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:

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

9.9 Small Changes and Approximations

Posted on April 22, 2020 by user

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9.9 Small Changes and Approximations

$If δ x is very small, \frac{δ y}{δ x} will be a good approximation of \frac{d y}{d x},$ ,

This is very useful information in determining an approximation of the change in one variable given the small change in the second variable.

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Example:

Given that y = 3x² + 2x – 4. Use differentiation to find the small change in y when x increases from 2 to 2.02.

Solution:

$\begin{array}{l} y = 3 x^{2} + 2 x - 4 \\ \frac{d y}{d x} = 6 x + 2 \end{array}$

The small change in y is denoted by δy while the small change in the second quantity that can be seen in the question is the x and is denoted by δx.

$\begin{array}{l} \frac{δ y}{δ x} \approx \frac{d y}{d x} \\ δ y = \frac{d y}{d x} \times δ x \\ δ y = (6 x + 2) \times (2.02 - 2) \to δ x = new x - original x \\ δ y = [6 (2) + 2] \times 0.02 \\ Substitute x with the original value of x, i .e . 2. \\ δ y = 0.28 \end{array}$

Related Rates of Change

Posted on April 22, 2020 by user

(A) Related Rates of Change

1. If two variables x and y are connected by the equation y = f(x)

Notes:

If x changes at the rate of 5 cms ^-1⇒

$\frac{d x}{d t} = 5$

Decreases/leaks/reduces Þ NEGATIVES values!!!

Example 1 (Rate of change of y and x)

Two variables, x and y are related by the equation

$y = 4 x + \frac{3}{x}$ . Given that y increases at a constant rate of 2 units per second, find the rate of change of x when x = 3.

Solution:

$\begin{array}{l} y = 4 x + \frac{3}{x} = 4 x + 3 x^{- 1} \\ \frac{d y}{d x} = 4 - 3 x^{- 2} = 4 - \frac{3}{x^{2}} \\ \frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t} \\ 2 = (4 - \frac{3}{x^{2}}) \times \frac{d x}{d t} \\ when x = 3 \\ 2 = (4 - \frac{3}{3^{2}}) \times \frac{d x}{d t} \\ 2 = \frac{11}{3} \times \frac{d x}{d t} \\ \frac{d x}{d t} = \frac{6}{11} unit s^{- 1} \end{array}$

(B) Rates of Change of Volume, Area, Radius, Height and Length

(C) Rate of Change of Any Combination of Two Variables

9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points

Posted on April 22, 2020 by user

(A) Second-Order Differentiation

1. When a function y = x³+ x²– 3x + 6 is differentiated with respect to x, the derivative

$\frac{d y}{d x} = 3 x^{2} + 2 x - 3$

2. The second function

$\frac{d y}{d x}$ can be differentiated again with respect to x. This is called the second derivative of y with respect to x and can be written as

$\frac{d^{2} y}{d x^{2}}$ .

3. Take note that

$\frac{d^{2} y}{d x^{2}} \neq {(\frac{d y}{d x})}^{2}$ .

For example,

If y = 4x³ – 7x² + 5x – 1,

The first derivative

$\frac{d y}{d x} = 12 x^{2} - 14 x + 5$

The second derivative

$\frac{d^{2} y}{d x^{2}} = 24 x - 14$

(B) Turning Points, Maximum and Minimum Points

(a) At Turning Points A and B,

(b) At Maximum Point A,

(c) At Minimum Point B,

9.6 Gradients of Tangents, Equations of Tangent and Normal

Posted on April 22, 2020 by user

9.6 Gradients of Tangents, Equations of Tangents and Normals

If A(x₁, y₁) is a point on a line y = f(x), the gradient of the line (for a straight line) or the gradient of the tangent of the line (for a curve) is the value of

$\frac{d y}{d x}$ when x = x₁.

(A) Gradient of tangent at A(x₁, y₁):

(B) Equation of tangent:

(C) Gradient of normal at A(x₁, y₁):

(D) Equation of normal :

Example 1 (Find the Equation of Tangent)

Given that

$y = \frac{4}{{(3 x - 1)}^{2}}$ . Find the equation of the tangent at the point (1, 1).

Solution:

$\begin{array}{l} y = \frac{4}{{(3 x - 1)}^{2}} = 4 {(3 x - 1)}^{- 2} \\ \frac{d y}{d x} = - 2.4 {(3 x - 1)}^{- 3} .3 \\ \frac{d y}{d x} = \frac{- 24}{{(3 x - 1)}^{3}} \\ At point (1, 1), \frac{d y}{d x} = \frac{- 24}{{[3 (1) - 1]}^{3}} = \frac{- 24}{8} = - 3 \\ Equation of tangent at point (1, 1) is, \\ y - 1 = - 3 (x - 1) \\ y - 1 = - 3 x + 3 \\ y = - 3 x + 4 \end{array}$

Example 2 (Find the Equation of Normal)

Find the gradient of the curve

$y = \frac{7}{3 x + 4}$ at the point (-1, 7). Hence, find the equation of the normal to the curve at this point.

Solution:

$\begin{array}{l} y = \frac{7}{3 x + 4} = 7 {(3 x + 4)}^{- 1} \\ \frac{d y}{d x} = - 7 {(3 x + 4)}^{- 2} .3 \\ \frac{d y}{d x} = \frac{- 21}{{(3 x + 4)}^{2}} \\ At point (- 1, 7), \frac{d y}{d x} = \frac{- 21}{{[3 (- 1) + 4]}^{2}} = - 21 \\ Gradient of the normal = \frac{1}{21} \\ Equation of the normal is \\ y - y_{1} = m (x - x_{1}) \\ y - 7 = \frac{1}{21} (x - (- 1)) \\ 21 y - 147 = x + 1 \\ 21 y - x - 148 = 0 \end{array}$

9.5 First Derivatives of Composite Function

Posted on April 22, 2020 by user

(A) Differentiate Composite Function using Chain Rule

Example:

Differentiate y = (x²– 1)⁸ .

Solution:

(B) Differentiate Composite Function using Alternative Method
- Easy Version

Example:

Differentiate y = (x² – 1)⁸ .

Solution:

$\begin{array}{l} y = {(x^{2} - 1)}^{8} \\ \frac{d y}{d x} = 8 {(x^{2} - 1)}^{7} \frac{d}{d x} (x^{2} - 1) \\ \frac{d y}{d x} = 8 {(x^{2} - 1)}^{7} (2 x) \\ \frac{d y}{d x} = 16 x {(x^{2} - 1)}^{7} \end{array}$

Practice 1:

$Given that y = \frac{1}{3 x - 7}, find \frac{d y}{d x}$

Solution:

$\begin{array}{l} y = \frac{1}{3 x - 7} = {(3 x - 7)}^{- 1} \\ \frac{d y}{d x} = - 1 {(3 x - 7)}^{- 2} .3 \\ \frac{d y}{d x} = \frac{- 3}{{(3 x - 7)}^{2}} \end{array}$

Practice 2:

$Given that y = \sqrt{2 x^{2} - 5 x + 1}, find \frac{d y}{d x}$

Solution:

$\begin{array}{l} y = \sqrt{2 x^{2} - 5 x + 1} = {(2 x^{2} - 5 x + 1)}^{\frac{1}{2}} \\ \frac{d y}{d x} = \frac{1}{2} {(2 x^{2} - 5 x + 1)}^{- \frac{1}{2}} (4 x - 5) \\ \frac{d y}{d x} = \frac{4 x - 5}{2 \sqrt{2 x^{2} - 5 x + 1}} \end{array}$