MATH 130 -- Tutorial Exercises -- Solutions

1999 D1 & E1, Mathematics IE

Week 8

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Calculus

Answer 1         (Question)

(i)

For f (x) = (1 + x²)(1 - x²), using the product rule we have that $${\frac{\dD f}{\dD x}}$$ = 2x(1 - x²) + (1 + x²)(-2x) = -2x³ - 2x³ = -4x³.
(Note that this is more easily obtained from f (x) = (1 + x²)(1 - x²) = 1 - x⁴.)

(ii)

To evaluate $${\frac{\dD f}{\dD x}}$$ for f (x) = $${\frac{\frac{4}x + x^2}{2x - \frac3{x^2}}}$$, it is best to first simplify: $${\frac{\frac{4}x + x^2}{2x - \frac3{x^2}}}$$     =    $${\frac{\frac{4+x^3}{x}}{\frac{2x^3-3}{x^2}}}$$     =    $${\frac{(4+x^3)x}{2x^3-3}}$$. Now use the quotient rule to get

$${\frac{\dD f}{\dD x}}$$     =    $${\frac{(4+4x^3)(2x^3-3)-(4x+x^4)(6x^2)}{(2x^3-3)^2}}$$     =    $${\frac{2x^6 -28x^3 -12}{(2x^3-3)^2}}$$     =    2$${\frac{x^6 - 14x^3 -6}{(2x^3-3)^2}}$$  .

(iii)

For f (x) = $\Bigl($$${\frac{4}{\sqrt{x}}}$$ + x$$\Bigr)$$$$\Bigl($$2$$\sqrt{x}$$ - $${\frac{3}{x^2}}$$$$\Bigr)$$ use the product rule to get

$${\frac{\dD f}{\dD x}} \Bigl( {\frac{{-}2}{x^{\frac32}}} \Bigr) \Bigl( \sqrt{x} {\frac{3}{x^2}} \Bigr) \Bigl( {\frac{4}{\sqrt{x}}} \Bigr) \Bigl( {\frac{1}{\sqrt{x}}} {\frac{6}{x^3}} \Bigr)$$ =

$${\frac{-4}{x}} \sqrt{x} {\frac{6}{x^{\frac72}}} {\frac{3}{x^2}} {\frac{4}{x}} \sqrt{x} {\frac{24}{x^{\frac72}}} {\frac{6}{x^2}} \sqrt{x} {\frac{3}{x^2}} {\frac{30}{x^{\frac72}}}$$ =

In fact this result is obtained more easily by first expanding: f (x) = 8 + 2x^{${\frac{3}{2}}$} - 3x^-1 - 12x^{- ${\frac{5}{2}}$}.

Answer 2         (Question)

The sensitivity of the eye to brightness of a light source is defined to be the rate of change of the area A(x) of the pupil to the light strength x.
With A(x) = $${\frac{40 + 24 x^{0.4}}{1 + 4 x^{0.4}}}$$, differentiate to get:

$${\frac{\dD A}{\dD x}} {\frac{24(0.4\,x^{-0.6})(1 + 4\,x^{0.4}) - (1.6\,x^{-0.6})(40 + 24\,x^{0.4})}{(1 + 4\,x^{0.4})^2}}$$ =

$${\frac{9.6\,x^{-0.6} + 38.4\,x^{-0.2} - 64.0\,x^{-0.6} - 38.4\,x^{-0.2}}{(1 + 4\,x^{0.4})^2}} {\frac{25.6\,x^{-0.6}}{(1 + 4\,x^{0.4})^2}}$$ =

A quicker way to get this is to rewrite the formula as A(x) = 6 + $${\frac{16}{1+4\,x^{0.4}}}$$, then differentiate.

Answer 3         (Question)

The curve y = $${\frac{x}{1+x^2}}$$ is called a serpentine.

(i)

The slope of the tangent line is given by $${\frac{\dD y}{\dD x}}$$ = $${\frac{1}{1+x^2}}$$ - $${\frac{2x^2}{(1+x^2)^2}}$$ = $${\frac{1-x^2}{(1+x^2)^2}}$$.
At (3, 0.3), which lies on the curve since y = $${\frac{3}{3^2+1}}$$ = 0.3, the slope is thus -0.08.
The equation of the tangent is $${\frac{y-0.3}{x-3}}$$ = -0.08; which is better expressed as y = 0.54 - 0.08 x.

(ii)

Here is part of the serpentine curve, which resembles the shape of a snake arching its back as it slithers forward; hence the name. The tangent is shown also.

$$\hbox{%% \ifx\pdfunknown\relax \WARMprocessMMA{w8q3}{eps}{m... ...arge\bfseries {serpentine curve\\ $y=x(1+x^2)^{-1}$} \end{xy}}$$

Algebra

Answer 4         (Question)

Rewrite (x - a)(x - b) - h² as x² - (a + b)x + (ab - h²). Solving (x - a)(x - b) - h² = 0 using the quadratic formula then gives x = ${\frac{1}{2}}$$\bigl($a + b$\pm$$\sqrt{(a+b)^2 - 4(ab-h^2)}$$\bigr)$.
The discriminant part (i.e. the bit under the square-root) is (a² + 2ab + b²) - 4(ab - h²) = (a - b)² + h², which is the sum of two squares, so cannot be negative.
Hence there will always be two real roots, which are equal when a = b and h = 0.

Answer 5         (Question)

Substitute t = x² within the equation x⁴ + 2x² - 8 = 0 to get 0 = t² + 2t - 8 = (t + 4)(t - 2).
Thus the solutions are x = $\pm$$\sqrt{2}$ and x = $\pm$2i, where i² = -1 is the imaginary unit.

Answer 6         (Question)

As the sum of a geometric series we have 1 + i² + i³ + ... + i¹⁰⁰ = $${\frac{1-i^{101}}{1-i}}$$ = $${\frac{1-i}{1-i}}$$ = 1,
since i¹⁰⁰ = $\bigl($i⁴$\bigr)^{25}_{}$ = 1²⁵ = 1.
Alternatively, collect the terms in groups of 4, as follows

1 + $$\bigl($$i + i² + i³ + i⁴$$\bigr)$$ + (i⁵ + ...) + ... + $$\bigl($$i⁹⁷ + i⁹⁸ + i⁹⁹ + i¹⁰⁰$$\bigr)$$     =    1 + 0 + 0 + ... + 0     =    1  .

Answer 7         (Question)

(a)

For z = ${\frac{1}{2}}$(1 + i) we have

(i)

$\bar{z}$ = ${\frac{1}{2}}$(1 - i);

(ii)

z² = ${\frac{1}{4}}$(1 + i)² = ${\frac{i}{2}}$;

(iii)

| z| = $\sqrt{\frac14(1^2+1^2)}$ = $${\frac{1}{\sqrt{2}}}$$;

(iv)

argz = ${\frac{1}{4}}$$\pi$;

(v)

z^-1 = $${\frac{\bar{z}}{\vert z\vert^2}}$$ = (1 - i).

(b)

In an Argand diagram, these complex numbers are located as follows:

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