10.2 The Cosine Rule

Posted on April 22, 2020 by user

10.2 The Cosine Rule

The cosine rule can be used when

(i) two sides and the included angle, or

(ii) three sides of a triangle are given.

(A) If you know 2 sides and 1 angle between them [included angle] ⇒ Cosine rule

Example:

Calculate the length of AC, x, in cm for the triangle above.

Solution:

$\begin{array}{l} b^{2} = a^{2} + c^{2} - 2 a c \cos B \\ x^{2} = 4^{2} + 7^{2} - 2 (4) (7) \cos 50^{\circ} \\ x^{2} = 16 + 49 - 56 (0.6428) \\ x^{2} = 65 - 35.997 \\ x^{2} = 29.003 \\ x = 5.385 cm \end{array}$

(B) If you know 3 sides ⇒ Cosine rule

Example:

Calculate ÐBAC for the triangle above.

Solution:

$\begin{array}{l} \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \\ \cos ∠ B A C = \frac{7^{2} + 6^{2} - 8^{2}}{2 (7) (6)} \\ \cos ∠ B A C = 0.25 \\ ∠ B A C = \cos^{- 1} 0.25 \\ ∠ B A C = {75.52}^{o} \end{array}$

10.1 The Sine Rule

Posted on April 22, 2020 by user

10.1 The Sine Rule

In a triangle ABC in which the sides BC, CA and AB are denoted by a, b, and c as shown, and A, B, C are used to denote the angles at the vertices A, B, C respectively,

The sine rule can be used when

(i) two sides and one non-included angle or

(ii) two angles and one opposite side are given.

(A) If you know 2 angles and 1 side ⇒ Sine rule

Example:

Calculate the length, in cm, of AB.

Solution:

∠ACB = 180^o – (50^o + 70^o) = 60^o

$\begin{array}{l} \frac{A B}{\sin 60^{o}} = \frac{4}{\sin 50^{o}} \\ A B = \frac{4 \times \sin 60^{o}}{\sin 50^{o}} \\ A B = 4.522 cm \end{array}$

(B) If you know 2 sides and 1 angle (but not between them) ⇒ Sine rule

Example:

Calculate ∠ACB.

Solution:

$\begin{array}{l} \frac{28}{\sin 54^{o}} = \frac{26}{\sin ∠ A C B} \\ \sin ∠ A C B = \frac{26 \times \sin 54^{o}}{28} \\ \sin ∠ A C B = 0.7512 \\ ∠ A C B = {48.7}^{o} \end{array}$

(C) Case of ambiguity (2 possible triangles)

Example

Calculate ∠ACB, θ.

Solution:

Two possible triangle with these measurement

AB = 26cm BC = 28 cm Ð BAC = 54^o

$\begin{array}{l} \frac{26}{\sin θ} = \frac{28}{\sin 54^{o}} \\ \sin θ = 0.7512 \\ θ = \sin^{- 1} 0.7512 \\ θ = {48.7}^{o}, 180^{o} - {48.7}^{o} \\ θ = {48.7}^{o} (Acute angle), {131.3}^{o} (Obtuse angle) \end{array}$

Long Questions (Question 2 & 3)

Posted on April 22, 2020 by user

Question 2:
Given the equation of a curve is:
y = x² (x – 3) + 1
(a) Find the gradient of the curve at the point where x = –1.
(b) Find the coordinates of the turning points.

Solution:
(a)

$\begin{array}{l} y = x^{2} (x - 3) + 1 \\ y = x^{3} - 3 x^{2} + 1 \\ \frac{d y}{d x} = 3 x^{2} - 6 x \\ When x = - 1 \\ \frac{d y}{d x} = 3 {(- 1)}^{2} - 6 (- 1) \\ = 9 \\ Gradient of the curve is 9 . \end{array}$

(b)
At turning points,

$\frac{d y}{d x} = 0$
3x² – 6x = 0
x² – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x² (x – 3) + 1
When x = 0, y = 1
When x = 2,
y = 2² (2 – 3) + 1
y = 4 (–1) + 1 = –3
Therefore, coordinates of the turning points are (0, 1) and (2, –3).

Question 3:
It is given the equation of the curve is y = 2x (1 – x)⁴ and the curve pass through T(2, 4).
Find
(a) the gradient of the curve at point T.
(b) the equation of the normal to the curve at point T.

