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Pergunta 1 Relatório
Find the equation of the normal to the curve y = 7x - 5x\(^2\) at x = 2
y = 7x - 5x\(^2\)
\(\frac{dy}{dx}\) = slope/gradient
\(\frac{dy}{dx}\) = 7 - 10x at x = 2
m\(_1\) = - 13
But,the equation of normal, we need m\(_2\), from m\(_1\) m\(_2\) = - 1
m\(_2\) = \(\frac{-1}{m_1}\) = \(\frac{-1}{-13}\) = \(\frac{1}{13}\)
Using, \(\frac{ y - y_1}{x - x_1}\) = \(\frac{1}{m_2}\)
\(\frac{ y - y_1}{x - x_1}\) = \(\frac{1}{m_2}\)
To find the value of y we put x = 2 into y = 7x - 5x\(^2\) = 14 - 20 = -6
\(\frac{ y - (-6)}{x - 2}\) = \(\frac{1}{m_2}\)
y + 6 = \(\frac{1}{m_2}\)(x - 2)
13y + 78 = x - 2
13y - x + 80 = 0
Thus, the equation of normal = 13y - x + 80 = 0
Detalhes da Resposta
y = 7x - 5x\(^2\)
\(\frac{dy}{dx}\) = slope/gradient
\(\frac{dy}{dx}\) = 7 - 10x at x = 2
m\(_1\) = - 13
But,the equation of normal, we need m\(_2\), from m\(_1\) m\(_2\) = - 1
m\(_2\) = \(\frac{-1}{m_1}\) = \(\frac{-1}{-13}\) = \(\frac{1}{13}\)
Using, \(\frac{ y - y_1}{x - x_1}\) = \(\frac{1}{m_2}\)
\(\frac{ y - y_1}{x - x_1}\) = \(\frac{1}{m_2}\)
To find the value of y we put x = 2 into y = 7x - 5x\(^2\) = 14 - 20 = -6
\(\frac{ y - (-6)}{x - 2}\) = \(\frac{1}{m_2}\)
y + 6 = \(\frac{1}{m_2}\)(x - 2)
13y + 78 = x - 2
13y - x + 80 = 0
Thus, the equation of normal = 13y - x + 80 = 0
Pergunta 2 Relatório
The table shows the distribution of distance (in km) of 60 villages from a state capital.
| Distance (in km) | 0 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 70 - 99 | 100 - 149 |
| Number of villages | 12 | 7 | 6 | 8 | 5 | 9 | 10 |
| Distance | Class size | Frequency | Frequency density | Class boundary |
| 0 - 19 | 20 | 12 | 0.6 | - 0.5 - 19.5 |
| 20 - 29 | 10 | 7 | 0.7 | 19.5 - 29.5 |
| 30 - 39 | 10 | 6 | 0.6 | 29.5 - 39.5 |
| 40 - 49 | 10 | 8 | 0.8 | 39.5 - 49.5 |
| 50 - 69 | 20 | 5 | 0.25 | 49.5 - 69.5 |
| 70 - 99 | 30 | 9 | 0.3 | 69.5 - 99.5 |
| 100 - 149 | 50 | 10 | 0.2 | 99.5 - 149.5 |
(NOTE): Frequency density = \(\frac{\text{frequency}}{\text{Class size}}\)
SEE THE DIAGRAM ABOVE
Detalhes da Resposta
| Distance | Class size | Frequency | Frequency density | Class boundary |
| 0 - 19 | 20 | 12 | 0.6 | - 0.5 - 19.5 |
| 20 - 29 | 10 | 7 | 0.7 | 19.5 - 29.5 |
| 30 - 39 | 10 | 6 | 0.6 | 29.5 - 39.5 |
| 40 - 49 | 10 | 8 | 0.8 | 39.5 - 49.5 |
| 50 - 69 | 20 | 5 | 0.25 | 49.5 - 69.5 |
| 70 - 99 | 30 | 9 | 0.3 | 69.5 - 99.5 |
| 100 - 149 | 50 | 10 | 0.2 | 99.5 - 149.5 |
(NOTE): Frequency density = \(\frac{\text{frequency}}{\text{Class size}}\)
SEE THE DIAGRAM ABOVE
Pergunta 3 Relatório
Three linear transformations, P, Q, and R in the oxy plane are defined by
P: (x, y) → (-4x - y, 2x)
Q: (x, y) → (y, 6x - 9y)
R: (x, y) → (x - 2y, 3x + 5y)
(a) write down the matrices of P, Q, and R
(b) Find:
(i) 2P - 3R + Q;
(ii) QR;
(iii) the inverse of the matrix R.
