5.3.3 Sketching Graphs of Trigonometric Functions (Part 2)

Posted on May 22, 2020 by Myhometuition

5.3.3 Sketching Graphs of Trigonometric Functions

Example 1:

Sketch the graph of each of the following trigonometric functions for π ≤ x ≤ 2π.

(g) y = –sin x
(h) y = –cos x

(i) y = | sin x |
(j) y = | cos x |

(k) y = tan 2x

Solution:

(g)

(h)

(i)

(j)

(k)

8.2b Standard Normal Distribution Tables (Example 2)

Posted on May 21, 2020 by Myhometuition

8.2b Standard Normal Distribution Tables (Example 2)

Example 2:

Find the value of each of the following probabilities by reading the standard normal distribution tables.

(a) P (0.4 < Z < 1.2)

(b) P (–1 < Z < 2.5)

Solution:

(a)

P (0.4 < Z < 1.2)

= Area P – Area Q

= Q (0.4) – Q (1.2) ← (reading from the standard normal distribution table for 0.4 and 1.2 are 0.3446 and 0.1151 respectively)

= 0.3446 – 0.1151

= 0.2295

(b)

P (–1 < Z < 2.5)

= 1– Area P – Area Q

= 1 – Q (1) – Q (2.5)

= 1 – 0.1587 – 0.00621 ← (reading from the standard normal distribution table for 1 and 2.5 are 0.1587 and 0.00621 respectively)

= 0.8351

(c)

P (–1.3 < Z < –0.5)

= Area P – Area Q

= Q (0.5) – Q (1.3)

= 0.3085 – 0.0968 ← (reading from the standard normal distribution table for 0.5 and 1.3 are 0.3085 and 0.0968 respectively)

= 0.2117

8.2c Probability of an Event

Posted on May 21, 2020 by Myhometuition

8.2c Probability of an Event

Example:

The masses of pears in a fruit stall are normally distributed with a mean of 220 g and a variance of 100 g. Find the probability that a pear that is picked at random has a mass

(a) of more than 230 g.

(b) between 210 g and 225 g.

Hence, find the value of h such that 90% of the pears weigh more than h g.

Solution:

µ = 220 g

σ = √100 = 10 g

Let X be the mass of a pear.

(a)

\begin{array}{l} P (X > 230) \\ = P (Z > \frac{230 - 220}{10}) \leftarrow \begin{array}{l} Convert to standard normal \\ distribution using Z = \frac{X - μ}{σ} \end{array} \\ = P (Z > 1) \\ = 0.1587 \end{array}

(b)

\begin{array}{l} P (210 < X < 225) \\ = P (\frac{210 - 220}{10} < Z < \frac{225 - 220}{10}) \leftarrow \begin{array}{l} Convert to standard normal \\ distribution using Z = \frac{X - μ}{σ} \end{array} \\ = P (- 1 < Z < 0.5) \\ = 1 - P (Z > 1) - P (Z > 0.5) \\ = 1 - 0.1587 - 0.3085 \\ = 0.5328 \end{array}

For 90% (probability = 0.9) of the pears weigh more than h g,

P (X > h) = 0.9

P (X < h) = 1 – 0.9

= 0.1

From the standard normal distribution table,

P (Z > 0.4602) = 0.1

P (Z < –0.4602) = 0.1

\begin{array}{l} \frac{h - 220}{10} = - 0.4602 \\ h - 220 = - 4.602 \\ h = 215.4 \end{array}

8.2b Standard Normal Distribution Tables (Example 3)

Posted on May 21, 2020 by Myhometuition

8.2b Standard Normal Distribution Tables (Example 3)

Example 3:

Find the value of k if

(a) P (Z > k) = 0.0480

(b) P (Z > k) = 0.8350

Solution:

(a)

From the standard normal distribution table, k = 1.665

Z	6	5 (Subtract)
1.6	.0485	5

(b)

From the standard normal distribution table,

k = –0.974 ← (Remember to put a negative sign at the value of k because it is on the left-hand side of the normal curve.)

Z	7	4 (Subtract)
0.9	.1660	10

8.2b Standard Normal Distribution Tables (Example 1)

Posted on May 21, 2020 by Myhometuition

8.2b Standard Normal Distribution Tables (Example 1)

Example 1:

Find the value of each of the following probabilities by reading the standard normal distribution tables.

(a) P (Z > 0.600)

(b) P (Z < –1.24)

(d) P (Z < 0.76)

Solution:

Standard Normal Distribution Table

*When reading the standard normal distribution tables, it involves subtraction of values.

(a)

From the standard normal distribution table, P (Z > 0.600) = 0.2743

(b)

P (Z < –1.24)

= P (Z > 1.24)

= Q (1.24)

= 0.1075 ← (reading from the standard normal distribution table)

(*In the standard normal distribution table, all the values of z are positive. As the curve is symmetrical about the vertical axis, the area of the shaded region in both of the graphs are the same.)