Solution:
(a)

$\begin{array}{l} y = 2 x {(1 - x)}^{4} \\ \frac{d y}{d x} = 2 x \times 4 {(1 - x)}^{3} (- 1) + {(1 - x)}^{4} \times 2 \\ = - 8 x {(1 - x)}^{3} + 2 {(1 - x)}^{4} \\ At T (2, 4), x = 2. \\ \frac{d y}{d x} = - 16 (- 1) + 2 (1) \\ = 16 + 2 \\ = 18 \end{array}$

(b)

$\begin{array}{l} Equation of normal: \\ y - y_{1} = - \frac{1}{\frac{d y}{d x}} (x - x_{1}) \\ y - 4 = - \frac{1}{18} (x - 2) \\ 18 y - 72 = - x + 2 \\ x + 18 y = 74 \end{array}$

Long Questions (Question 1)

Posted on April 22, 2020 by user

Question 1:

The curve y = x³ – 6x² + 9x + 3 passes through the point P (2, 5) and has two turning points, A (3, 3) and B.

Find

(a) the gradient of the curve at P.

(b) the equation of the normal to the curve at P.

(c) the coordinates of B and determine whether B is a maximum or the minimum point.

Solution:

(a)

y = x³ – 6x² + 9x + 3

dy/dx = 3x²– 12x + 9

At point P (2, 5),

dy/dx = 3(2)² – 12(2) + 9 = –3

Gradient of the curve at point P = –3.

(b)

Gradient of normal at point P = 1/3

Equation of the normal at P (2, 5):

y – y₁= m (x – x₁)

y – 5 = 1/3 (x – 2)

3y – 15 = x – 2

3y = x + 13

(c)

At turning point, dy/dx = 0.

3x² – 12x + 9 = 0

x² – 4x + 3 = 0

(x – 1)( x – 3) = 0

x – 1 = 0 or x – 3 = 0

x = 1 x = 3 (Point A)

Thus at point B:

x = 1

y = (1)³– 6(1)² + 9(1) + 3 = 7

Thus, coordinates of B = (1, 7)

$\begin{array}{l} when x = 1, \\ \frac{d^{2} y}{d x^{2}} = 6 x - 12 \\ \frac{d^{2} y}{d x^{2}} = 6 (1) - 12 \\ \frac{d^{2} y}{d x^{2}} = - 6 < 0 \\ Since \frac{d^{2} y}{d x^{2}} < 0, B is a maximum point . \end{array}$

Short Questions (Question 22 – 25)

Posted on April 22, 2020 by user

Question 23:

Given that

$y = 15 x + \frac{24}{x^{3}}$ ,

(a) Find the value of

$\frac{d y}{d x}$ when x = 2,

(b) Express in terms of k, the approximate change in y when x changes from 2 to

2 + k, where k is a small change.

Solution:
(a)

$\begin{array}{l} y = 15 x + \frac{24}{x^{3}} \\ y = 15 x + 24 x^{- 3} \\ \frac{d y}{d x} = 15 - 72 x^{- 4} \\ \frac{d y}{d x} = 15 - \frac{72}{x^{4}} \\ When x = 2 \\ \frac{d y}{d x} = 15 - \frac{72}{2^{4}} = 10.5 \end{array}$

(b)

$\begin{array}{l} Approximate change in y to x in terms of k, \\ \frac{δ y}{δ x} \approx \frac{d y}{d x} \\ δ y = \frac{d y}{d x} \times δ x \\ δ y = 10.5 \times (2 + k - 2) \\ δ y = 10.5 k \end{array}$

Question 24:

If the radius of a circle increases from 4 cm to 4.01 cm, find the approximate increase in the area.

Solution:

$\begin{array}{l} Area of circle, A = π r^{2} \\ \frac{d A}{d r} = 2 π r \\ Approximate increase in the area to radius, \\ \frac{δ A}{δ r} \approx \frac{d A}{d r} \\ δ A = \frac{d A}{d r} \times δ r \\ δ A = (2 π r) \times (4.01 - 4) \\ δ A = [2 π (4)] \times (0.01) \\ δ A = 0.08 π {cm}^{2} \end{array}$

Question 25:

Given that y =3t+ 5t² and x = 5t –1.

(a) Find

$\frac{d y}{d x}$ in terms of x,

(b) If x increases from 5 to 5.01, find the small increase in t.