Given:
P: (x, y) → (-4x - y, 2x)
Q: (x, y) → (y, 6x - 9y)
R: (x, y) → (x - 2y, 3x + 5y)
(a) the matrix of P, Q, and R
P = \(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\), Q = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\), R = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
(b) (i) 2P - 3R + Q
= 2\(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\) - 3\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -8 & -2 \\ 4 & 0 \end{pmatrix}\) - \(\begin{pmatrix} 3 & -6 \\ 9 & 15 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -11 & 5 \\ 1 & -24 \end{pmatrix}\)
(ii) QR = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)
= \(\begin{pmatrix} 0 + 3 & 0+ 5 \\ 6 - 27 & -12 - 45 \end{pmatrix}\)
= \(\begin{pmatrix} 3 & 5 \\ -21 & -57 \end{pmatrix}\)
(iii) the inverse of the matrix R. \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
|R| = (1 x 5) - ( -2 x 3) = 5 + 6 = 11
R\(^{-1}\) = \(\frac{Adj (R)}{|R|}\) = \(\frac{1}{11}\)\(\begin{pmatrix} 5 & 2 \\ -3 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{-3}{11} & \frac{1}{11} \end{pmatrix}\)
Detalhes da Resposta
Given:
P: (x, y) → (-4x - y, 2x)
Q: (x, y) → (y, 6x - 9y)
R: (x, y) → (x - 2y, 3x + 5y)
(a) the matrix of P, Q, and R
P = \(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\), Q = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\), R = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
(b) (i) 2P - 3R + Q
= 2\(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\) - 3\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -8 & -2 \\ 4 & 0 \end{pmatrix}\) - \(\begin{pmatrix} 3 & -6 \\ 9 & 15 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -11 & 5 \\ 1 & -24 \end{pmatrix}\)
(ii) QR = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)
= \(\begin{pmatrix} 0 + 3 & 0+ 5 \\ 6 - 27 & -12 - 45 \end{pmatrix}\)
= \(\begin{pmatrix} 3 & 5 \\ -21 & -57 \end{pmatrix}\)
(iii) the inverse of the matrix R. \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
|R| = (1 x 5) - ( -2 x 3) = 5 + 6 = 11
R\(^{-1}\) = \(\frac{Adj (R)}{|R|}\) = \(\frac{1}{11}\)\(\begin{pmatrix} 5 & 2 \\ -3 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{-3}{11} & \frac{1}{11} \end{pmatrix}\)
Pergunta 4 Relatório
The data shows the ordered marks scored by students in a test: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). Given that the median is 13\(\frac{1}{2}\) and y is greater than x by 1, find:
(a) the values of x and y
(b) correct to three significant figures, the standard deviation of the distribution.
