How many SCSA Mathematics Methods questions cover Further differentiation and applications?

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SCSA Mathematics Methods Further differentiation and applications

5 sample questions with marking guides and sample answers · Avg. score: 60.3%

2024

QCAA

Paper 1

1 mark

Determine $\frac{dy}{dx}$ for the function $y = e^{\sin(x)}$

$\cos(x) e^{\sin(x)}$

$\sin(x) e^{\cos(x)}$

$e^{\sin(x)}$

$e^{\cos(x)}$

2024

QCAA

Paper 1

1 mark

Differentiate $y = \ln(x) \cos(x)$ with respect to $x$ .

$\frac{\cos(x)}{x}$

$-\frac{\sin(x)}{x}$

$\frac{\cos(x)}{x} + \ln(x) \sin(x)$

$\frac{\cos(x)}{x} - \ln(x) \sin(x)$

Q15

2020

SCSA

Paper 2

9 marks

Q15

A chef needs to use an oven to boil 100 mL of water in five minutes for a new experimental recipe. The temperature of the water must reach 100 °C in order to boil. The temperature, $T$ , of 100 mL of water $t$ minutes after being placed in an oven set to $T_0$ °C can be modelled by the equation below.

$T(t) = T_0 - 175e^{-0.07t}$

In a preliminary experiment, the chef placed a 100 mL bowl of water into an oven that had been heated to $T_0 = 200$ °C.

Q15a

1 mark

What is the temperature of the water at the moment it is placed into the oven?

Descriptor	Marks
states correct temperature	1

Q15b

1 mark

What is the temperature of the water five minutes after being placed in the oven?

Descriptor	Marks
states correct temperature	1

Q15c

2 marks

What change could be made to the temperature at which the oven is set in order to achieve the five-minute boiling requirement?

$100 = T_0 - 175e^{-0.07(5)}$
$T_0 = 100 + 175e^{-0.07(5)} \approx 223 \ ^\circ\text{C}$

Marking Criteria

Descriptor	Marks
states correct equation to be solved	1
solves for $T_0$ , giving changed temperature	1

Q15d

2 marks

Assume that $T_0$ is still 200 °C.

Determine the rate of increase in temperature of the water five minutes after being placed in the oven. Give your answer rounded to two decimal places.

$T'(t) = 12.25e^{-0.07t}$
$T'(5) = 12.25e^{-0.07(5)} = 8.63 \ ^\circ\text{C/min}$

Marking Criteria

Descriptor	Marks
states correct derivative of $T$ with respect to $t$	1
calculates correct rate	1

Q15e

3 marks

Explain what happens to the rate of change in the temperature of the water as time increases and how this relates to the temperature of the water.

As time increases, the rate of change in the temperature of the water $\rightarrow 0$ .
The temperature of the water $\rightarrow$ the constant value of $T_0$ .

Marking Criteria

Descriptor	Marks
states that the rate of change in the temperature $\rightarrow 0$	1
states the water temperature approaches a constant	1
states the water temperature approaches $T_0$	1

Q16

2021

SCSA

Paper 2

12 marks

Q16

An analyst was hired by a large company at the beginning of 2021 to develop a model to predict profit. At that time, the company’s profit was ＄4 million. The model developed by the analyst was:

$P(x)=\dfrac{20\ln(x+a)}{x+5}$ ,

where $P(x)$ is the profit in millions of dollars after $x$ weeks and $a$ is a constant.

Q16a

2 marks

Show that $a=e$ .

Descriptor	Marks
recognises $P(0) = 4$	1
obtains $\ln(a) = 1$	1

Q16b

1 mark

What does the model predict the profit will be after five weeks?

$P(5) = \frac{20 \ln(5 + e)}{5 + 5}$
$= 4.087 \text{ (3 d.p.)}$

Profit will be approximately ＄4 087 000

Marking Criteria

Descriptor	Marks
states the profit	1

Q16c

3 marks

Showing use of the quotient rule, determine an equation that, when solved, will give the time when the model predicts the profit will be maximised.

$P'(x) = \frac{(x + 5) \frac{20}{(x + e)} - 20 \ln(x + e)}{(x + 5)^2}$

For maximum profit we require :
$\frac{(x + 5) \frac{20}{(x + e)} - 20 \ln(x + e)}{(x + 5)^2} = 0$

Marking Criteria

Descriptor	Marks
demonstrates use of the quotient rule	1
writes correct expression for $P'(x)$	1
equates $P'(x)$ to zero	1

Q16d

2 marks

What is this maximum profit and during which week will it occur?

Maximum profit is approximately ＄4 436 000
This occurs when $x \approx 1.79$ , so during the second week.

Marking Criteria

Descriptor	Marks
states maximum profit	1
states it occurs in the second week	1

Q16e

1 mark

According to the model, during which week will the company’s profit fall below its value at the beginning of 2021?

$4 = \frac{20 \ln(x + e)}{x + 5}$
$x = 0 \text{ or } 5.581$

The model predicts during the $6^{\text{th}}$ week

Marking Criteria

Descriptor	Marks
determines when the profit falls below the 2021 value	1

Q16f

3 marks

The model proved accurate and after 10 weeks the company implemented some changes. From this time the analyst used a new model to predict the profit:

$N(y)=2e^{b(10+y)}$ ,

where $N(y)$ is the profit in millions of dollars $y$ weeks from this point in time and $b$ is a constant.

The company is projecting its profit to exceed ＄5 million. During which week does the new model suggest this will happen?

$P(10) = 3.39072 = N(0)$
$3.39072 = 2e^{10b}$
$b = 0.05279$
$5 = 2e^{b(10+y)}$
$y \approx 7.36$

Profit should exceed ＄5 million during the $8^{\text{th}}$ week after the changes.

Marking Criteria

Descriptor	Marks
determines $P(10)$	1
determines the value of the constant $b$	1
determines the week when the profit exceeds ＄5 million	1

Q14

2021

SCSA

Paper 2

5 marks

Q14

5 marks

Question 14

The displacement in metres, $x(t)$ , of a power boat $t$ seconds after it was launched is given by:

$x(t)=\dfrac{5t(t^2-15t+48)}{6},\quad t\ge 0$

How far has the power boat travelled before its acceleration is zero?

$x(t) = \frac{5t(t^2 - 15t + 48)}{6}, \quad t \ge 0$

$v(t) = \frac{dx}{dt} = \frac{5t^2 - 50t + 80}{2}$

$a(t) = \frac{d^2x}{dt^2} = 5t - 25$

$5t - 25 = 0 \implies \therefore t = 5$

$\text{Distance travelled} = \int_0^5 \left| \frac{5t^2 - 50t + 80}{2} \right| dt = \frac{245}{3} \approx 81.7 \text{ metres}$

Marking Criteria

Descriptor	Marks
determines an expression for velocity	1
determines an expression for acceleration	1
equates acceleration to zero and determines $t$	1
shows integration expression for distance travelled	1
determines how far the power boat has travelled	1

SCSA Mathematics Methods Further differentiation and applications

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