(c)

P (Z > –1.1)

= 1 – Area P

= 1 – Q (–1.1)

= 1 – 0.1357 ← (reading from the standard normal distribution table)

= 0.8643

(d)

P (Z < 0.76)

= 1 – Area P

= 1 – Q (0.76)

= 1 – 0.2236 ← (reading from the standard normal distribution table)

= 0.7764

4.3.1a Addition of Vectors (Examples)

Posted on May 21, 2020 by Myhometuition

Example 1:

Find

(a) The resultant vector of the addition of the two parallel vectors above

(b) The magnitude of the resultant vector.

Solution:

(a)

Resultant vector

= addition of the two vectors

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

(b)

Magnitude of the resultant vector

\begin{array}{l} = | \vec{P Q} | + | \vec{R S} | \\ = | \sqrt{6^{2} + 8^{2}} | + | \sqrt{6^{2} + 8^{2}} | \\ = 10 + 10 \\ = 20 units \end{array}

Example 2:

The diagram above shows a parallelogram OABC. M is the midpoint of BC. Vector

\vec{O A} = \underset{˜}{a} and \vec{O C} = \underset{˜}{c} .

Find the following vectors in terms of

\underset{˜}{a} and \underset{˜}{c} .

\begin{array}{l} (a) \vec{O B} \\ (b) \vec{M B} \\ (c) \vec{O M} \end{array}

Solution:

(a)

\begin{array}{l} \vec{O B} = \vec{O A} + \vec{A B} \leftarrow Triangle Law of addition \\ = \vec{O A} + \vec{O C} \leftarrow \vec{A B} = \vec{O C} \\ = \underset{˜}{a} + \underset{˜}{c} \end{array}

(b)

\begin{array}{l} \vec{M B} = \frac{1}{2} \vec{C B} \leftarrow M is the midpoint of C B \\ = \frac{1}{2} \vec{O A} \\ = \frac{1}{2} \underset{˜}{a} \end{array}

(c)

\begin{array}{l} \vec{O M} = \vec{O C} + \vec{C M} \leftarrow Triangle Law of addition \\ = \vec{O C} + \vec{M B} \leftarrow \vec{C M} = \vec{M B} \\ = \underset{˜}{c} + \frac{1}{2} \underset{˜}{a} \end{array}

3.6.1 Integration as the Summation of Volumes – Examples

Posted on May 21, 2020 by Myhometuition

Example 1:

Find the volume generated for the following diagram when the shaded region is revolved through 360° about the x-axis.

Solution:
Volume generated, V_x

\begin{array}{l} V_{x} = π \int_{a}^{b} y^{2} d x \\ V_{x} = π \int_{2}^{4} {(3 x - \frac{8}{x})}^{2} d x \\ V_{x} = π \int_{2}^{4} (3 x - \frac{8}{x}) (3 x - \frac{8}{x}) d x \\ V_{x} = π \int_{2}^{4} (9 x^{2} - 48 + \frac{64}{x^{2}}) d x \\ V_{x} = π {[\frac{9 x^{3}}{3} - 48 x + \frac{64 x^{- 1}}{- 1}]}_{2}^{4} \\ V_{x} = π {[3 x^{3} - 48 x - \frac{64}{x}]}_{2}^{4} \\ V_{x} = π [(3 {(4)}^{3} - 48 (4) - \frac{64}{4}) - (3 {(2)}^{3} - 48 (2) - \frac{64}{2})] \\ V_{x} = π (- 16 + 104) \\ V_{x} = 88 π u n i t^{3} \end{array}

Example 2:

Find the volume generated for the following diagram when the shaded region is revolved through 360° about the y-axis.

Solution:
Volume generated, V_y

\begin{array}{l} V_{y} = π \int_{a}^{b} x^{2} d y \\ V_{y} = π \int_{1}^{2} {(\frac{2}{y})}^{2} d y \\ V_{y} = π \int_{1}^{2} (\frac{4}{y^{2}}) d y \\ V_{y} = π \int_{1}^{2} 4 y^{- 2} d y \\ V_{y} = π {[\frac{4 y^{- 1}}{- 1}]}_{1}^{2} = π {[- \frac{4}{y}]}_{1}^{2} \\ V_{y} = π [(- \frac{4}{2}) - (- \frac{4}{1})] \\ V_{y} = 2 π u n i t^{3} \end{array}

3.5.1 Integration as the Summation of Areas – Examples

Posted on May 21, 2020 by Myhometuition

Example 1
Find the area of the shaded region.

Solution:

\begin{array}{l} Area of the shaded region \\ = \int_{a}^{b} y d x \\ = \int_{0}^{4} (6 x - x^{2}) d x \\ = {[\frac{6 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{4} \\ = [3 {(4)}^{2} - \frac{{(4)}^{3}}{3}] - 0 \\ = 26 \frac{2}{3} {unit}^{2} \end{array}

Example 2
Find the area of the shaded region.

