changing the integral activity · web views 10 = -200 10+1 2 + 200 10+1 = -200 121 + 200 11 = -200...

NESA exemplar question solutions

C3.1 Areas and volumesSolutions for questions from the NESA topic guidance related to areas and volumes.

1. Sketch the region bounded by the curve y=x2 and the lines y=4 and y=9. Evaluate the area of this region.

Note: The region in desmos is shaded using y ≥x2 {4 ≤ y ≤9}

Method 1: The shaded area is twice the area between the curve and the y-axis where x≥0

Area=2∫4

9

y12 dy¿2[ 2 y3

32 ]4

9

¿2[ 2 (9 )3

32−2 (4 )3

32 ]¿2[ 2⋅273 −2 ⋅8

3 ]¿2[543 −16

3 ]¿2[383 ]¿ 763 =25 1

3unit s2

Method 2: The shaded area is twice the area between x=0∧x=3 Shaded area between x=0∧x=2 is 2 ⋅5=10unit s2

Shaded area between x=2∧x=3 is:

A=9−∫2

3

x2dx¿9−[ x33]2

3

¿9−[ 333

−23

3]¿ 83 unit s2

© NSW Department of Education, 2019 1

https://education.nsw.gov.au/about-us/copyright

Total area=2(10+ 83 )¿25 13 unit s2

2. The graphs of the curves y=x2 and y=12−2 x2 are shown in the diagram.

a. Find the points of intersection of the two curves.

Solve y=x2andy=12−2x2simultaneously.

x2=12−2 x23 x2=12x2=4x=±2

When x=2 , y=22=4When x=−2 , y=(−2 )2=4

Points of intersection: (−2,4) and (2,4)

2 ME-C3 NESA sample questions

b. The shaded region between the curves and the y-axis is rotated about the y-axis. By splitting the shaded region into two parts, or otherwise, find the volume of the solid formed.

V=π∫a

b

{ f ( y ) }2. dy

Volume 1: The volume of the curve rotated about the y-axis between 4 and 12

y=x2

x=±√ y=± y12

V 1=π∫4

12 ( y 12)2

. dy

¿ π∫4

12

y .dy

¿ π [ y22 ]4

12

¿ π [1222 −42

2 ]¿64 π unit s3

Volume 2: The volume of the curve rotated about the y-axis between 0 and 4

y=12−2 x2

x=±√6− y2=±(6− y

2 )12

V 2=π∫0

4

((6− y2 )

12 )2

. dy¿ π∫0

4

(6− y2 ). dy¿ π [6 y− y2

4 ]0

4

¿ π [(6 ⋅ 4−424 )−(6 ⋅0− 024 )]¿ π [ (24−4 )−(0 ) ]¿20 πunit s3

Total volume=64 π+20π¿84 π unit s3



3. The region bounded by the curve y= (x−1 ) (3−x ) and the x-axis is rotated about the line x=3 to form a solid. When the region is rotated, the horizontal line segment at height y sweeps out an annulus.

a. Find the area of the annulus as a function of y.

Area=π (R2−r2 )r=3−x2∧R=3−x1Where x1 is the smaller x value and x2 is the larger x value for a given height y.

y= (x−1 ) (3−x )y=−x2+4 x−3x2−4 x=−3− y

Complete the square.

x2−4 x+4=−3− y+4( x−2 )2=1− yx=2±√1− y

x1=2−√1− y and x2=2+√1− y

r=3−(2+√1− y )¿1−√1− y

R=3−(2−√1− y )¿1+√1− y

Area=π ( (1+√1− y )2−(1−√1− y )2 )¿ π ( (1+2√1− y+(1− y))−(1−2√1− y+(1− y )))¿ π ( (2− y+2√1− y )−(2− y−2√1− y ) )¿4 π √1− y


b. Find the volume of the solid.

Sum of all annuluses from y=0 to y=1

V=∫0

1

4 π √1− y dy¿4 π∫0

1

(1− y )12dy¿4 π [−2 (1− y )

32

3 ]0

1

¿4 π [(−2 (1−1 )32

3 )−(−2 (1−0 )32

3 )]¿4 π [ (0 )−(−23 )]¿ 8π3 unit s3

4. The region enclosed by the curve y=4√ x and the x-axis between x=0 and x=4 is rotated about the x-axis. Find the volume of the solid of revolution.

