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The MEDIAN score is C+. 49% of students get B or greater21% of students get better than 33/40. Losing any marks will have a big impact on your Study Score. This was about the same as previous years.23/40 is the MEDIAN grade.
Exam 1 Assessors Report
Hand writing and general setting out of work a concern.Use logical connectors between lines.
6 + xx2 + 4
=6
x2 + 4+
xx2 + 4
6x2 + 4
+x
x2 + 4∫ ⋅dx
=3⋅Tan−1 x2
⎛⎝⎜
⎞⎠⎟+12⋅ln x2 + 4 +c
Most students who recognised to split the fraction into two parts performed quite well.
Not very well done. Average 0.9/3. Many student omitted cos(x)=0 as a possible solution. Many students only gave solutions for 0≤x≤2π.
Real coefficients so look for conjugate pairThird solution must have real root.
Many students attempted long division and were unable to find the correct solutions.
Many students incorrectly put the solution z=3 at z=2 confusing the question with a root of a single complex number and not an expression.
Conjugate is a reflection in the Re axisMultiply by i is a rotation of pi/4 anti-clockwise, in this case 3 timesThis is the same result as C.
Perpendicular bisector of origin and z=-2iLine z=-i. This is BNote: Im(z) must be R, C is incorrectly represented.
Use calculator program or other to show y(0.2)=1.1995In this region cos(x) is concave down OR gradient is decreasingAnswer will be an over estimate: B
Area=π8
r22 −r1
2( )
Area=π8
22 −12( ) =3π8
units2
z=cisπ12
⎛⎝⎜
⎞⎠⎟
Re z( ) =0, when Arg z( ) =π2
or−π2
⇒ zn ∀n=6 or −6
General soln n=6+12k ,k∈J
z=±i
17 tan−1
0
19
∫πt6
⎛⎝⎜
⎞⎠⎟⋅dt
120×25
−1100
145t−t2( )+ 25⎛⎝⎜
⎞⎠⎟⋅dt
0
20
∫
Need to have completed an earlier section to get first part. Most errors were because of incorrect terminals
3637m use CASDon’t round each answer this may lead to errors, add separately and then round.
400 km
When directly above P, i component is zero, which makes sin=0. When sin=0, cos=1. 6800-6400=400 km
a(t)=&&r t( ) =−1.32π 2 ⋅6800sin π 1.3t−0.1( )( )i −1.32π 2 ⋅6800cos π 1.3t−0.1( )( ) j
a t( )⊥v t( ) as
v(t)×a(t) =0where −Acosθ sinθ + Acosθ sinθ =0
speed=1.3×π ×6800=27,772 m/ s
x =6800sin π 1.3t−0.1( )( )
y=6800cos π 1.3t−0.1( )( )−6400
circlemoved← 6400
⇒ x2 + y+6400( )2 =68002
Use Pythagora’s Thm and a substitution of the time function:
x2 + y2 =10002
6800sinθ( )2 + 6800cosθ −6400( )
2 =10002
whereθ =π 1.3t−0.1( )−π ≤θ ≤2πgivesθ =−0.139, 0.139, 6.144t=0.04, 0.11, 1.58,...
t=1.58 is outside the domain, only two answers.
x =12
u+v( )t
10 =12
0 + 6( )t
t=103
sec
3.3 sec scores zero. Do not give approximate answers unless specified.
u =10m/ s x=ut+12
at2
v= −6 =10t−4.9t2
a=−9.8m/ s2 t=2.5 sec
x=−6m(asit falls)t=
dv
dt=− 196−v2
10dtdv
=−10
196−v2
⇒ t=−10⋅Sin−1 v14
⎛⎝⎜
⎞⎠⎟+c
whent=0, v=7
0 =−10⋅Sin−1 12
⎛⎝⎜
⎞⎠⎟+c
c=5π3
5π30
−t10
=Sin−1 v14
⎛⎝⎜
⎞⎠⎟
v=14sinπ6−
t10
⎛⎝⎜
⎞⎠⎟
or 14cosπ3+
t10
⎛⎝⎜
⎞⎠⎟
v=0, 0 =14sinπ6−
t10
⎛⎝⎜
⎞⎠⎟
t=5π3
sec
dx
dt=14⋅Sin
π6−
t10
⎛⎝⎜
⎞⎠⎟
x=140⋅Cosπ6−
t10
⎛⎝⎜
⎞⎠⎟+c
whent=0, x=0
⇒ c=−70 3
x=140⋅Cosπ6−
t10
⎛⎝⎜
⎞⎠⎟−70 3
whent=5π3
x=18.8m OR x= 14⋅Sinπ6−
t10
⎛⎝⎜
⎞⎠⎟⋅dt=18.8m
0
5π3
∫
y=kx⋅e2x dydx
=k(2x+1)⋅e2x d2ydx2 =4k(x+1)⋅e2x
⇒ 4k(x+1)−2k(2x+1)+ 5kx=15x+6⇒ 4kx+ 4k−4kx−2k+5kx=15x+6k=3
Many students did not use the product rule to find dy/dx.