Given: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). median = 13\(\frac{1}{2}\) = \(\frac{27}{2}\)
Median = \(\frac{(2x + y) + (x + 2y)}{2}\) = \(\frac{27}{2}\)
\(\frac{3x + 3y}{2}\) = \(\frac{27}{2}\)
3x + 3y = 27
x + y = 9 - - - - - - - -(i)
also, y = x + 1 - - - - -(ii)
put y = x + 1 into eqn (ii)
x + x + 1 = 9
2x = 9 - 1 = 8
x = \(\frac{8}{2}\) = 4
y = x + 1 = 4 + 1 = 5
(b) Since x = 4 and y = 5
11, 12, (2(3) + 5), (4 + (2(5)), 14, (5\(^2\) - 2(4))
11, 12, 13, 14, 14, 17
Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) = \(\frac{11 + 12 + 13 + 14 + 14 + 17}{6}\) = 13.5
| x | f | fx | (x - \(\overline{x}\)) | (x - \(\overline{x}\))\(^2\) | f(x - \(\overline{x}\))\(^2\) |
| 11 | 1 | 11 | - 2.5 | 6.25 | 6.25 |
| 12 | 1 | 12 | - 1.5 | 2.25 | 2.25 |
| 13 | 1 | 13 | - 0.5 | 0.25 | 0.25 |
| 14 | 2 | 28 | 0.5 | 0.25 | 0.50 |
| 17 | 1 | 17 | 3.5 | 12.5 | 12.25 |
\(\sum f(x - \overline{x})^2\) = 21.50
S.D = \(\sqrt{\frac{\sum f(x - \overline{x})^2}{\sum {f}}}\) = \(\sqrt{\frac{21.50}{6}}\) = 1.89
Detalhes da Resposta
Given: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). median = 13\(\frac{1}{2}\) = \(\frac{27}{2}\)
Median = \(\frac{(2x + y) + (x + 2y)}{2}\) = \(\frac{27}{2}\)
\(\frac{3x + 3y}{2}\) = \(\frac{27}{2}\)
3x + 3y = 27
x + y = 9 - - - - - - - -(i)
also, y = x + 1 - - - - -(ii)
put y = x + 1 into eqn (ii)
x + x + 1 = 9
2x = 9 - 1 = 8
x = \(\frac{8}{2}\) = 4
y = x + 1 = 4 + 1 = 5
(b) Since x = 4 and y = 5
11, 12, (2(3) + 5), (4 + (2(5)), 14, (5\(^2\) - 2(4))
11, 12, 13, 14, 14, 17
Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) = \(\frac{11 + 12 + 13 + 14 + 14 + 17}{6}\) = 13.5
| x | f | fx | (x - \(\overline{x}\)) | (x - \(\overline{x}\))\(^2\) | f(x - \(\overline{x}\))\(^2\) |
| 11 | 1 | 11 | - 2.5 | 6.25 | 6.25 |
| 12 | 1 | 12 | - 1.5 | 2.25 | 2.25 |
| 13 | 1 | 13 | - 0.5 | 0.25 | 0.25 |
| 14 | 2 | 28 | 0.5 | 0.25 | 0.50 |
| 17 | 1 | 17 | 3.5 | 12.5 | 12.25 |
\(\sum f(x - \overline{x})^2\) = 21.50
S.D = \(\sqrt{\frac{\sum f(x - \overline{x})^2}{\sum {f}}}\) = \(\sqrt{\frac{21.50}{6}}\) = 1.89
Pergunta 5 Relatório
The magnitude of two vectors u and v are 10N and 12N respectively. If the magnitude of their resultant is 15N, calculate the angle between them.