Note: The region in desmos is shaded using 0≤ y≤4 √x {0≤ x≤4 }

V=π∫a

b

{ f (x ) }2 . dx¿ π∫0

4

(4√ x )2 . dx¿ π∫0

4

16 x .dx¿ π [8x2 ]04¿ π [ (8 ⋅42 )−(8⋅ 02 ) ]¿ π [ (128 )−(0 ) ]¿128π unit s3



5. A curved funnel has a shape formed by rotating part of the parabola y=2√ x about the y-axis, where x and y are given in cm. The funnel is 4 cm deep. Find the volume of liquid which the funnel will hold if it is sealed at the bottom.

Note: The diagram shows an example of the region to be rotated about the y-axis. If the lower bound is y, then the upper bound is y+4 as the funnel is 4 cm deep.

y=2√ x

x= y2

4

V=π∫a

b

{ f ( y ) }2. dy¿ π ∫y

y+4

( y24 )2

. dy¿ π ∫y

y+4

( y 416 ) . dy¿ π [ y5

80 ]y

y+4

¿ π [( ( y+4 )5

80 )−( y5

80 )]¿ ( y+4 )5− y5

80π unit s3

If the solution is expanded:

¿ π [1 ⋅ y5⋅ 40+5⋅ y 4⋅ 41+10 ⋅ y3 ⋅42+10 ⋅ y2 ⋅43+5 ⋅ y1⋅ 44+1 ⋅ y0 ⋅4580−y5

80 ]¿ π [20 y4+160 y3+640 y2+1280 y+102480 ]¿( y44 +2 y3+8 y2+16 y+ 64

5 )π unit s3


6.a. Sketch the region bounded by the curve y=sin x+cos x and the coordinate axes in the

first quadrant, taking the upper limit of x as 3π4 .

Note: The region in desmos is shaded using 0≤ y≤ sin x+cos x {0≤x ≤3π4

}

b. Show the intercepts on the axes, and calculate the area of the region.

y-intercept, substitute x=0

y=sin 0+cos0¿0+1¿1

x-intercept, substitute y=0

0=sin x+cos x−cos x=sin x

−cos xcos x

= sin xcos x

−1=tan x

x=3π4 (limiting the x value to the upper bound of

3π4 )

Area=∫0

3π4

(sin x+cos x )dx¿ [−cos x+sin x ]03 π4 ¿(−cos 3π4 +sin 3 π

4 )−(−cos0+sin 0 )

¿(√22 + √22 )− (−1+0 )¿1+√2unit s2

c. Find the volume of the solid formed if the region is rotated about the x-axis to form a solid of revolution.

V=π∫0

3π4

(sin x+cos x )2. dx¿ π∫0

3 π4

( sin2 x+2sin xcos x+cos2 x ) . dx¿ π∫0

3 π4

(1+sin 2 x ) . dx

¿ π [ x−12 cos2 x]03π4 ¿ π [( 3π4 −1

2cos(2 ⋅ 3π4 ))−(0− 12 cos (2 ⋅0 ))]¿ π [( 3π4 −1

2⋅ 0)−(0−12 )]

¿ π [3 π4 + 12 ]¿ 3π24 + π

2unit s3

Note: This volume can be check using online calculators such as symbolab. It is

entered as volume y=sinx+cosx ,x=0 , x=3π4



C3.2 Differential equations part 1Solutions for questions from NESA’s topic guidance related to differential equations.

1. If a product which has just been launched is judged by the market to be of poor quality, sales will decline as people try the product but do not continue to buy it. For a certain product, the

rate of weekly sales is modelled by S' (t )= 400

(t+1 )3− 200

(t+1 )2, where S is the number of sales in

millions and t is the number of weeks since the launch of the product.

a. Find the function that describes the weekly sales.

S' (t )= 400( t+1 )3

− 200( t+1 )2

S(t )=∫( 400( t+1 )3−200

( t+1 )2 )dtS(t )=∫ (400 (t+1 )−3−200 (t+1 )−2 )dt

S (t )= 400 (t+1 )−2

−2−200 ( t+1 )−1

−1+C

S (t )=−200(t+1 )2

+ 200( t+1)

+C

S (0 )=00=−200(0+1 )2

+ 200(0+1)

+C0=−200+200+CC=0

S (t )=−200(t+1 )2

+ 200( t+1)

a. Find the number of sales for the first week and for the tenth week.

