4.1 Vector

Posted on April 22, 2020 by user

(A) Definition of Vector and Vector Notation

1. Vector is a quantity that has both magnitude and direction.

2. Scalar is a quantity that has magnitude only.

3. A vector can be presented by a line segment with an arrow, known as a directed line segment.

(B) Equality of Two Vectors

1. Negative vector of

\vec{A B}

has the same magnitude as

\vec{A B}

but its direction is opposite to that of

\vec{A B}

2. A zero vector is a vector whose magnitude is zero. It is denoted by

\underset{˜}{0}

3. Two vectors are equal if both the vectors have the same magnitude and direction.

Long Question 4

Posted on April 22, 2020 by user

Question 4:
Diagram below shows quadrilateral ABCD. The straight line AC intersects the straight line BD at point E.

\begin{array}{l} It is given that B E : E D = 2 : 3, \vec{A B} = 10 \underset{˜}{x}, \vec{A D} = 25 \underset{˜}{y} and \vec{B C} = - \underset{˜}{x} + 15 \underset{˜}{y} . \\ (a) Express in terms of \underset{˜}{x} and \underset{˜}{y}, \\ (i) \vec{B D} \\ (ii) \vec{A E} \\ (b) Find the ratio A E : E C . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{B D} = \vec{B A} + \vec{A D} \\ = \vec{A D} - \vec{A B} \\ = 25 \underset{˜}{y} - 10 \underset{˜}{x} \end{array}

(a)(ii)

\begin{array}{l} \vec{A E} = \vec{A B} + \vec{B E} \\ = \vec{A B} + \frac{2}{5} \vec{B D} \\ = 10 \underset{˜}{x} + \frac{2}{5} (25 \underset{˜}{y} - 10 \underset{˜}{x}) \\ = 10 \underset{˜}{x} + \frac{2}{5} (25 \underset{˜}{y} - 10 \underset{˜}{x}) \\ = 10 \underset{˜}{x} + 10 \underset{˜}{y} - 4 \underset{˜}{x} \\ = 6 \underset{˜}{x} + 10 \underset{˜}{y} \\ = 2 (3 \underset{˜}{x} + 5 \underset{˜}{y}) \end{array}

(b)

\begin{array}{l} \vec{E C} = \vec{E B} + \vec{B C} \\ = \vec{B C} - \vec{B E} \\ = \vec{B C} - \frac{2}{3} \vec{E D} \\ = \vec{B C} - \frac{2}{3} (\vec{E A} + \vec{A D}) \\ = - \underset{˜}{x} + 15 \underset{˜}{y} - \frac{2}{3} (- 6 \underset{˜}{x} - 10 \underset{˜}{y} + 25 \underset{˜}{y}) \\ = - \underset{˜}{x} + 15 \underset{˜}{y} - \frac{2}{3} (- 6 \underset{˜}{x} + 15 \underset{˜}{y}) \\ = - \underset{˜}{x} + 15 \underset{˜}{y} + 4 \underset{˜}{x} - 10 \underset{˜}{y} \\ = 3 \underset{˜}{x} + 5 \underset{˜}{y} \\ \frac{A E}{E C} = \frac{2 (3 \underset{˜}{x} + 5 \underset{˜}{y})}{1 (3 \underset{˜}{x} + 5 \underset{˜}{y})} \\ A E : E C = 2 : 1 \end{array}

4.4 Expression of a Vector as the Linear Combination of a Few Vectors

Posted on April 22, 2020 by user

4.4 Expression of a Vector as the Linear Combination of a Few Vectors

1. Polygon Law for Vectors

\vec{P Q} = \vec{P U} + \vec{U T} + \vec{T S} + \vec{S R} + \vec{R Q}

2. To prove that two vectors are parallel, we must express one of the vectors as a scalar multiple of the other vector.

For example,

\vec{A B} = k \vec{C D} or \vec{C D} = h \vec{A B} .

3. To prove that points P, Q and R are collinear, prove one of the following.

\begin{array}{l} • \vec{P Q} = k \vec{Q R} or \vec{Q R} = h \vec{P Q} \\ • \vec{P R} = k \vec{P Q} or \vec{P Q} = h \vec{P R} \\ • \vec{P R} = k \vec{Q R} or \vec{Q R} = h \vec{P R} \end{array}

Example:

Diagram below shows a parallelogram ABCD. Point Q lies on the straight line AB and point S lies on the straight line DC. The straight line AS is extended to the point T such that AS = 2ST.