From the diagram above, using cosine's law
Cos \(\theta\) = \(\frac{10^2 + 12^2 - 15^2}{2 \times 10 \times 12}\)
Cos \(\theta\) = \(\frac{100 + 144 - 225}{2 \times 10 \times 12}\) = \(\frac{244 - 225}{240}\)
= \(\frac{19}{240}\) = 0.0791667
\(\theta\) = Cos\(^{-1}\)(0.0791667) = 85.459º
But the angle between the two vectors = 180º - \(\theta\) = 180º - 85.459º = 94.54º
Detalhes da Resposta
From the diagram above, using cosine's law
Cos \(\theta\) = \(\frac{10^2 + 12^2 - 15^2}{2 \times 10 \times 12}\)
Cos \(\theta\) = \(\frac{100 + 144 - 225}{2 \times 10 \times 12}\) = \(\frac{244 - 225}{240}\)
= \(\frac{19}{240}\) = 0.0791667
\(\theta\) = Cos\(^{-1}\)(0.0791667) = 85.459º
But the angle between the two vectors = 180º - \(\theta\) = 180º - 85.459º = 94.54º
Pergunta 6 Relatório
If tan x = \(\frac{1}{3}\), where 180º < x < 270º
evaluate \(\frac{sin2 x - cos x}{2 tan x + sin 2x}\), leaving the answer in surd form (radicals)
hyp\(^2\) = 3\(^2\) + 1\(^2\) (Pythagoras's theorem)
hyp = \(\sqrt{10}\)
Given: tan x = \(\frac{1}{3}\) ( sin and cos will be in the third quadrant and are both negative where 180º < x < 270º)
sin x = \(\frac{-1}{\sqrt{10}}\), cos x = \(\frac{-3}{\sqrt{10}}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{2sinx cos x - cos x}{2 tan x + 2sin x cos x}\) but 2snx cox = \(\frac{3}{5}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{2(\frac{1}{3}) + \frac{3}{5}}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{\frac{19}{15}}\)
= (\(\frac{3}{5}\) + \(\frac{3\sqrt{10}}{10}\)) \(\div\) \(\frac{19}{15}\) = \(\frac{15}{19}\)[\(\frac{3}{5}\) + \(\frac{3}{\sqrt{10}}\)]
= \(\frac{9}{19}\) + \(\frac{45}{19\sqrt{10}}\)
= \(\frac{9}{19}\) + \(\frac{45 \sqrt{10}}{19(10)}\)
= \(\frac{9}{19}\) + \(\frac{9 \sqrt{10}}{19(2)}\)
= \(\frac{9}{19}\)[1 + \(\frac{\sqrt{10}}{2}\)]
Detalhes da Resposta
hyp\(^2\) = 3\(^2\) + 1\(^2\) (Pythagoras's theorem)
hyp = \(\sqrt{10}\)
Given: tan x = \(\frac{1}{3}\) ( sin and cos will be in the third quadrant and are both negative where 180º < x < 270º)
sin x = \(\frac{-1}{\sqrt{10}}\), cos x = \(\frac{-3}{\sqrt{10}}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{2sinx cos x - cos x}{2 tan x + 2sin x cos x}\) but 2snx cox = \(\frac{3}{5}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{2(\frac{1}{3}) + \frac{3}{5}}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{\frac{19}{15}}\)
= (\(\frac{3}{5}\) + \(\frac{3\sqrt{10}}{10}\)) \(\div\) \(\frac{19}{15}\) = \(\frac{15}{19}\)[\(\frac{3}{5}\) + \(\frac{3}{\sqrt{10}}\)]
= \(\frac{9}{19}\) + \(\frac{45}{19\sqrt{10}}\)
= \(\frac{9}{19}\) + \(\frac{45 \sqrt{10}}{19(10)}\)
= \(\frac{9}{19}\) + \(\frac{9 \sqrt{10}}{19(2)}\)
= \(\frac{9}{19}\)[1 + \(\frac{\sqrt{10}}{2}\)]
Pergunta 7 Relatório
Two events M and N are such that P(M) = \(\frac{1}{2}\), P(N) = \(\frac{9}{20}\) and P(M ∩ N) = \(\frac{11}{50}\)
(a) P(M ∩ N'): (b) P(M' ∩ N)
Given that: P(M) = \(\frac{1}{2}\), P(N) = \(\frac{9}{20}\) and P(M ∩ N) = \(\frac{11}{50}\)
Using the diagram above
(a) P(M ∩ N') = \(\frac{1}{2} - \frac{11}{50}\) = \(\frac{25 - 11}{50}\) = \(\frac{14}{50}\) = \(\frac{7}{25}\)
(b) P(M' ∩ N) = \(\frac{9}{20} - \frac{11}{50}\) = \(\frac{45 - 22}{100}\) = \(\frac{23}{100}\).