Solution:

$\begin{array}{l} y = 3 t + 5 t^{2} \\ \frac{d y}{d t} = 3 + 10 t \\ x = 5 t - 1 \\ \frac{d x}{d t} = 5 \end{array}$

(a)

$\begin{array}{l} \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} \\ \frac{d y}{d x} = (3 + 10 t) \times \frac{1}{5} \\ \frac{d y}{d x} = \frac{3 + 10 (\frac{x + 1}{5})}{5} \leftarrow \begin{array}{l} x = 5 t - 1 \\ t = \frac{x + 1}{5} \end{array} \\ \frac{d y}{d x} = \frac{3 + 2 x + 2}{5} \\ \frac{d y}{d x} = \frac{5 + 2 x}{5} \end{array}$

(b)

$\begin{array}{l} Small increase in t to x, \\ δ t = \frac{d t}{d x} \times δ x \\ δ t = \frac{1}{5} \times (5.01 - 5) \\ δ t = 0.002 \end{array}$

Short Questions (Question 20 – 22)

Posted on April 22, 2020 by user

Question 20:

The volume of water V cm³, in a container is given by

$V = \frac{1}{5} h^{3} + 7 h$ , where h cm is the height of the water in the container. Water is poured into the container at the rate of 15cm³s^-1. Find the rate of change of the height of water in cms^-1, at the instant when its height is 3cm.

Solution:

$\begin{array}{l} V = \frac{1}{5} h^{3} + 7 h \\ \frac{d V}{d h} = \frac{3}{5} h^{2} + 7 = \frac{3 h^{2} + 35}{5} \\ Given \frac{d V}{d t} = 15, h = 3 \\ Rate of change of the height of water = \frac{d h}{d t} \\ \frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t} \leftarrow Chain rule \\ \frac{d h}{d t} = \frac{5}{3 h^{2} + 35} \times 15 \\ \frac{d h}{d t} = \frac{75}{62} {cms}^{- 1} \end{array}$

Question 21:

A wire of length 88 cm is bent to form a circle. When the wire is heated, the length increases at the rate of 0.3 cms^-1.

(a) Calculate the rate of change in the radius of the circle.

(b) Hence, calculate the radius of the circle after 5s.

Solution:

$\begin{array}{l} Length of circumference of a circle, \\ L = 2 π r \\ \frac{d L}{d r} = 2 π \end{array}$

(a)

$\begin{array}{l} Given \frac{d L}{d t} = 0.3 \\ Rate of change in the radius of the circle = \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{d r}{d L} \times \frac{d L}{d t} \\ \frac{d r}{d t} = \frac{1}{2 π} \times 0.3 \\ \frac{d r}{d t} = 0.0477 {cms}^{- 1} \end{array}$

(b)

$\begin{array}{l} 2 π r = 88 \\ r = \frac{88}{2 π} = \frac{44}{π} \\ Hence, the radius of the circle after 5 s \\ = \frac{44}{π} + 5 (0.0477) \\ = 14.24 cm \end{array}$

Question 22:

The diagram shows a conical container with diameter 0.8m and height 0.6m. Water is poured into the container at a constant rate of 0.02m³s^-1. Calculate the rate of change of the height of the water level when the height of water level is 0.5m.

Solution:

$\begin{array}{l} Let, \\ h = height of the water level \\ r = radius of the water surface \\ V = volume of the water \\ \frac{r}{h} = \frac{0.4}{0.6} \leftarrow Concept of similar triangles \\ \frac{r}{h} = \frac{2}{3} \\ r = \frac{2}{3} h \\ Volume of water, V = \frac{1}{3} π r^{2} h \\ V = \frac{1}{3} π {(\frac{2}{3} h)}^{2} h \\ V = \frac{4}{27} π h^{3} \\ \frac{d V}{d h} = (3) \frac{4}{27} π h^{2} \\ \frac{d V}{d h} = \frac{4}{9} π h^{2} \\ The rate of change of the height of the water level \\ when the height of water level is 0 .5 m = \frac{d h}{d t} . \\ \frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t} \leftarrow Chain rule \\ \frac{d h}{d t} = \frac{9}{4 π h^{2}} \times 0.02 \leftarrow Given \frac{d V}{d t} = 0.02 \\ \frac{d h}{d t} = \frac{9}{4 π {(0.5)}^{2}} \times 0.02 \\ \frac{d h}{d t} = 0.0572 m s^{- 1} \end{array}$

Short Questions (Question 14 – 16)

Posted on April 22, 2020 by user

Question 14:

Given y = x (6 – x), express

$y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18$ in terms of x in the simplest form.