It is given that AQ : QB = 3 : 1, DS : SC = 3 : 1,

\vec{A Q} = 6 \underset{˜}{a} and \vec{A D} = \underset{˜}{b}

(a) Express, in terms of

\underset{˜}{a} and \underset{˜}{b} :

(i) \vec{A S} (ii) \vec{Q C}

(b) Show that the points Q, C and T are collinear.

Solution:

\begin{array}{l} (a)(i) \vec{A S} = \vec{A D} + \vec{D S} \\ = \vec{A D} + \vec{A Q} \leftarrow \begin{array}{l} A Q : Q B = 3 : 1 and \\ D S : S C = 3 : 1 \\ ∴ \vec{A Q} = \vec{D S} \end{array} \\ = \underset{˜}{b} + 6 \underset{˜}{a} \\ = 6 \underset{˜}{a} + \underset{˜}{b} \end{array}

\begin{array}{l} (a)(ii) \vec{Q C} = \vec{Q B} + \vec{B C} \\ = \frac{1}{3} \vec{A Q} + \vec{A D} \leftarrow \begin{array}{l} A Q : Q B = 3 : 1 \\ \frac{A Q}{Q B} = \frac{3}{1} \Rightarrow Q B = \frac{1}{3} A Q \\ and for parallelogram, \\ B C / / A D, B C = A D \end{array} \\ = \frac{1}{3} (6 \underset{˜}{a}) + \underset{˜}{b} \\ = 2 \underset{˜}{a} + \underset{˜}{b} \end{array}

\begin{array}{l} (b) \vec{Q T} = \vec{Q A} + \vec{A T} \\ = \vec{Q A} + \frac{3}{2} \vec{A S} \leftarrow \begin{array}{l} A S = 2 S T \\ A T = 3 S T = \frac{3}{2} A S \end{array} \\ = - 6 \underset{˜}{a} + \frac{3}{2} (6 \underset{˜}{a} + \underset{˜}{b}) \\ = 3 \underset{˜}{a} + \frac{3}{2} \underset{˜}{b} \\ = \frac{3}{2} (2 \underset{˜}{a} + \underset{˜}{b}) \\ = \frac{3}{2} \vec{Q C} \\ ∴ Points Q, C and T are collinear . \end{array}

Long Question 1

Posted on April 22, 2020 by user

Question 1:

The above diagram shows triangle OAB. The straight line AP intersects the straight line OQ at R. It is given that

O P = \frac{1}{4} O B, A Q = \frac{1}{4} A B, \vec{O P} = 4 \underset{˜}{b} and \vec{O A} = 8 \underset{˜}{a} .

(a) Express in terms of

\underset{˜}{a} and/ or \underset{˜}{b} :

\begin{array}{l} (i) \vec{A P} \\ (ii) \vec{O Q} \end{array}

(b)(i) Given that

\vec{A R} = h \vec{A P}

, state

\vec{A R}

in terms of h,

\underset{˜}{a} and \underset{˜}{b} .

(ii) Given that

\vec{R Q} = k \vec{O Q},

state in terms of k,

\underset{˜}{a} and \underset{˜}{b} .

(c) Using

\vec{A Q} = \vec{A R} + \vec{R Q},

find the value of h and of k.

Solution:
(a)(i)

\begin{array}{l} \vec{A P} = \vec{A O} + \vec{O P} \\ \vec{A P} = - \vec{O A} + \vec{O P} \\ \vec{A P} = - 8 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}

(a)(ii)

\begin{array}{l} \vec{O Q} = \vec{O A} + \vec{A Q} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} \vec{A B} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (\vec{A O} + \vec{O B}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 \vec{O P}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 (4 \underset{˜}{b})) \\ \vec{O Q} = 8 \underset{˜}{a} - 2 \underset{˜}{a} + 4 \underset{˜}{b} \\ \vec{O Q} = 6 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}

(b)(i)

\begin{array}{l} \vec{A R} = h \vec{A P} \\ \vec{A R} = h (- 8 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{A R} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} \end{array}

(b)(ii)

\begin{array}{l} \vec{R Q} = k \vec{O Q} \\ \vec{R Q} = k (6 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{R Q} = 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \end{array}