Detalhes da Resposta
Given that: P(M) = \(\frac{1}{2}\), P(N) = \(\frac{9}{20}\) and P(M ∩ N) = \(\frac{11}{50}\)
Using the diagram above
(a) P(M ∩ N') = \(\frac{1}{2} - \frac{11}{50}\) = \(\frac{25 - 11}{50}\) = \(\frac{14}{50}\) = \(\frac{7}{25}\)
(b) P(M' ∩ N) = \(\frac{9}{20} - \frac{11}{50}\) = \(\frac{45 - 22}{100}\) = \(\frac{23}{100}\).
Pergunta 8 Relatório
(a) Express \(\frac{9x}{(2x + 1)(x^2 + 1)}\) in partial fraction
(b) If \(^{2m}P_2\) - 10 = \(^m P_2\), find the positive value of m.
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{A}{2x + 1}\) + \(\frac{Bx + C}{x^2 + 1}\) = \(\frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(2x + 1)(x^2 + 1)}\) - - - - -- - - -(i)
Using the cover-up method, put x = \(\frac{-1}{2}\)
A = \(\frac{9x}{x^2 + 1}\) = \(\frac{9(\frac{-1}{2})}{(\frac{-1}{2})^2 + 1}\) = \(\frac{\frac{-9}{2}}{\frac{5}{4}}\) = \(\frac{-18}{5}\)
From equation (i)
9x = A(\(x^2\) + 1) + (Bx + C)(x + 1)
9x = A\(x^2\) + A + 2B\(x^2\) + Bx + 2x + C
9x = (A + B)\(x^2\) + (B + 2C)x + (A + C)
comparing coefficients(x\(^2\)
A + 2B = 0
but A = \(\frac{-18}{5}\)
- A = 2B
-(\(\frac{-18}{5}\)) = 2B
18 = 10B
B = \(\frac{18}{10}\) = \(\frac{9}{5}\).
comparing coefficients (constant terms)
A + C = 0
C = - A = -(\(\frac{-18}{5}\))
C = \(\frac{18}{5}\)
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{\frac{-18}{5}}{2x + 1}\) + \(\frac{\frac{9x}{5} + \frac{18}{5}}{x^2 + 1}\)
= \(\frac{-18}{5(2x + 1)}\) + \(\frac{9x}{5(x^2 + 1)}\) + \(\frac{18}{5(x^2 + 1)}\)
= \(\frac{18}{5}\)[\(\frac{1}{x^2 + 1}\) + \(\frac{1}{2(x^2 + 1)}\) - \(\frac{1}{2x + 1}\)]
2(b) \(^{2m}P_2\) - 10 = \(^m P_2\)
\(\frac{(2m)!}{(2m - 2)!}\) - 10 = \(\frac{m!}{(m - 2)!}\)
\(\frac{(2m)(2m - 1) \times (2m - 2)!}{(2m - 2)!}\) - 10 = \(\frac{m \times (m - 1) \times (m - 2)!}{(m - 2)!}\)
⇒ 2m(2m - 1) - 10 = m(m - 1)
4m\(^2\) - 2m - 10 = m\(^2\) - m
4m\(^2\) - m\(^2\) - 2m + m - 10 = 0
3 m\(^2\) - m - 10 = 0
(3m + 5)(m - 2) = 0
m = 2( +ve value only)
Detalhes da Resposta
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{A}{2x + 1}\) + \(\frac{Bx + C}{x^2 + 1}\) = \(\frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(2x + 1)(x^2 + 1)}\) - - - - -- - - -(i)
Using the cover-up method, put x = \(\frac{-1}{2}\)
A = \(\frac{9x}{x^2 + 1}\) = \(\frac{9(\frac{-1}{2})}{(\frac{-1}{2})^2 + 1}\) = \(\frac{\frac{-9}{2}}{\frac{5}{4}}\) = \(\frac{-18}{5}\)
From equation (i)
9x = A(\(x^2\) + 1) + (Bx + C)(x + 1)
9x = A\(x^2\) + A + 2B\(x^2\) + Bx + 2x + C
9x = (A + B)\(x^2\) + (B + 2C)x + (A + C)
comparing coefficients(x\(^2\)
A + 2B = 0
but A = \(\frac{-18}{5}\)
- A = 2B
-(\(\frac{-18}{5}\)) = 2B
18 = 10B
B = \(\frac{18}{10}\) = \(\frac{9}{5}\).
