Applications of differentiation

Find the equations of tangents and normals to a curve

  • Theory
    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 \(x=x_0\)
  • Examples
    Find the equations of the tangents to the curve \(y = - {x^2} + 6x + 2\) where the tangents pass through (-1, 4).
  • Solutions
    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

  • Theory
    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)
  • Examples
    For the graph of the function \(y = \frac{x}{{{{({x^2} + 3)}^2}}}\), find the turning point(s) and point(s) of inflexion.
  • Solutions

    \(y = \frac{x}{{{{({x^2} + 3)}^2}}}\)
    \(\begin{array}{1}\frac{{dy}}{{dx}} = \frac{{{{({x^2} + 3)}^2} - x(2)({x^2} + 3)(2x)}}{{{{({x^2} + 3)}^4}}}\\
    = {\frac{{({x^2} + 3)({x^2} + 3 - 4{x^2})}}{{{{({x^2} + 3)}^4}}}^{}}\\
    = - {\frac{{3(x + 1)(x - 1)}}{{{{({x^2} + 3)}^3}}}^{}}\end{array}\)
    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

  • Theory
  • Examples
    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.
  • Solutions
    (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

  • Theory
  • Examples
    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]
  • Solutions
    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]