {\rm{ }}4)^2} + 6( - {\rm{ }}4) + 2\\ = - {\rm{ }}{38^{}}\end{array}$$
∴ (-4, -38) is a point of contact.
Slope of the tangent at $$\begin{array}{c}( - {\rm{ }}4{\rm{ }},{\rm{ }} - {\rm{ }}38) = - {\rm{ }}2( - {\rm{ }}4) + 6\\ = 14\end{array}$$
∴ The equation of the tangent at (-4, -38) is
$$\begin{array}{c}y + 38 = 14(x + 4)\\14x - y + 18 = 0\end{array}$$[can't underline]

When $${x_1} = 2$$,
$$\begin{array}{c}{y_{\rm{1}}} = - {\rm{ }}{(2)^2} + 6(2) + 2\\ = 10\end{array}$$
∴ (2, 10) is another point of contact.
Slope of the tangent at $$\begin{array}{c}(2{\rm{ }},{\rm{ }}10) = - {\rm{ }}2(2) + 6\\ = 2\end{array}$$
∴ The equation of the tangent at (2, 10) is
$$\begin{array}{c}y - 10 = 2(x - 2)\\2x - y + 6 = 0\end{array}$$[can't underline]

#### Find maxima and minima

• Derivative of a function, f(x), defines the slope of that function at any point. Particularly, when
• $$f(x) > 0$$ over $$x_1 < x < x_0$$, f(x) is increasing
• $$f(x) < 0$$ over $$x_0 < x < x_2$$, f(x) is decreasing
• $$x_0$$ is the local maxima of f(x)
• $$f(x) < 0$$ over $$x_1 < x < x_0$$, f(x) is decreasing
• $$f(x) > 0$$ over $$x_0 < x < x_2$$, f(x) is increasing
• $$x_0$$ is the local minima of f(x)
• For the graph of the function $$y = \frac{x}{{{{({x^2} + 3)}^2}}}$$, find the turning point(s) and point(s) of inflexion.

• $$y = \frac{x}{{{{({x^2} + 3)}^2}}}$$
$$\begin{array}{1}\frac{{dy}}{{dx}} = \frac{{{{({x^2} + 3)}^2} - x(2)({x^2} + 3)(2x)}}{{{{({x^2} + 3)}^4}}}\\ = {\frac{{({x^2} + 3)({x^2} + 3 - 4{x^2})}}{{{{({x^2} + 3)}^4}}}^{}}\\ = - {\frac{{3(x + 1)(x - 1)}}{{{{({x^2} + 3)}^3}}}^{}}\end{array}$$
Take $$\frac{{dy}}{{dx}} = 0$$, we have $$x = - 1$$ or $$x = 1$$.

∴ The maximum point is $$(1{\rm{ }},{\rm{ }}\frac{1}{{16}})$$ and the minimum point is $$( - 1{\rm{ }},{\rm{ }} - {\rm{ }}\frac{1}{{16}})$$.
$$\begin{array}{1}\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{{({x^2} + 3)}^3}(3)[(x + 1) + (x - 1)] - 3(x + 1)(x - 1)(3){{({x^2} + 3)}^2}(2x)}}{{{{({x^2} + 3)}^6}}}\\ = - {\frac{{6x{{({x^2} + 3)}^2}[({x^2} + 3) - 3(x + 1)(x - 1)]}}{{{{({x^2} + 3)}^6}}}^{}}\\ = {\frac{{12x({x^2} - 3)}}{{{{({x^2} + 3)}^4}}}^{}}\end{array}$$
Take $$\frac{{{d^2}y}}{{d{x^2}}} = 0$$, we have $$x = 0$$ or $$x = \pm \sqrt 3$$.
∴ The points of inflexion are $$( - {\rm{ }}\sqrt 3 {\rm{ }},{\rm{ }} - {\rm{ }}\frac{{\sqrt 3 }}{{36}})$$, (0, 0) and $$(\sqrt 3 {\rm{ }},{\rm{ }}\frac{{\sqrt 3 }}{{36}})$$.