comparing coefficients (constant terms)
A + C = 0
C = - A = -(\(\frac{-18}{5}\))
C = \(\frac{18}{5}\)
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{\frac{-18}{5}}{2x + 1}\) + \(\frac{\frac{9x}{5} + \frac{18}{5}}{x^2 + 1}\)
= \(\frac{-18}{5(2x + 1)}\) + \(\frac{9x}{5(x^2 + 1)}\) + \(\frac{18}{5(x^2 + 1)}\)
= \(\frac{18}{5}\)[\(\frac{1}{x^2 + 1}\) + \(\frac{1}{2(x^2 + 1)}\) - \(\frac{1}{2x + 1}\)]
2(b) \(^{2m}P_2\) - 10 = \(^m P_2\)
\(\frac{(2m)!}{(2m - 2)!}\) - 10 = \(\frac{m!}{(m - 2)!}\)
\(\frac{(2m)(2m - 1) \times (2m - 2)!}{(2m - 2)!}\) - 10 = \(\frac{m \times (m - 1) \times (m - 2)!}{(m - 2)!}\)
⇒ 2m(2m - 1) - 10 = m(m - 1)
4m\(^2\) - 2m - 10 = m\(^2\) - m
4m\(^2\) - m\(^2\) - 2m + m - 10 = 0
3 m\(^2\) - m - 10 = 0
(3m + 5)(m - 2) = 0
m = 2( +ve value only)
Pergunta 9 Relatório
A body of mass 40 kg is placed on a rough inclined plane which makes an angle of 30\(^0\) with the horizontal. If a force of 420 N is applied upwards parallel to the plane. find the:
(a) maximum friction force that will keep the body in equilibrium;
(b) coefficient of friction.[Take g = 10ms\(^{-1}\)
From the diagram above;
fr + mg sin \(\theta\) = 420
fr = 420 - 40 x 10 sin 30º
fr = 420 - 200 = 220N
(b) coefficient of friction(μ) = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)
μ = \(\frac{fr}{mg cos \theta}\) = \(\frac{220}{400 cos 30}\) = 0.63508
Therefore, μ = 0.635.
Detalhes da Resposta
From the diagram above;
fr + mg sin \(\theta\) = 420
fr = 420 - 40 x 10 sin 30º
fr = 420 - 200 = 220N
(b) coefficient of friction(μ) = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)
μ = \(\frac{fr}{mg cos \theta}\) = \(\frac{220}{400 cos 30}\) = 0.63508
Therefore, μ = 0.635.
Pergunta 10 Relatório
SECTION B (PART 1)
A curve is given by y = 8x + \(\frac{27}{2x^2}\),
FIND:
(a) an expression for \(\frac{dy}{dx}\),
(b) the coordinates of the stationary point on the curve and the nature of the stationary point;
(c) the equation of the normal to the curve at (2, 2).
Given: y = 8x + \(\frac{27}{2x^2}\) = y = 8x + \(\frac{27}{2}\)x\(^{-2}\),
(a) \(\frac{dy}{dx}\) = 8x\(^0\) + \(\frac{27}{2}\)(-2)x\(^{-3}\)
= 8 - 27x\(^{-3}\) = 8 - \(\frac{27}{x^3}\)
(b) \(\frac{dy}{dx}\) = 8 - \(\frac{27}{x^3}\) = 0
= 8 = \(\frac{27}{x^3}\)
= 8x\(^3\) = 27
x\(^3\) = \(\frac{27}{8}\)
x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)
x = \(\frac{3}{2}\)
y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)
y = 4(3) + \(\frac{27}{2}\)(\(\frac{4}{9}\))
y = 12 + 6 = 18
Thus, stationary points are (\(\frac{3}{2}\), 18)
\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0
For all x, the stationary point is a minimum.