Hence, find the value of x which satisfies the equation

$y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0$

Solution:

$\begin{array}{l} y = x (6 - x) = 6 x - x^{2} \\ \frac{d y}{d x} = 6 - 2 x \\ \frac{d^{2} y}{d x^{2}} = - 2 \\ ∴ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = (6 x - x^{2}) (- 2) + x (6 - 2 x) + 18 \\ = - 12 x + 2 x^{2} + 6 x - 2 x^{2} + 18 \\ = - 6 x + 18 \\ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0 \\ - 6 x + 18 = 0 \\ x = 3 \end{array}$

Question 15:

Find the coordinates of the point on the curve, y = (4x – 5)² such that the gradient of the normal to the curve is

$\frac{1}{8}$ .

Solution:

y = (4x – 5)²

$\frac{d y}{d x}$ = 2(4x – 5).4 = 32x – 40

Given the gradient of the normal is 1/8, therefore the gradient of the tangent is –8.

$\frac{d y}{d x}$ = –8

32x – 40 = –8

32x = 32

x = 1

y = (4(1) – 5)²= 1

Hence, the coordinates of the point on the curve, y = (4x – 5)² is (1, 1).

Question 16:

A curve has a gradient function of kx² – 7x, where k is a constant. The tangent to the curve at the point (1, 4) is parallel to the straight line y + 2x–1 = 0. Find the value of k.

Solution:

Gradient function of kx²– 7x is parallel to the straight line y + 2x–1 = 0

$\frac{d y}{d x}$ = kx²– 7x

y + 2x –1 = 0, y = –2x + 1, gradient of the straight line = –2

Therefore kx²– 7x = –2

At the point (1, 4),

k(1)² – 7(1) = –2

k – 7 = –2

k = 5

Short Questions (Question 11 – 13)

Posted on April 22, 2020 by user

Question 11:
Given that the graph of function

$f (x) = h x^{3} + \frac{k}{x^{2}}$ has a gradient function

$f' (x) = 12 x^{2} - \frac{258}{x^{3}}$ such that h and k are constants. Find the values of h and k.

Solution:

$\begin{array}{l} f (x) = h x^{3} + \frac{k}{x^{2}} = h x^{3} + k x^{- 2} \\ f' (x) = 3 h x^{2} - 2 k x^{- 3} \\ f' (x) = 3 h x^{2} - \frac{2 k}{x^{3}} \\ But it is given that f' (x) = 12 x^{2} - \frac{258}{x^{3}} \\ Hence, by comparison, \\ 3 h = 12 and 2 k = 258 \\ h = 4 k = 129 \end{array}$

Question 12:

$\begin{array}{l} Given that y = \frac{x^{2}}{x + 3}, show that \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} \\ Find \frac{d^{2} y}{d x^{2}} in the simplest form . \end{array}$

Solution:

$\begin{array}{l} y = \frac{x^{2}}{x + 3} \\ \frac{d y}{d x} = \frac{(x + 3) (2 x) - x^{2} .1}{{(x + 3)}^{2}} = \frac{2 x^{2} + 6 x - x^{2}}{{(x + 3)}^{2}} \\ \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} (shown) \\ \frac{d^{2} y}{d x^{2}} = \frac{{(x + 3)}^{2} (2 x + 6) - (x^{2} + 6 x) .2 (x + 3)}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{(x + 3) [(x + 3) (2 x + 6) - 2 (x^{2} + 6 x)]}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{[2 x^{2} + 6 x + 6 x + 18 - 2 x^{2} - 12 x]}{{(x + 3)}^{3}} \\ \frac{d^{2} y}{d x^{2}} = \frac{18}{{(x + 3)}^{3}} \end{array}$

Question 13:

$If y = x^{2} + 4 x, show that x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y = 0.$

Solution:

$\begin{array}{l} y = x^{2} + 4 x \\ \frac{d y}{d x} = 2 x + 4 \\ \frac{d^{2} y}{d x^{2}} = 2 \\ x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y \\ = x^{2} (2) - 2 x (2 x + 4) + 2 (x^{2} + 4 x) \\ = 2 x^{2} - 4 x^{2} - 8 x + 2 x^{2} + 8 x \\ = 0 (Shown) \end{array}$

Short Questions (Question 5 – 7)

Posted on April 22, 2020 by user

Question 5:

Given that f (x) = 3x²(4x²– 1)⁷, find f’(x).