S (1 )=−200(1+1 )2

+ 200(1+1 )¿

−2004

+ 2002 ¿−50+100¿50

Number of sales∈week1=50000000

S (10 )= −200(10+1 )2

+ 200(10+1 )¿

−200121

+ 20011

¿−200121

+ 2200121

¿ 2000121

∨16.528…

Number of sales∈week10=16528925.619…



b. Comment on your results in the context of the given information.Based on the assumption given, this is a poor quality product as sales have declined as people try the product but do not continue to buy. The model only approximates products sold as only a whole number of products can be sold.

2. Which of the following direction fields best represents the differential equation dydx

=x− y?

Answer: C. When x= y , dydx

=x− y=0 , i.e. the slope of the line segments will be zero.


C3.2 Differential equations part 2This contains solutions for the questions from the NESA sample unit for the content:

model and solve differential equations including but not limited to the logistic equation that will arise in situations where rates are involved, for example in chemistry, biology and economics (ACMSM132) AAM

1. Gardeners are concerned about the spread of a particular pest. All specimens detected so far lie within a circular region with radius 25 km and it is suggested that the increase of the radius

rkm might be modelled by a differential equation drdt

=16 √r, where t denotes the time in

months. What does this model predict for the radius of the region affected by the pest after t months?

drdt

=16 √r 6dr

√r=1×dt6 r

12 dr=1 ⋅dt∫6 r

12dr=∫ 1⋅dt4 r

32=t+C

r=25whent=04 ⋅2532=0+C4 ⋅125=CC=500

4 r32=t+500r

32= t+500

4r=( t+5004 )

23

2. Water is slowly leaking from a tank. The depth of the water after t hours is h metres and the

variables are related by the equation dhdt

=−a e−0.1 t. Initially the depth of water is 6 metres and

after 2 hours it has fallen to 5 metres. At what depth will the level eventually settle?

dhdt

=−a e−0.1 t

h=∫−ae−0.1 t dt

h=10ae−0.1 t+C where C is the height the water will eventually settle as limt→∞

10ae−0.1 t=0

h=6when t=0

6=10ae−0.1 ⋅0+C6=10a+C

a=6−c10

h=5whent=2

5=10ae−0.1⋅2+C

5=10⋅ 6−c10

⋅e−0.2+C

5=6−ce0.2

+C

5= 6e0.2

− ce0.2

+C



5− 6e0.2

=C (−1e0.2+1)5e0.2−6e0.2

=C ( e0.2−1e0.2 )

C=

5e0.2−6e0.2

e0.2−1

e0.2

C=5 e0.2−6

e0.2−1∨1.0175…m

The water will settle at a depth of approximately 1.02 m

3. When a ball is dropped from the roof of a tall building the greatest speed that it can reach is ums-1. One model for its speed v ms-1 when it has fallen a distance x m is given by the

differential equation dvdx

=c u2−v2

v where c is a positive constant. Find an expression for v in

terms of x.

dvdx

=c u2−v2

vv

u2−v2dv=c ⋅dx∫ v

u2−v2dv=∫ c ⋅dx−1

2 ∫ −2 vu2−v2

dv=∫c ⋅ dx−12 ln|u2−v2|=cx+C

−12ln (u2−v2 )=cx+C as u≥v

v=0 , x=0−12ln (u2−02 )=c ⋅0+C

−12ln (u2 )=C

−12⋅2 ln (u )=C

C=−ln (u )

Substitute C into the function −12ln (u2−v2 )=cx+C

−12ln (u2−v2 )=cx−ln (u )ln (u2−v2 )=−2cx+2 ln (u )ln (u2−v2 )=−2cx+ ln (u2 )u2−v2=e−2cx+ln (u2 )

u2−v2=e−2cx ⋅e ln (u2 )u2−v2=e−2cx ⋅u2u2−v2=u2e−2 cxv2=u2−u2 e−2cxv=±√u2−u2 e−2cx

v=√u2−u2 e−2cx as v ( speed )is positive.

Note: The solution could be expressed in a range of alternate forms such as:

v=u (1−e−2cx )12 . The solution can be checked by finding the derivative.


4. The number of cane toads in a colony after t months can be modelled by ( t )= 400001+3e−t .

a. Sketch the function C ( t )= 400001+3 e−t

.

a. What is the initial cane toad population?Substitute t=0

C (0 )= 400001+3e−0

¿ 400001+3 ¿10000

The initial cane toad population is 10000.

b. What is the population after 2 months, correct to 3 significant figures?

