MATH 130 -- Tutorial Exercises -- Solutions

1999 D1 & E1, Mathematics IE

Week 6

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Calculus

Answer 1         (Question)

(i)

y = x³ - 6x² - 15x + 7 has stationary points where
0 = $${\frac{\dD y}{\dD x}}$$ = 3x² - 12x - 15 = 3(x² - 4x - 5) = 3(x - 5)(x + 1).
These are where x = 5 and where x = -1; i.e. at points (5, - 93) and (-1, 15).

At values of x just a bit less than 5 the slope (as given by $${\frac{\dD y}{\dD x}}$$) is negative, whereas it is positive just greater than 5.
Thus the shape is roughly \ $\underline{\phantom{N}}$/; i.e., the point (5, - 93) is a local minimum for y.

Similarly for values of x just a bit less than -1 the slope is positive, but negative for values a bit greater than -1; roughly / $\overline{\phantom{N}}$\. Thus (-1, 15) is a local maximum.
These stationary points can be seen on the red curve in the first plot below.

(ii)

y = x³ - 6x² + 12x - 9 has stationary points where
0 = $${\frac{\dD y}{\dD x}}$$ = 3x² - 12x + 12 = 3(x² - 4x + 4) = 3(x - 2)².
There is a single stationary point where x = 2; i.e., at (2, 11).
Since the slope is positive for values of x near to 2, on either side, the stationary point (2, 11) is a flat point of inflection, as can be seen on the blue curve in the first plot below.

(iii)

y = x¹⁰ - 5x⁸ + 6 has stationary points where $${\frac{\dD y}{\dD x}}$$ = 0, which is where the x-coordinate satisfies
10x⁹ - 40x⁷ = 20x⁷(x² - 4) = 20x⁷(x - 2)(x + 2) = 0.
There are stationary points where x = 0, x = 2 and where x = -2; namely at points (0, 6), (-2,-250) and (2,-250).
Notice that at x = 0 the vanishing is to a high order; making the curve very flat near this point, as in the second plot below.

Figure: Curves y = x3 - 6x2 - 15x + 7 and y = x3 - 6x2 + 12x - 9.

Figure: Curve y = x10 - 5x8 + 6.

Figure: Intervals and curve y = x3 - 3x2 - 9x + 20.

Answer 2         (Question)

The function y = x³ - 3x² - 9x + 20 has stationary points where 0 = $${\frac{\dD y}{\dD x}}$$ = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x - 3)(x + 1).
These are the points (-1, 25) and (3,-7).
As x becomes large, y grows to be very large and positive. Similarly, as x becomes large and negative y also becomes large and negative.

Look at this function y on the following intervals:

(i)

The interval [-3, 4] includes the stationary points at x = -1 and x = 3, but neither occur at an end of the interval. Hence to get the global maximum and minimum, we need to evaluate y at the end-points: (-3,-7) and (4, 0).

Looking at the y values for the critical points: (-3,-7), (-1, 25), (3,-7) and (4, 0), we can deduce that (-1, 25) is a global maximum on this interval, and that both (-3,-7) and (3,-7) share the global minimal value.
Strictly speaking, neither is the global minimum point.
This can be seen clearly in the third plot above.

(ii)

The interval [-2, 0] contains the local maximum for y at the point (-1, 25).
This will also be a global maximum, with the global minimum being the lesser of the values at the endpoints; namely, (-2, 18) and (0, 20).
Hence (-2, 18) is the global minimum, as can be seen in a strip of the plot above.

(iii)

The interval [0, 4] contains the local minimum at (3,-7), so this will be a global minimum, with global maximum at an end-point; that is, look at (0, 20) and (4, 0) to deduce the global maximum being where x = 0.
This can be clearly seen in a strip of the plot above.

(iv)

On the open interval [0,$\infty$), the function grows unboundedly; there is no global maximum, but a global minimum occurs at the locally minimal point (3,-7).

(v)

On the complete real line (-$\infty$,$\infty$) there are neither global maximum, nor global minimum points.

Answer 3         (Question)

From the diagram, one deduces the volume of the box, and area of the metal rectangle are given by:

volume = 2x²h = 1   cubic metre

area = 2(x + h) x 2(h + x) = 4(x + h)²   square metres

Thus, putting h = $${\frac{1}{2x^2}}$$ the area is area = 4(x + ${\frac{1}{2}}$x^-2)² = 4x² + ${\frac{4}{x}}$ + x^-4.
Stationary points occur where 0 = 8x - 4x^-2 - 4x^-5 = 4x^-5(2x⁶ - x³ - 1) = 4x^-5(2x³ + 1)(x³ - 1), hence where x³ = 1, or x³ = - ${\frac{1}{2}}$. Only x = 1 is physically reasonable, and this can be verified to be a local minimum.