(c)

\begin{array}{l} \vec{A Q} = \vec{A R} + \vec{R Q} \\ \vec{A Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + (6 k \underset{˜}{a} + 4 k \underset{˜}{b}) \\ \vec{A O} + \vec{O Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \\ - 8 \underset{˜}{a} + 6 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 6 k \underset{˜}{a} + 4 h \underset{˜}{b} + 4 k \underset{˜}{b} \\ - 2 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 6 k \underset{˜}{a} + 4 h \underset{˜}{b} + 4 k \underset{˜}{b} \\ - 2 = - 8 h + 6 k \\ - 1 = - 4 h + 3 k \to (1) \\ 4 = 4 h + 4 k \\ 1 = h + k \\ k = 1 - h \to (2) \\ Substitute (2) into (1), \\ - 1 = - 4 h + 3 (1 - h) \\ - 1 = - 4 h + 3 - 3 h \\ - 4 = - 7 h \\ h = \frac{4}{7} \\ From (2), \\ k = 1 - \frac{4}{7} = \frac{3}{7} \end{array}

Long Question 3

Posted on April 22, 2020 by user

Question 3:

In diagram below, PQRS is a quadrilateral. PTS and TUR are straight lines.

It is given that

\vec{P Q} = 20 \underset{˜}{x}, \vec{P T} = 8 \underset{˜}{y}, \vec{S R} = 25 \underset{˜}{x} - 24 \underset{˜}{y}, \vec{P T} = \frac{1}{4} \vec{P S} and \vec{T U} = \frac{3}{5} \vec{T R}

(a) Express in terms of

\underset{˜}{x}

and/or

\underset{˜}{y}

(i) $\vec{Q S}$
(ii) $\vec{T R}$
(b) Show that the points Q, U and S are collinear.

(c) If

| \underset{˜}{x} |

= 2 and

| \underset{˜}{y} |

= 3, find

| \vec{Q S} |

Solution:
(a)(i)

\begin{array}{l} \vec{Q S} = \vec{Q P} + \vec{P S} \\ \vec{Q S} = - 20 \underset{˜}{x} + 32 \underset{˜}{y} \leftarrow \begin{array}{l} Given \vec{P T} = \frac{1}{4} \vec{P S} \\ \vec{P S} = 4 \vec{P T} = 4 (8 \underset{˜}{y}) = 32 \underset{˜}{y} \end{array} \end{array}

(a)(ii)

\begin{array}{l} \vec{T R} = \vec{T S} + \vec{S R} \\ \vec{T R} = \frac{3}{4} \vec{P S} + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = \frac{3}{4} (32 \underset{˜}{y}) + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = 24 \underset{˜}{y} + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = 25 \underset{˜}{x} \end{array}

(b)

\begin{array}{l} \vec{Q U} = \vec{Q P} + \vec{P T} + \vec{T U} \\ \vec{Q U} = - 20 \underset{˜}{x} + 8 \underset{˜}{y} + \frac{3}{5} (25 \underset{˜}{x}) \leftarrow \begin{array}{l} Given \\ \vec{T U} = \frac{3}{5} \vec{T R} \end{array} \\ \vec{Q U} = - 20 \underset{˜}{x} + 8 \underset{˜}{y} + 15 \underset{˜}{x} \\ \vec{Q U} = - 5 \underset{˜}{x} + 8 \underset{˜}{y} \\ From (a)(i) \vec{Q S} = - 20 \underset{˜}{x} + 32 \underset{˜}{y} \\ \frac{\vec{Q S}}{\vec{Q U}} = \frac{- 20 \underset{˜}{x} + 32 \underset{˜}{y}}{- 5 \underset{˜}{x} + 8 \underset{˜}{y}} \\ \frac{\vec{Q S}}{\vec{Q U}} = \frac{4 (- 5 \underset{˜}{x} + 8 \underset{˜}{y})}{(- 5 \underset{˜}{x} + 8 \underset{˜}{y})} \\ \frac{\vec{Q S}}{\vec{Q U}} = 4 \\ \vec{Q S} = 4 \vec{Q U} \\ ∴ Q, U and S are collinear . \end{array}

(c)

\begin{array}{l} \vec{P S} = 32 \underset{˜}{y} \\ | \vec{P S} | = 32 | \underset{˜}{y} | \\ | \vec{P S} | = 32 \times 3 = 96 \\ \vec{P Q} = 20 \underset{˜}{x} \\ | \vec{P Q} | = 20 | \underset{˜}{x} | \\ | \vec{P Q} | = 20 \times 2 = 40 \\ ∴ | \vec{Q S} | = \sqrt{96^{2} + 40^{2}} \\ | \vec{Q S} | = 104 \end{array}

Integration as the Inverse of Differentiation

Posted on April 22, 2020 by user

3.4c Integration as the Inverse of Differentiation

Example:

\begin{array}{l} Shows that \frac{d}{d x} [\frac{2 x + 5}{x^{2} - 3}] = \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} \\ Hence, find the value of \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x \end{array}

