MATH 130 -- Assignment 3 -- Solutions

1999 D1 & E1, Mathematics IE

Due: Thursday 6 May

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Calculus

Answer 1         (Question)

(i)

Using the Chain Rule, the derivative of f (x) = (1 - 2x²)³ is $${\frac{\dD f}{\dD x}}$$ = 3(1 - 2x²)²(- 4x) = -12x(1 - 2x²)².

(ii)

Using the Product Rule, the derivative of f (x) = (4 + 2x²)³(2 - 3x⁴)⁵ is

$${\frac{\dD f}{\dD x}} \bigl( \bigr) \bigl( \bigr)$$ =

$$\bigl[ \bigr]$$ =

= (24x - 36x⁵ - 240x³ - 120x⁵)(4 + 2x²)²(2 - 3x⁴)⁴

= 12x(2 - 20x² - 13x⁴)(4 + 2x²)²(2 - 3x⁴)⁴  .

.

(iii)

Using the Chain Rule, the derivative of f (x) = $\sqrt[3]{2+3x^2}$ is $${\frac{\dD f}{\dD x}}$$ = $${\textstyle\frac{1}{3}}$$(2 + 3x²)^-2/3(6x) = $${\frac{2x}{(\sqrt[3]{2+3x^2})^2}}$$.

(iv)

Using the Product Rule and Chain Rule, the derivative of f (x) = $${\frac{(3-2x^2)^2}{\sqrt[3]{1+4x^2}}}$$ is

$${\frac{\dD f}{\dD x}} \bigl( \bigr) \bigl( {\textstyle\frac{1}{3}} \bigr)$$ =

$$\bigl[ {\textstyle\frac{1}{3}} \bigr]$$ =

$${\textstyle\frac{8}{3}}$$ =

$${\textstyle\frac{8}{3}} {\frac{6+10x^2}{(\sqrt[3]{(1+4x^2)})^4}}$$ =

Answer 2         (Question)

For the ball of mass m, thrown from the origin with speed V, in the direction of the positive x-axis, at an angle $\theta$ to the x-axis, the vertical distance at any instant of time t is given by y = Vtsin$\theta$ - ${\frac{gt^2}{2m}}$.

(i)

The vertical speed, at time t, is then the derivative

$${\frac{\dD y}{\dD t}}$$ = Vsin$$\theta$$ - $${\frac{gt}{m}}$$  .

This (vertical) speed is zero at time t = $${\frac{mV}{g}}$$sin$$\theta$$, so this is when the maximal height is attained.

When t = $${\frac{2mV}{g}}$$ then the ball has returned to having y = 0. (Note that the ball is thus rising for half the time and falling for an equal length of time.)

(ii)

The ball lands on the x-axis a distance d = $${\frac{2mV^2}{g}}$$cos$$\theta$$ sin$$\theta$$ from the origin, which depends on the ascending angle $\theta$ at which it was thrown.

Rewrite this as d = $${\frac{mV^2}{g}}$$$$\bigl($$2sin$$\theta$$cos$$\theta$$$$\bigr)$$ = $${\frac{mV^2}{g}}$$sin(2$$\theta$$).
Now it is clear that the maximum distance is $${\frac{mV^2}{g}}$$, which is attained when the ball is thrown at an angle of ${\frac{1}{4}}$$\pi$ (or 45^o).

Algebra

Answer 3         (Question)

(a)

Since 1 + i + i² + i³ = 0, and i³ = -i then (1 + i + i²)¹⁸ = (-i)¹⁸ = i¹⁸ = i^{16 + 2} = i² = -1.

(b)

For z = -10$\sqrt{3}$ + 10i, we have:

(i)

$\bar{z}$ = -10$\sqrt{3}$ - 10i

(ii)

Im z² = Im 100(3 - 2$\sqrt{3}$i - 1) = -200$\sqrt{3}$

(iii)

$\bigl\vert$z$\bigr\vert$ = 10$\bigl\vert$$\sqrt{3}$ - i$\bigr\vert$ = 10$\sqrt{3+1}$ = 20

(iv)

argz = ${\frac{5}{6}}$$\pi$

(v)

$${\frac{z}{1+3i}}$$ = -10($$\sqrt{3}$$ - i)$${\frac{(1-3i)}{1^2+3^2}}$$ = -$$\bigl($$3 + $$\sqrt{3}$$ - i(1 + 3$$\sqrt{3}$$)$$\bigr)$$.

Answer 4         (Question)

(a)

Cross-multiplying to equalise denominators, $${\frac{5-x}{2-x}}$$ - $${\frac{3-2x}{2x}}$$ = 1 becomes $${\frac{2x(5-x) - (2-x)(3-2x)}{2x(2-x)}}$$ = 1, equivalently 10x - 2x² - 6 + 7x - 2x² = 2x(2 - x), so that 17x - 4x² - 6 = 4x - 2x².
This is 2x² - 13x + 6 = 0 which factorises as (2x - 1)(x - 6) = 0, so that the solutions are x = 6 and x = ${\frac{1}{2}}$.

(Check these: ${\frac{{-}1}{{-}4}}$ - ${\frac{{-}9}{12}}$ = ${\frac{1}{4}}$ + ${\frac{3}{4}}$ = 1 for x = 6; and ${\frac{4\frac12}{1\frac12}}$ - ${\frac{2}{1}}$ = 3 - 2 = 1 for x = ${\frac{1}{2}}$.)

(b)

Rearrange 1 - $\sqrt{x-1}$ = x as $\sqrt{x-1}$ = 1 - x; now square both sides to get x - 1 = (1 - x)², equivalently x² - 3x + 2 = 0, which factorises as (x - 2)(x - 1) = 0.
Putting x = 1 gives a solution to the original equation; however x = 2 does not give a solution, when interpreting $\sqrt{1}$ as +1, not -1.
Thus x = 1 is the only solution of the given equation.

(c)

To solve x³ + 2x² + 2x + 1 = 0 first try some simple negative (else all terms are positive) integer values for x.
Indeed x = -1 is easily seen to be a solution.
Next factorise the polynomial as x³ + 2x² + 2x + 1 = (x + 1)(x² + x + 1).
All the solutions are then (i) x = -1 and (ii) x = ${\frac{1}{2}}$$\bigl($-1 + $\sqrt{3}$i$\bigr)$ and (ii) x = -${\frac{1}{2}}$$\bigl($1 + $\sqrt{3}$i$\bigr)$.

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