Solution:

y = x -----(1)

x = 8y – y²-----(2)

Substitute (1) into (2),

y = 8y – y²

y²– 7y = 0

y (y – 7) = 0

y = 0 or 7

From (1), x = 0 or 7

Therefore the intersection points of the curve and the straight line is (0, 0) and (7, 7).

Intersection point of the curve and y-axis is,

x = 8y – y²

At y-axis, x = 0

0 = 8y – y²

y (y – 8) = 0

y = 0, 8

Area of shaded region = (A₁) Area of triangle + (A₂) Area under the curve from y = 7 to y = 8.

\begin{array}{l} = \frac{1}{2} \times base \times height + \int_{7}^{8} x d y \\ = \frac{1}{2} \times (7) (7) + \int_{7}^{8} (8 y - y^{2}) d y \\ = \frac{49}{2} + {[\frac{8 y^{2}}{2} - \frac{y^{3}}{3}]}_{7}^{8} \\ = 24 \frac{1}{2} + [4 {(8)}^{2} - \frac{{(8)}^{3}}{3}] - [4 {(7)}^{2} - \frac{{(7)}^{3}}{3}] \\ = 24 \frac{1}{2} + 85 \frac{1}{3} - 81 \frac{2}{3} \\ = 28 \frac{1}{6} {unit}^{2} \end{array}

3.4b Laws of Definite Integrals

Posted on May 21, 2020 by Myhometuition

3.4b Laws of Definite Integrals

Example:
Given that

\int_{3}^{7} f (x) d x = 5

, find the values for each of the following:

\begin{array}{l} (a) \int_{3}^{7} 6 f (x) d x \\ (b) \int_{3}^{7} [3 - f (x)] d x \\ (c) \int_{7}^{3} 2 f (x) d x \\ (d) \int_{3}^{4} f (x) d x + \int_{4}^{5} f (x) d x + \int_{3}^{7} f (x) d x \\ (e) \int_{3}^{7} \frac{f (x) + 7}{2} d x \end{array}

Solution:

\begin{array}{l} (a) \int_{3}^{7} 6 f (x) d x = 6 \int_{3}^{7} f (x) d x \\ = 6 (5) = 30 \\ (b) \int_{3}^{7} [3 - f (x)] d x = \int_{3}^{7} 3 d x - \int_{3}^{7} f (x) d x \\ = {[3 x]}_{3}^{7} - 5 \\ = [3 (7) - 3 (3)] - 5 \\ = 7 \\ (c) \int_{7}^{3} 2 f (x) d x = - \int_{3}^{7} 2 f (x) d x \\ = - 2 \int_{3}^{7} f (x) d x \\ = - 2 (5) \\ = - 10 \\ (d) \int_{3}^{4} f (x) d x + \int_{4}^{5} f (x) d x + \int_{3}^{7} f (x) d x \\ = \int_{3}^{7} f (x) d x \\ = 5 \\ (e) \int_{3}^{7} \frac{f (x) + 7}{2} d x = \int_{3}^{7} [\frac{1}{2} f (x) + \frac{7}{2}] d x \\ = \int_{3}^{7} \frac{1}{2} f (x) d x + \int_{3}^{7} \frac{7}{2} d x \\ = \frac{1}{2} \int_{3}^{7} f (x) d x + {[\frac{7 x}{2}]}_{3}^{7} \\ = \frac{1}{2} (5) + [\frac{7 (7)}{2} - \frac{7 (3)}{2}] \\ = \frac{5}{2} + 14 \\ = 16 \frac{1}{2} \end{array}

Short Questions (Question 10)

Posted on May 18, 2020 by Myhometuition

Question 10 (4 marks):
Diagram 10 shows the position of three campsites A, B and C at a part of a riverbank drawn on a Cartesian plane, such that A and B lie on the same straight riverbank.

Diagram 10

Sam wants to cross the river from campsite C to the opposite riverbank where the campsites A and B are located.
Find the shortest distance, in m, that he can take to cross the river. Give your answer correct to four decimal places.

Solution:

\begin{array}{l} Let the shortest distance from campsite C \\ to opposite riverbank is C D . \\ Area of △ A B C \\ = \frac{1}{2} | \begin{array}{l} - 2 7     4 \\ 35 - 3 \end{array} \begin{array}{l} - 2 \\ 3 \end{array} | \\ = \frac{1}{2} | (- 2) (5) + (7) (- 3) + (4) (3) \\ - (3) (7) - (5) (4) - (- 3) (- 2) | \\ = \frac{1}{2} | - 66 | \\ = 33 {units}^{2} \\ Distance of A B \\ = \sqrt{{(- 2 - 7)}^{2} + {(3 - 5)}^{2}} \\ = \sqrt{85} m \\ Area of △ A B C = 33 \\ \frac{1}{2} (A B) (C D) = 33 \\ \frac{1}{2} (\sqrt{85}) (C D) = 33 \\ C D = \frac{33 (2)}{\sqrt{85}} \\ C D = 7.1587 (4 d .p .) \\ Thus, the shortest distance is 7 .1587 m . \end{array}