Solution:

f (x) = 3x²(4x²– 1)⁷

f’(x) = 3x². 7(4x²– 1)⁶. 8x + (4x²– 1)⁷. 6x

f’(x) = 168x³ (4x²– 1)⁶ + 6x (4x²– 1)⁷

f’(x) = 6x (4x²– 1)⁶ [28x²+ (4x²– 1)]

f’(x) = 6x (4x²– 1)⁶ (32x²– 1)

Question 6:

Given that y = (1 + 4x)³(3x²– 1)⁴, find

$\frac{d y}{d x}$ .

Solution:

y = (1 + 4x)³(3x²– 1)⁴

$\frac{d y}{d x}$

= (1 + 4x)³. 4(3x²– 1)³.6x + (3x²– 1)⁴. 3(1 + 4x)².4

= 24x (1 + 4x)³(3x²– 1)³ + 12 (3x² – 1)⁴(1 + 4x)²

= 12 (1 + 4x)²(3x²– 1)³ [2x (1 + 4x) + (3x²– 1)]

= 12 (1 + 4x)²(3x²– 1)³ [2x + 8x² + 3x²– 1]

= 12 (1 + 4x)²(3x²– 1)³ [11x² + 2x – 1]

Question 7:

$Given that f (x) = 3 x \sqrt{4 x^{2} - 1}, find f' (x) .$ .

Solution:

$\begin{array}{l} f (x) = 3 x \sqrt{4 x^{2} - 1} = 3 x {(4 x^{2} - 1)}^{\frac{1}{2}} \\ f' (x) = 3 x . \frac{1}{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} .8 x + {(4 x^{2} - 1)}^{\frac{1}{2}} .3 \\ f' (x) = 12 x^{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} + 3 {(4 x^{2} - 1)}^{\frac{1}{2}} \\ f' (x) = 3 {(4 x^{2} - 1)}^{- \frac{1}{2}} [4 x^{2} + (4 x^{2} - 1)] \\ f' (x) = \frac{3 (8 x^{2} - 1)}{\sqrt{(4 x^{2} - 1)}} \end{array}$

Short Questions (Question 1 – 4)

Posted on April 22, 2020 by user

Question 1:

Differentiate the expression 2x (4x² + 2x – 5) with respect to x.

Solution:

2x (4x² + 2x – 5) = 8x³ + 4x²– 10x

$\frac{d}{d x}$ (8x³ + 4x² – 10x)

= 24x + 8x –10

Question 2:

$Given that y = \frac{x^{3} + 2 x^{2} + 1}{3 x}, find \frac{d y}{d x} .$ .

Solution:

$\begin{array}{l} y = \frac{x^{3} + 2 x^{2} + 1}{3 x} \\ y = \frac{x^{3}}{3 x} + \frac{2 x^{2}}{3 x} + \frac{1}{3 x} \\ y = \frac{x^{2}}{3} + \frac{2 x}{3} + \frac{1}{3} x^{- 1} \\ \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3} x^{- 2} \\ \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3 x^{2}} \end{array}$

Question 3:

$Given that y = \frac{3}{\sqrt{5 x + 1}}, find \frac{d y}{d x}$

Solution:

$\begin{array}{l} y = \frac{3}{\sqrt{5 x + 1}} = 3 {(5 x + 1)}^{- \frac{1}{2}} \\ \frac{d y}{d x} = - \frac{1}{2} .3 {(5 x + 1)}^{- \frac{3}{2}} (5) \\ \frac{d y}{d x} = \frac{- 15}{2 {[{(5 x + 1)}^{3}]}^{\frac{1}{2}}} \\ \frac{d y}{d x} = \frac{- 15}{2 \sqrt{{(5 x + 1)}^{3}}} \end{array}$

Question 4:

Given that

$y = \frac{3}{5} u^{5}$ , where u = 4x + 1. Find

$\frac{d y}{d x}$ in terms of x.

Solution:

$\begin{array}{l} y = \frac{3}{5} u^{5}, u = 4 x + 1 \\ y = \frac{3}{5} {(4 x + 1)}^{5} \\ \frac{d y}{d x} = 5. \frac{3}{5} {(4 x + 1)}^{4} .4 \\ \frac{d y}{d x} = 12 {(4 x + 1)}^{4} \end{array}$