smc-7132-20-3+ Applications

Calculus, 2ADV C4 2024 HSC 22

The graph of the function \(f(x) = \ln(1 + x^{2})\) is shown.

Prove that \(f(x)\) is concave up for \(-1 < x < 1\). (3 marks)
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A table of function values, correct to 4 decimal places, for some \(x\) values is provided.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
\rule{0pt}{2.5ex} \ln(1+x^2) \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 0.0606 & 0.2231 & 0.4463 & 0.6931 \\
\hline
\end{array}

Using the function values provided and the trapezoidal rule, estimate the shaded area in the diagram. (2 marks)
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Is the answer to part (b) an overestimate or underestimate? Give a reason for your answer. (1 mark)
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a. \(f(x)= \ln(1+x^2)\)

\(f^{′}(x)=\dfrac{2x}{1+x^2}\)

\(f^{″}(x)\)	\(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\)
	\(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\)
	\(=\dfrac{2(1-x^2)}{(1+x^2)^2}\)

\(\text{Consider domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)

\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)

\(\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) \)

\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)

b. \(\text{Total shaded area}\ \approx 0.5383\ \text{(4 d.p.)}\)

c. \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)

\(\text{Since the graph is concave up in the given domain, the trapezoidal rule will}\)

\(\text{overestimate the area.}\)

Show Worked Solution

a. \(f(x)= \ln(1+x^2)\)

\(f^{′}(x)=\dfrac{2x}{1+x^2}\)

\(f^{″}(x)\)	\(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\)
	\(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\)
	\(=\dfrac{2(1-x^2)}{(1+x^2)^2}\)

\(\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)

\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)

\(\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) \)

\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)

♦ Mean mark (a) 47%.

b. \(\text{Total shaded area}\)

\(\approx 2 \times \dfrac{h}{2}[ y_0 + 2(y_1+y_2+y_3) + y_4] \)

\(\approx 2 \times \dfrac{0.25}{2}[ 0 + 2(0.0606+0.2231+0.4463) + 0.6931] \)

\(\approx 0.538275 \)

\(\approx 0.5383\ \text{(4 d.p.)}\)

♦ Mean mark (b) 50%.

c. \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)

\(\text{Since the graph is concave up in the given domain, the trapezoidal rule will}\)

\(\text{overestimate the area.}\)

♦ Mean mark (c) 49%.

Calculus, 2ADV C4 2022 HSC 29

The diagram shows the graph of `y=2^{-x}`. Also shown on the diagram are the first 5 of an infinite number of rectangular strips of width 1 unit and height `y=2^{-x}` for non-negative integer values of `x`. For example, the second rectangle shown has width 1 and height `(1)/(2)`.

The sum of the areas of the rectangles forms a geometric series.
Show that the limiting sum of this series is 2. (1 mark)

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Show that `int_(0)^(4)2^(-x)\ dx=(15)/(16 ln 2)`. (2 marks)

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Use parts (a) and (b) to show that `e^(15) < 2^(32)`. (2 marks)

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a. `text{Proof (See Worked Solutions)}`

b. `text{Proof (See Worked Solutions)}`

c. `text{Proof (See Worked Solutions)}`

Show Worked Solution

a. `text{Consider the rectangle heights:}`

`2^0=1, \ 2^(-1)=1/2, \ 2^(-2)= 1/4, \ 2^(-3)= 1/8, …`

`=>\ text{Rectangle Areas}\ = 1, \ 1/2, \ 1/4, \ 1/8, …`

`a=1,\ \ r=1/2`

`S_oo=a/(1-r)=1/(1-1/2)=2\ \ text{… as required}`

b. `text{Show}\ \ int_0^4 2^(-x)\ dx = 15/(16ln2)`

`int_0^4 2^(-x)\ dx `	`=(-1)/ln2[2^(-x)]_0^4`
	`=(-1)/ln2(1/16-1)`
	`=1/ln2-1/(16ln2)`
	`=(16-1)/(16ln2)`
	`=15/(16ln2)\ \ text{… as required}`

Mean mark (b) 56%.

c. `text{Show}\ \ e^15<2^32`

`text{Area under curve < Sum of rectangle areas}`

`15/(16ln2)`	`<2`
`15`	`<32ln2`
`15/32`	`<ln2`
`e^(15/32)`	`<e^(ln2)`
`root(32)(e^15)`	`<2`
`e^15`	`<2^32\ \ text{… as required}`

♦♦♦ Mean mark (c) 9%.