C (2 )= 400001+3e−2

¿28449.383…≈28400 (correct to 3 significant figures)

The cane toad population after two months is approximately 28400, correct to 3 significant figures.

c. When does the population reach 30 000?Substitute C ( t )=30000

30000= 400001+3e−t

30000 (1+3e−t )=40000

1+3e−t=43

3e−t=43−1

3e−t=13

e−t=19



−t=ln 19

t=−ln 19

¿ ln 9¿2.1972…months

The population will reach 30000 after approximately 2.2months∨¿ 2months∧6days

d. Show that C ' (t )=120000 et

(e t+3 )2

C (t )= 400001+3e−t

¿40000 (1+3e−t )−1

C ' ( t )=−1⋅−1 ⋅3e−t ⋅40000 (1+3e−t )−2

¿ 120000et (1+3e−t )2

¿ 120000

et (1+ 3et )2

¿ 120000

et ( e t+3et )2

¿ 120000e t

(et )2(e t+3 )2

¿ 1200001e t

(e t+3 )2

¿ 120000 et

(et+3 )2as required .

e. Find the maximum growth rate of the cane toad population.

The maximum growth rate of the population is at the time corresponding to the maximum value of the derivative. i.e. Solve C ' ' ( t )=0This corresponds to the point of inflexion in the graph of the original function, which is the C ( t ) value halfway between the asymptotes. i.e. Solve C ( t )=20000

Find the time of the maximum growth rate of the population:

Option 1: Solve C ' ' ( t )=0


C ' (t)=120000 et

(e t+3 )2

C ' (t)=120000 e t (e t+3 )−2

Apply the product rule:

u=120000 et , u'=120000 e t

v=(e t+3 )−2 , v '=−2et (et+3 )−3

C ' ' ( t )=u' v+v ' u

C ' ' ( t )=120000et (et+3 )−2−2 e t (e t+3 )−3⋅120000 e t

¿120000 e t( 1(e t+3 )2

− 2e t

(e t+3 )3 )¿120000 e t( (e t+3 )−2et

(et+3 )3 )¿120000 e t( 3−e

t

(e t+3 )3 )SolveC ' ' ( t )=0

0=120000 et ( 3−et

(e t+3 )3 )0=3−et

e t=3t=ln3¿1.098…

Check for change in concavity either side of t .

C ' ' (1 )=491.466…

C ' ' (1.2 )=−505.205…∴ t=ln3=1.098… is time of the maximum growth rate in the cane toad population.Option 2: Solve C ( t )=20000

20000= 400001+3e−t

20000 (1+3e−t )=400001+3e−t=23e−t=2−13e−t=1

e−t=13



−t=ln 13

t=− ln 13

¿ ln 3¿1.098…

Find the maximum growth rate of the population:

C ' (ln 3 )is the maximum value of the derivative and the maximum growth rate in the cane toad population.

C ' (ln 3 )=120000 et

(e t+3 )2

¿ 120000 eln 3

(eln 3+3 )2

¿ 120000 ⋅3(3+3 )2

¿ 36000036

¿10000

The maximum growth rate in the cane toad population is 10000 toads per month.


5. The population of a town is decreasing at a rate proportional to the square root of the population at that time.

a. Write a differential equation to describe the situation.

Let P' (t) be the rate of change in the population at time t .

P ' (t )∝√tP ' (t )=k √ t where k is a constant.

a. If the population was initially 2500 and decreased to 2025 after 10 years, find the population after 20 years.

P ' ( t )=k √ tP (t )=∫ k √ t ⋅dt

P (t )=∫ k t12 ⋅dt

P (t )=2k t32

3 +C

Use the initial conditions to find C and k :

P (0 )=2500

2500=2k ⋅032

3 +C

2500=C

P (t )=2k t32

3 +2500

P (10 )=2025

2025=2k ⋅1032

3 +2500

−475=2k ⋅1032

3

−1425=2k ⋅1032

−1425

2 ⋅1032

=k

k= −14252√1000

P (t )=2 ⋅ −14252√1000

⋅ t32

3+2500



P (t )=−1425 ⋅ t32

3√1000+2500

Find the population after 20 years:

P (20 )=−1425 ⋅2032

3√1000+2500¿1156.497…

The population of the town after 20 years is 1156.497…


changing the integral activity · web views 10 = -200 10+1 2 + 200 10+1 = -200 121 + 200 11 = -200...

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