Solution:

\begin{array}{l} \frac{d}{d x} [\frac{2 x + 5}{x^{2} - 3}] = \frac{(x^{2} - 3) (2) - (2 x + 5) (2 x)}{{(x^{2} - 3)}^{2}} \\ = \frac{2 x^{2} - 6 - 4 x^{2} - 10 x}{{(x^{2} - 3)}^{2}} \\ = \frac{- 2 x^{2} - 10 x - 6}{{(x^{2} - 3)}^{2}} \\ = \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} \\ \int_{0}^{2} \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = {[\frac{2 x + 5}{x^{2} - 3}]}_{0}^{2} \\ - 2 \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = {[\frac{2 x + 5}{x^{2} - 3}]}_{0}^{2} \\ \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = - \frac{1}{2} [(\frac{2 (2) + 5}{2^{2} - 3}) - (\frac{2 (0) + 5}{0^{2} - 3})] \\ = - \frac{1}{2} [9 - (- \frac{5}{3})] \\ = - \frac{1}{2} \times \frac{32}{3} \\ = - \frac{16}{3} \\ = - 5 \frac{1}{3} \end{array}

3.6 Integration as the Summation of Volumes

Posted on April 22, 2020 by user

3.6 Integration as the Summation of Volumes

(1).

The volume of the solid generated when the region enclosed by the curve y = f(x), the x-axis, the line x = a and the line x = b is revolved through 360° about the x-axis is given by

V_{x} = π \int_{a}^{b} y^{2} d x

(2).

The volume of the solid generated when the region enclosed by the curve x = f(y), the y-axis, the line y = a and the line y = b is revolved through 360° about the y-axis is given by

V_{y} = π \int_{a}^{b} x^{2} d y

Long Question 1 & 2

Posted on April 22, 2020 by user

Question 1:

A curve with gradient function

5 x - \frac{5}{x^{2}}

has a turning point at (m, 9).

(a) Find the value of m.

(b) Determine whether the turning point is a maximum or a minimum point.
(c) Find the equation of the curve.

Solution:
(a)

\begin{array}{l} \frac{d y}{d x} = 5 x - \frac{5}{x^{2}} \\ At turning point (m, 9), \frac{d y}{d x} = 0. \\ 5 m - \frac{5}{m^{2}} = 0 \\ \frac{5}{m^{2}} = 5 m \\ m^{3} = 1 \\ m = 1 \end{array}

(b)

\begin{array}{l} \frac{d y}{d x} = 5 x - \frac{5}{x^{2}} = 5 x - 5 x^{- 2} \\ \frac{d^{2} y}{d x^{2}} = 5 + \frac{10}{x^{3}} \\ When x = 1, \frac{d^{2} y}{d x^{2}} = 15 (> 0) \\ Thus, (1, 9) is a minimum point . \end{array}

(c)

\begin{array}{l} y = \int (5 x - 5 x^{- 2}) d x \\ y = \frac{5 x^{2}}{2} + \frac{5}{x} + c \\ At turning point (1, 9), x = 1 and y = 9. \\ 9 = \frac{5 {(1)}^{2}}{2} + \frac{5}{1} + c \\ c = \frac{3}{2} \\ Equation of the curve: y = \frac{5 x^{2}}{2} + \frac{5}{x} + \frac{3}{2} \end{array}

Question 2:

A curve has a gradient function kx²– 7x, where k is a constant. The tangent to the curve at the point (1, 3 ) is parallel to the straight line y + x– 4 = 0.

Find

(a) the value of k,

(b) the equation of the curve.

Solution:
(a)

y + x – 4 = 0

y = – x + 4

m = –1

f ’(x) = kx² – 7x

Given tangent to the curve at the point (1, 3) parallel to the straight line

k (1)² – 7 (1) = –1

k – 7 = –1

k = 6

(b)

\begin{array}{l} f' (x) = 6 x^{2} - 7 x \\ f (x) = \int (6 x^{2} - 7 x) d x \\ f (x) = \frac{6 x^{3}}{3} - \frac{7 x^{2}}{2} + c \\ 3 = 2 {(1)}^{3} - \frac{7 {(1)}^{2}}{2} + c at point (1, 3) \\ c = \frac{9}{2} \\ ∴ f (x) = 2 x^{3} - \frac{7 x^{2}}{2} + \frac{9}{2} \end{array}

Long Question 5

Posted on April 22, 2020 by user

Question 5:
In Diagram below, the straight line WY is normal to the curve

y = \frac{1}{2} x^{2} + 1

at B (2, 4). The straight line BQ is parallel to the y–axis.