Calculus, 2ADV C4 2020 HSC 20

Kenzo is driving his car along a road while his friend records the velocity of the car, `v(t)`, in km/h every minute over a 5-minute period. The table gives the velocity `v(t)` at time `t` hours.

The distance covered by the car over the 5-minute period is given by

`int_0^(5/60) v(t)\ dt`.

Use the trapezoidal rule and the velocity at each of the six time values to find the approximate distance in kilometres the car has travelled in the 5-minute period. Give your answer correct to one decimal place. (2 marks)

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`5.4\ text(km)`

Show Worked Solution

`int_0^(5/60) v(t)\ dt`	`~~ 1/2 xx 1/60 [60 + 2(55 + 65 + 68 + 70) + 67]`
	`~~ 1/120 (643)`
	`~~ 5.358…`
	`~~ 5.4\ text(km)`

Calculus, 2ADV C4 2014* HSC 16a

Use the Trapezoidal rule with five function values to show that

`int_(- pi/3)^(pi/3) sec x\ dx ~~ pi/6 (3 + 4/sqrt3)`. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

♦ Mean mark below 50%. BE CAREFUL!

`A`	`~~ h/2 [y_0 + 2(y_1 + y_2 + y_3) + y_5]`
	`~~ pi/12 [2 + 2(2/sqrt3 + 1 + 2/sqrt3) + 2]`
	`~~ pi/12 [6 + 8/sqrt3]`
	`~~ pi/6 (3 + 4/sqrt3)\ text(u² … as required)`

Calculus, 2ADV C4 2018* HSC 15c

The shaded region is enclosed by the curve `y = x^3-7x` and the line `y = 2x`, as shown in the diagram. The line `y = 2x` meets the curve `y = x^3-7x` at `O(0, 0)` and `A(3, 6)`. Do NOT prove this.

Use integration to find the area of the shaded region. (2 marks)

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Use the Trapezoidal rule and four function values to approximate the area of the shaded region. (2 marks)

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The point `P` is chosen on the curve `y = x^3 − 7x` so that the tangent at `P` is parallel to the line `y = 2x` and the `x`-coordinate of `P` is positive

Show that the coordinates of `P` are `(sqrt 3, -4 sqrt 3)`. (2 marks)

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Using the perpendicular distance formula `|ax_1 + by_1 + c|/sqrt(a^2 + b^2)`, find the area of `Delta OAP`. (2 marks)

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i. `81/4\ text(units)^2`

ii. `18\ text(u)^2`

iii. `text(Proof)\ \ text{(See Worked Solutions)}`

iv. `9 sqrt 3\ text(units)^2`

Show Worked Solution

i.	`text(Area)`	`= int_0^3 2x-(x^3-7x)\ dx`
		`= int_0^3 9x-x^3\ dx`
		`= [9/2 x^2-1/4 x^4]_0^3`
		`= [(9/2 xx 3^2-1/4 xx 3^4)-0]`
		`= 81/2-81/4`
		`= 81/4\ text(units)^2`

ii. `f(x) = 9x-x^3`

`text(Area)~~ 1/2[0 + 2(8 + 10) + 0]~~ 1/2(36)~~ 18\ text(u)^2`

iii. `y = x^3-7x`

`(dy)/(dx) = 3x^2-7`

`text(Find)\ x\ text(such that)\ \ (dy)/(dx) = 2:`

`3x^2-7`	`= 2`
`3x^2`	`= 9`
`x^2`	`= 3`
`x`	`= sqrt 3 qquad (x > 0)`

`y= (sqrt 3)^3-7 sqrt 3= 3 sqrt 3-7 sqrt 3= -4 sqrt 3`

`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`

iv.