(c) gradient = 8 - \(\frac{27}{x^3}\) = 8 - \(\frac{27}{2^3}\) = 8 - \(\frac{27}{8}\) = \(\frac{37}{8}\)
Equation of normal, at (2, 2)
but m\(_2\) = \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)
y - y\(_1\) = m\(_2\)[x - x\(_1\)]
y - 2 = \(\frac{- 8}{37}\)[x - 2]
37y - 74 = - 8x + 16
37y + 8x - 90 = 0
Detalhes da Resposta
Given: y = 8x + \(\frac{27}{2x^2}\) = y = 8x + \(\frac{27}{2}\)x\(^{-2}\),
(a) \(\frac{dy}{dx}\) = 8x\(^0\) + \(\frac{27}{2}\)(-2)x\(^{-3}\)
= 8 - 27x\(^{-3}\) = 8 - \(\frac{27}{x^3}\)
(b) \(\frac{dy}{dx}\) = 8 - \(\frac{27}{x^3}\) = 0
= 8 = \(\frac{27}{x^3}\)
= 8x\(^3\) = 27
x\(^3\) = \(\frac{27}{8}\)
x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)
x = \(\frac{3}{2}\)
y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)
y = 4(3) + \(\frac{27}{2}\)(\(\frac{4}{9}\))
y = 12 + 6 = 18
Thus, stationary points are (\(\frac{3}{2}\), 18)
\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0
For all x, the stationary point is a minimum.
(c) gradient = 8 - \(\frac{27}{x^3}\) = 8 - \(\frac{27}{2^3}\) = 8 - \(\frac{27}{8}\) = \(\frac{37}{8}\)
Equation of normal, at (2, 2)
but m\(_2\) = \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)
y - y\(_1\) = m\(_2\)[x - x\(_1\)]
y - 2 = \(\frac{- 8}{37}\)[x - 2]
37y - 74 = - 8x + 16
37y + 8x - 90 = 0
Pergunta 11 Relatório
(a) A boy runs in a line and his displacement at time t seconds after leaving the start point O is X metres, where 20X = 4t\(^2\) + t\(^3\). Find the:
(i) velocity of the body when t = 15 seconds (ii) value of t for which the acceleration of the body is 8 times his initial acceleration
(b) A body of mass 6 kg moves with a velocity of 7 ms\(^{-1}\). It collides with a second body moving in the opposite direction with a velocity of 5 ms\(^{-1}\). After collision, the two bodies move together with a velocity of 4 ms\(^{-1}\). Find the mass of the second body.
(a)i X = \(\frac{1}{5}\)t\(^2\) + \(\frac{1}{20}\)t\(^3\) ( after dividing thru by 20)
\(\frac{dX}{dt}\) = velocity = \(\frac{2}{5}\)t + \(\frac{3}{20}\))t\(^2\)
at t = 15, v = \(\frac{2}{5}\)(15) + \(\frac{3}{20}\)(15)\(^2\) = 6 + 33.75 = 39.75m/s
Acceleration = \(\frac{dV}{dt}\) = \(\frac{2}{5}\)t + \(\frac{3}{20}\)t\(^2\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t
for initial acceleration, t = 0
a = \(\frac{2}{5}\)ms\(^2\)
(8) \(\frac{2}{5}\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t
40 + 30t = 320
30t = 320 - 40 = 280
t = \(\frac{280}{30}\) = 9\(\frac{1}{3}\)secs.
(b) m\(_1\)u\(_1\) + m\(_2\)u\(_2\) = v\(_c\)(m\(_1\) + m\(_2\))
m\(_1\) = 6kg, u\(_1\) = 7m/s, m\(_2\) = ?, u\(_2\) = - 5 m/s
6(7) + m\(_2\)(-5) = 4(6 + m\(_2\))
42 - 5m\(_2\) = 24 + 4m\(_2\)
42 - 24 = 4m\(_2\) + 5m\(_2\)
9m\(_2\) = 18
m\(_2\) = \(\frac{18}{2}\) = 2 kg.