The minimal area of metal occurs with x = 1 and h = ${\frac{1}{2}}$.
For this the area required is 4 x (${\frac{3}{2}}$)² = 9 square metres, of which 2 x ${\frac{1}{2}}$ + 4 x (${\frac{1}{2}}$)² = 2 square metres is wasted.

Algebra

Answer 1         (Question)

(a)

tan$${\frac{3\pi}{4}}$$ + sin$${\frac{5\pi}{3}}$$ + cos$${\frac{7\pi}{6}}$$ = -tan$${\frac{\pi}{4}}$$ - sin$${\frac{\pi}{3}}$$ - cos$${\frac{\pi}{6}}$$ = -1 - $${\textstyle\frac{1}{2}}$$$$\sqrt{3}$$ - $${\textstyle\frac{1}{2}}$$$$\sqrt{3}$$ = -(1 + $$\sqrt{3}$$).

(b)

A triangle ABC has the angle at A of ${\frac{\pi}{4}}$ and the angle at B of ${\frac{\pi}{6}}$.
The length of the side opposite the angle C is 3cm.
The angle at C is then $\pi$ - ${\frac{\pi}{4}}$ - ${\frac{\pi}{6}}$ = ${\frac{7\pi}{12}}$.
We will need that sin${\frac{7\pi}{12}}$ = sin${\frac{5\pi}{12}}$ = sin${\frac{\pi}{4}}$cos${\frac{\pi}{6}}$ + cos${\frac{\pi}{4}}$sin${\frac{\pi}{6}}$.
This can be calculated exactly as $${\frac{1}{\sqrt 2}}$$$$\bigl($$$${\textstyle\frac{1}{2}}$$+$${\frac{\sqrt3}{2}}$$$$\bigr)$$ = $${\frac{1+\sqrt3}{2\sqrt2}}$$ = ${\frac{1}{4}}$($\sqrt{2}$ + $\sqrt{6}$).

Using the sin rule, the remaining side-lengths are calculated from

$${\frac{\sin A}{BC}}$$ = $${\frac{\sin B}{AC}}$$ = $${\frac{\sin C}{AB}}$$ = $${\frac{\sin \frac{7\pi}{12}}{3}}$$ = $${\frac{1+\sqrt3}{6\sqrt 2}}$$  .

So BC = sin${\frac{\pi}{4}}$$\bigl($${\frac{1+\sqrt3}{6\sqrt 2}}$$\bigr)^{-1}_{}$ = ${\frac{3}{1+\sqrt3}}$ = ${\frac{3}{2}}$($\sqrt{3}$ - 1),
and AC = sin${\frac{\pi}{6}}$$\bigl($${\frac{1+\sqrt3}{6\sqrt 2}}$$\bigr)^{-1}_{}$ = ${\frac{3\sqrt 2}{1+\sqrt3}}$ = ${\frac{3}{2}}$$\sqrt{2}$($\sqrt{3}$ - 1).

(c)

$${\frac{1}{1-\cos x}}$$ - $${\frac{\cos x}{1+\cos x}}$$ = $${\frac{1+\cos x}{1-\cos^2 x}}$$ - $${\frac{\cos x (1-\cos x)}{1-\cos^2 x}}$$ = $${\frac{1 + \cos^2 x}{\sin^2 x}}$$.
But 2csc²x - 1 = $${\frac{2 - \sin^2 x}{\sin^2 x}}$$ = $${\frac{1 + (1 - \sin^2 x)}{\sin^2 x}}$$ which reduces to the same thing.

(d)

The expression sin 7xcos 2x - cos 7xsin 2x can be written as the sin of an angle sum as follows.
sin 7xcos 2x - cos 7xsin 2x = sin 7xcos(-2x) + cos 7xsin(-2x) = sin(7x - 2x) = sin 5x.

(e)

In the trigonometric equation sinxcos 2x - cosxsin 2x = ${\frac{\sqrt 3}{2}}$, the LHS can be expressed differently, as sinxcos(-2x) + cosxsin(-2x) = sin(x - 2x) = sin(-x).
But we know that ${\frac{\sqrt 3}{2}}$ = sin${\frac{\pi}{3}}$ = sin${\frac{2\pi}{3}}$.
Hence it could be that x is -${\frac{\pi}{3}}$ or -${\frac{2\pi}{3}}$. Other possible values for x are obtained by adding 2$\pi$ to these.
All solutions for x are thus the values (2k - ${\frac{1}{3}}$)$\pi$ and (2k - ${\frac{2}{3}}$)$\pi$, where k may be any integer.

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