#### Sketch curves of polynomial functions and rational functions

• A curve $$y = \frac{{2{x^3}}}{{{{(x - 2)}^2}}}$$ is given.
(a) Find $$\frac{{dy}}{{dx}}$$ and $$\frac{{{d^2}y}}{{d{x^2}}}$$.
(b) Find the turning point(s) and point(s) of inflexion of the curve.
(c) Find the asymptote(s) of the curve.
(d) Sketch the curve.
• (a) $$y = \frac{{2{x^3}}}{{{{(x - 2)}^2}}}$$
$$\begin{array}{1}\frac{{dy}}{{dx}} = \frac{{{{(x - 2)}^2}(6{x^2}) - 2{x^3}(2)(x - 2)}}{{{{(x - 2)}^4}}}\\ = \underline{\underline {\frac{{2{x^2}(x - 6)}}{{{{(x - 2)}^3}}}}} \end{array}$$
$$\begin{array}{1}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{{(x - 2)}^3}(6{x^2} - 24x) - 2{x^2}(x - 6)(3){{(x - 2)}^2}}}{{{{(x - 2)}^6}}}\\ = \underline{\underline {\frac{{48x}}{{{{(x - 2)}^4}}}}} \end{array}$$

(b) Take $$\frac{{dy}}{{dx}} = 0$$, we have $$x = 0$$ or 6.
∴ The minimum point is (6, 27). Take $$\frac{{{d^2}y}}{{d{x^2}}} = 0$$, we have $$x = 0$$.
∴ The point of inflexion is (0, 0). (c)∵ $$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{2{x^3}}}{{{{(x - 2)}^2}}} = \infty$$
∴ $$x = 2$$ is a vertical asymptote.
∵ $$\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^3}}}{{{{(x - 2)}^2}}}$$ and $$\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^3}}}{{{{(x - 2)}^2}}}$$ do not exist.
∴ $$y = \frac{{2{x^3}}}{{{{(x - 2)}^2}}}$$ has no horizontal asymptotes.
Also, $$\begin{array}{1}y = \frac{{2{x^3}}}{{{{(x - 2)}^2}}}\\ = \frac{{8(3x - 4)}}{{{{(x - 2)}^2}}} + 2x + 8\end{array}$$
∵ $$\begin{array}{1}\mathop {\lim }\limits_{x \to \pm \infty } \frac{{8(3x - 4)}}{{{{(x - 2)}^2}}} = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{8({\textstyle{3 \over x}} - {\textstyle{4 \over {{x^2}}}})}}{{{{(1 - {\textstyle{2 \over x}})}^2}}}\\ = \frac{{8(0 - 0)}}{{{{(1 - 0)}^2}}}\\ = {0^{}}\end{array}$$
∴ $$y = 2x + 8$$ is an oblique asymptote.

(d) The sketch of the graph of $$y = \frac{{2{x^3}}}{{{{(x - 2)}^2}}}$$ is as follows:

#### Solve the problems relating to rate of change, maximum and minimum

• The figure shows a container of 3 m long with a uniform cross-section of an isosceles triangle, where $$AB = AC$$ and $$\angle BAC = 120^\circ$$. If water is flowing at a rate of $$1{\rm{ }}{{\rm{m}}^{\rm{3}}}/\min$$ into the container, find the depth of water when the water level is rising at a rate of $$\frac{1}{{\sqrt 3 }}{\rm{ m}}/\min$$. [GraphMissing M2E08_2 Q155]
• Let $$V{\rm{ }}{{\rm{m}}^{\rm{3}}}$$ be the volume of water in the container and h m be the depth of water.
Consider the following.
[GraphMissing M2E08_2 Q155]
$$\begin{array}{1}\angle MAY = \frac{{120^\circ }}{2}\\ = 60^\circ \\ \tan 60^\circ = {\frac{{MY}}{{MA}}^{}}\\ MY = \sqrt 3 {h^{^{^{}}}}\end{array}$$
∴ $$XY = 2MY = 2\sqrt 3 h$$
$$\begin{array}{1}V = \frac{1}{2}(2\sqrt 3 h)h \times 3\\ = 3\sqrt 3 {h^2}^{^{}}\end{array}$$
$$\begin{array}{1}\frac{{dV}}{{dt}} = 3\sqrt 3 (2h)\frac{{dh}}{{dt}}\\ = 6\sqrt 3 h{\frac{{dh}}{{dt}}^{^{}}}\end{array}$$
When $$\frac{{dV}}{{dt}} = 1$$ and $$\frac{{dh}}{{dt}} = \frac{1}{{\sqrt 3 }}$$,
$$\begin{array}{1}1 = 6\sqrt 3 h(\frac{1}{{\sqrt 3 }})\\ h = \frac{1}{6}\end{array}$$
∴ The depth of water is $$\frac{1}{6}{\rm{ m}}$$ when the water level is rising at a rate of $$\frac{1}{{\sqrt 3 }}{\rm{ m}}/\min$$.[Cant Underline]