`text(dist)\ OA= sqrt((3-0)^2 + (6-0)^2)= sqrt 45= 3 sqrt 5`

`text(Find)\ _|_\ text(distance of)\ P\ text(from)\ OA:`

`P(sqrt 3, -4 sqrt 3),\ \ 2x-y=0`

`_|_\ text(dist)= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|= (6 sqrt 3)/sqrt 5`

`:.\ text(Area)= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5= 9 sqrt 3\ text(units)^2`

Calculus, 2ADV C4 2011* HSC 5c

The table gives the speed `v` of a jogger at time `t` in minutes over a 20-minute period. The speed `v` is measured in metres per minute, in intervals of 5 minutes.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ t\ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ &\ \ \ 5\ \ \ &\ \ \ 10\ \ \ &\ \ \ 15\ \ \ &\ \ \ 20\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ \ v\ \ \ \rule[-1ex]{0pt}{0pt} & 173 & 81 & 127 & 195 & 168 \\
\hline
\end{array}

The distance covered by the jogger over the 20-minute period is given by `int_0^20 v\ dt`.

Use the Trapezoidal rule and the speed at each of the five time values to find the approximate distance the jogger covers in the 20-minute period. (3 marks)

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`text(2867.5 metres)`

Show Worked Solution

`int_0^20 v\ dt`	`~~ 5/2[173 + 2(81 + 127 + 195) + 168]`
	`~~ 5/2(1147)`
	`~~ 2867.5\ text(metres)`

Calculus, 2ADV C4 2005* HSC 6a

Five values of the function `f(x)` are shown in the table.

Use the Trapezoidal rule with the five values given in the table to estimate

`int_0^20 f(x)\ dx`. (3 marks)

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`387.5`

Show Worked Solution

`:. int_0^20 f(x)\ dx`	`~~ 5/2[15 + 2(25 + 22 + 18) + 10]`
	`~~ 5/2(155)`
	`~~ 387.5`

Calculus, 2ADV C4 2012* HSC 12d

At a certain location a river is 12 metres wide. At this location the depth of the river, in metres, has been measured at 3 metre intervals. The cross-section is shown below.

Use the Trapezoidal rule with the five depth measurements to calculate the approximate area of the cross-section. (3 marks)

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The river flows at 0.4 metres per second.
Calculate the approximate volume of water flowing through the cross-section in 10 seconds. (1 mark)
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a. `30.9 \ text(m)^2`

b. `123.6 \ text(m)^3`

Show Worked Solution

`A`	`~~ 3/2[0.5 + 2(2.3 + 2.9 + 3.8) + 2.1]`
	`~~ 3/2(20.6)`
	`~~ 30.9\ text(m)^2`

♦ Mean mark (b) 49%.

b. `text(Distance water flows)= 0.4 xx 10= 4 \ text(metres)`

`text(Volume flow in 10 seconds)~~ 4 xx 30.9 ~~123.6 text(m)^3`

Calculus, 2ADV C4 2016* HSC 14a

The diagram shows the cross-section of a tunnel and a proposed enlargement.

The heights, in metres, of the existing section at 1 metre intervals are shown in Table `A.`

The heights, in metres, of the proposed enlargement are shown in Table `B.`

Use the Trapezoidal rule with the measurements given to calculate the approximate increase in area. (3 marks)

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`1.3\ text(m)^2`

Show Worked Solution

`text(Consider the shaded area distances:)`

`A`	`~~ 1/2[0 + 2(0.4 + 0.5 + 0.4) + 0]`
	`~~ 1/2(2.6)`
	`~~ 1.3\ text(m)^2`