Detalhes da Resposta
(a)i X = \(\frac{1}{5}\)t\(^2\) + \(\frac{1}{20}\)t\(^3\) ( after dividing thru by 20)
\(\frac{dX}{dt}\) = velocity = \(\frac{2}{5}\)t + \(\frac{3}{20}\))t\(^2\)
at t = 15, v = \(\frac{2}{5}\)(15) + \(\frac{3}{20}\)(15)\(^2\) = 6 + 33.75 = 39.75m/s
Acceleration = \(\frac{dV}{dt}\) = \(\frac{2}{5}\)t + \(\frac{3}{20}\)t\(^2\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t
for initial acceleration, t = 0
a = \(\frac{2}{5}\)ms\(^2\)
(8) \(\frac{2}{5}\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t
40 + 30t = 320
30t = 320 - 40 = 280
t = \(\frac{280}{30}\) = 9\(\frac{1}{3}\)secs.
(b) m\(_1\)u\(_1\) + m\(_2\)u\(_2\) = v\(_c\)(m\(_1\) + m\(_2\))
m\(_1\) = 6kg, u\(_1\) = 7m/s, m\(_2\) = ?, u\(_2\) = - 5 m/s
6(7) + m\(_2\)(-5) = 4(6 + m\(_2\))
42 - 5m\(_2\) = 24 + 4m\(_2\)
42 - 24 = 4m\(_2\) + 5m\(_2\)
9m\(_2\) = 18
m\(_2\) = \(\frac{18}{2}\) = 2 kg.
Pergunta 12 Relatório
Using the trapezium rule with seven ordinates, evaluate, correct to three decimal places, \(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx
\(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx using trapezium rule
| x | 2.4 | 2.6 | 2.8 | 3.0 | 3.2 | 3.4 | 3.6 |
| x\(^2\) - 2 | 3.76 | 4.76 | 5.84 | 7.00 | 8.24 | 9.56 | 10.96 |
| \(\sqrt{x^2 - 2}\) |
1.9391 | 2.1817 | 2.4166 | 2.6458 | 2.8705 | 3.0919 | 3.3106 |
| \(\frac{1}{\sqrt{x^2 - 2}}\) | 0.5157 | 0.4583 | 0.4138 | 0.3780 | 0.3484 | 0.3234 | 0.3021 |
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | \(y_5\) | \(y_6\) | \(y_7\) |
= \(\frac{1}{2}\)(h)[[(\(y_1\) + \(y_7\)] + 2[\(y_2\) + \(y_3\) + \(y_4\) + \(y_5\) + \(y_6\)]]
= \(\frac{1}{2}\)(0.2)[0.5157 + 0.3021 ] + 2[0.4583 + 0.4138 + 0.3780 + 0.3484 + 0.3234]
= (0.1)[0.8178 + 3.8440] = 0.46618 ≈ 0.466 to 3 dp.
Detalhes da Resposta
\(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx using trapezium rule
| x | 2.4 | 2.6 | 2.8 | 3.0 | 3.2 | 3.4 | 3.6 |
| x\(^2\) - 2 | 3.76 | 4.76 | 5.84 | 7.00 | 8.24 | 9.56 | 10.96 |
| \(\sqrt{x^2 - 2}\) |
1.9391 | 2.1817 | 2.4166 | 2.6458 | 2.8705 | 3.0919 | 3.3106 |
| \(\frac{1}{\sqrt{x^2 - 2}}\) | 0.5157 | 0.4583 | 0.4138 | 0.3780 | 0.3484 | 0.3234 | 0.3021 |
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | \(y_5\) | \(y_6\) | \(y_7\) |
= \(\frac{1}{2}\)(h)[[(\(y_1\) + \(y_7\)] + 2[\(y_2\) + \(y_3\) + \(y_4\) + \(y_5\) + \(y_6\)]]
= \(\frac{1}{2}\)(0.2)[0.5157 + 0.3021 ] + 2[0.4583 + 0.4138 + 0.3780 + 0.3484 + 0.3234]
= (0.1)[0.8178 + 3.8440] = 0.46618 ≈ 0.466 to 3 dp.
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