Find
(a) the value of t,
(b) the area of the shaded region,
(c) the volume generated, in terms of π, when the region bounded by the curve, the y–axis
and the straight line y = 4 is revolved through 360° about the y-axis.

Solution:
(a)

\begin{array}{l} y = \frac{1}{2} x^{2} + 1 \\ Gradient of tangent, \frac{d y}{d x} = 2 (\frac{1}{2} x) = x \\ At point B, \frac{d y}{d x} = 2 \\ Gradient of normal, m_{2} = - \frac{1}{2} \\ \frac{4 - 0}{2 - t} = - \frac{1}{2} \\ 8 = - 2 + t \\ t = 10 \end{array}

(b)

\begin{array}{l} Area of the shaded region \\ = Area under the curve + Area of triangle B Q Y \\ = \int_{0}^{2} (\frac{1}{2} x^{2} + 1) d x + \frac{1}{2} (10 - 2) (4) \\ = {[\frac{x^{3}}{6} + x]}_{0}^{2} + 16 \\ = [\frac{8}{6} + 2] - 0 + 16 \\ = 19 \frac{1}{3} {unit}^{2} \end{array}

(c)

\begin{array}{l} At y - axis, x = 0, y = \frac{1}{2} (0) + 1 = 1 \\ y = \frac{1}{2} x^{2} + 1, x^{2} = 2 y - 2 \\ Volume generated \\ = π \int x^{2} d y \\ = π \int_{1}^{4} (2 y - 2) d y \\ = π {[y^{2} - 2 y]}_{1}^{4} \\ = π [(16 - 8) - (1 - 2)] \\ = 9 π {unit}^{3} \end{array}

Short Question 1

Posted on April 22, 2020 by user

Question 1:

\begin{array}{l} Find the integral of each of the following . \\ (a) \int (\frac{3}{x^{2}} - \frac{5}{2 x^{3}} + 2) d x \\ (b) \int x^{2} (x^{5} + 2 x) d x \\ (c) \int \frac{3 x^{4} + 2 x}{x^{3}} d x \\ (d) \int \frac{(7 + x) (7 - x)}{x^{4}} d x \\ (e) \int {(5 x - 1)}^{3} d x \\ (f) \int \frac{3}{{(4 x + 7)}^{8}} d x \end{array}

Solution:
(a)

\begin{array}{l} \int (\frac{3}{x^{2}} - \frac{5}{2 x^{3}} + 2) d x \\ = \int (3 x^{- 2} - \frac{5 x^{- 3}}{2} + 2) d x \\ = - 3 x^{- 1} + \frac{5 x^{- 2}}{4} + 2 x + c \\ = - \frac{3}{x} + \frac{5}{4 x^{2}} + 2 x + c \end{array}

(b)

\begin{array}{l} \int x^{2} (x^{5} + 2 x) d x \\ = \int (x^{7} + 2 x^{3}) d x \\ = \frac{x^{8}}{8} + \frac{2 x^{4}}{4} + c \\ = \frac{x^{8}}{8} + \frac{x^{4}}{2} + c \end{array}

(c)

\begin{array}{l} \int \frac{3 x^{4} + 2 x}{x^{3}} d x \\ = \int (\frac{3 x^{4}}{x^{3}} + \frac{2 x}{x^{3}}) d x = \int (3 x + 2 x^{- 2}) d x \\ = \frac{3 x^{2}}{2} - \frac{2}{x} + c \end{array}

(d)

\begin{array}{l} \int \frac{(7 + x) (7 - x)}{x^{4}} d x = \int (\frac{49 - x^{2}}{x^{4}}) d x \\ = \int (\frac{49}{x^{4}} - \frac{1}{x^{2}}) d x = \int (49 x^{- 4} - x^{- 2}) d x \\ = \frac{49 x^{- 3}}{- 3} + \frac{1}{x} + c \\ = - \frac{49}{3 x^{3}} + \frac{1}{x} + c \end{array}

(e)

\begin{array}{l} \int {(5 x - 1)}^{3} d x \\ = \frac{{(5 x - 1)}^{4}}{(4) (5)} + c \\ = \frac{1}{20} {(5 x - 1)}^{4} + c \end{array}

(f)

\begin{array}{l} \int \frac{3}{{(4 x + 7)}^{8}} d x = \int 3 {(4 x + 7)}^{- 8} d x \\ = \frac{3 {(4 x + 7)}^{- 7}}{(- 7) (4)} + c \\ = - \frac{3}{28 {(4 x + 7)}^{7}} + c \end{array}