F =0∑considerhorizontal onlyT1sin30 =T2 sin60
T1 =T2 3
T2 =T1
3
⇒ T2 ≤ 98
∴T1 ≤ 3 × 98
F = 0 vertical only∑T1 cos30 +T2 cos60 ≥ mg
98 3 ×3
2+ 98 ×
1
2≥ mg
m ≤196
g= 20 kg
y=cosec2 πx6
⎛⎝⎜
⎞⎠⎟
43=cosec2 πx
6⎛⎝⎜
⎞⎠⎟
±32
=sinπx6
⎛⎝⎜
⎞⎠⎟
πx6
=π3,2π3
,4π6
,5π6
,...
x=2,4,8,10
pts⇒ 2,43
⎛⎝⎜
⎞⎠⎟, 4,
43
⎛⎝⎜
⎞⎠⎟, 8,
43
⎛⎝⎜
⎞⎠⎟, 10,
43
⎛⎝⎜
⎞⎠⎟
Many students miss the last two points that come from the negative sine angle.
z=x−iyz =x+ iy1z=
1x+ iy
×x−iyx−iy
1z=
x−iyx2 + y2
Same angle but closer, as it will be divided by magnitude of z. C
A Perpendicular vectorsB Equal vectorsC Perpendicular diagonalsD Diagonals same lengthE Vectors equal and opposite direction
−dx
dy=2 y −1( )x −1( )
−dy
dx=
x −1( )2 y −1( )
dy
dx+
x −1( )2 y −1( )
= 0
Quite a difficult question. Only 25% correct.
When x=0, gradient is constant and positive. Omit EWhen y=0, gradient is constant and negative. Omit CWhen y=x, m is negative. Omit DWhen y=-x m is positive. Omit BLeaves A
t =0, xo =5
xn+1 =xn +h⋅f ' tn( )
⇒ x1 =xo +h⋅f ' t0( )x1 =5−0.5⋅1.0 =4.5
t=0.5, xo =4.5
x2 =4.5−0.5⋅109.5
=3.97
Could also use CAS program to find the answer.Be careful with the negative gradient.
v= a⋅dt∫v(t) =−9.8tk+c
whent=0,v=35i +5 j +24.5k=c
v(t) =35i +5 j + 24.5−9.8t( ) k
r t( ) = v⋅dt∫whent=0,r(t) =0⇒ c=0
r(t) =35ti +5tj + 24.5t−4.9t2( ) k
Not very well done for a easy question. Many answer lacking care and details. Marks were lost for omitting the tilda’s on the vectors.
when dist k =0⇒ 24.5 =4.9tt=5.0sec
r(5) =175i +25 j +0k
asholeis200i
position fromhole=25i +25 j
25i +25 j =35m
2gsin 2θ( )−T =2a
T −3gsinθ =3a
⇒ T =2gsin 2θ( )−2a=3a+ 3gsinθ⇒ T =2gsin 2θ( )−3gsinθ =5a
but sin 2x( ) =2sin x( )cos x( )
4gsin θ( )cos θ( )−3gsinθ =5a
a=gsin θ( )
54cos θ( )−3( )
⇒ a = 0
4cos θ( ) − 3 = 0
θ = cos−13
4⎛⎝⎜
⎞⎠⎟= 41.40
F =ma∑5a=2gsin2θ −3gsinθ −μN5a=2gsin60−3gsin30−μ3gcosθ
5a=2g32
−3g12−μ3g 3
210a=2g 3 −3g−μ3g 3
a=0.2ms−2
Conc.=mass
volume=
x10 +10t
volume starts at 10 and increases by 10 every min
dx
dt=Input−Output
dxdt
=20e−0.2t −10x
10 +10tdxdt
=20e−0.2t −x
1+ t
⇒dxdt
+x
1+ t=20e−0.2t
must show dx/dt=Input-Output
dx
dt=20 t2 + 7t+ 31( )e−0.2t
t+1( )2 −
600t+1( )
2
& x(t) =600t+1
−100e−0.2t t+6( )
t+1
ifdxdt
+x
t+1=20e−0.2t
⇒20 t2 + 7t+ 31( )e−0.2t
t+1( )2 −
600t+1( )
2 +1
t+1⎛⎝⎜
⎞⎠⎟
600t+1
−100e−0.2t t+6( )
t+1⎛
⎝⎜⎞
⎠⎟
⇒20t2 +140t+620( )e−0.2t
t+1( )2 + −
e−0.2t 100t+600( )t+1( )
2
⎛
⎝⎜
⎞
⎠⎟
⇒20t2 + 40t+20( )e−0.2t
t+1( )2 =
20 t2 +2t+1( )e−0.2t
t+1( )2
=20e−0.2t
Initial cond.
x(0)=600−600 =0thisneedstobeshown.
Must have turning point. I would have end point, it needed to be shown in the correct square. Marks were lost if the graph did not show correct concavity.
AB=−a+b
= −i −2 j −2k( )+ −i + 3 j + 4k( )
= −2i + j +2k( )
a⋅b=a bcosθ−2 +2 + 4 =3⋅3cosθ
⇒ cosθ =49
Must show step 2 to get mark.
Area=12
a⋅b⋅sinθ
if cosθ =49
sinθ =659
Area=123⋅3⋅
659
=652
units2
Many students were unable to start this problem. Either ignoring the instruction hence, or unable to find sin(x) given cos(x)=4/9.
Need to identify the Difference of Perfect Squares.
area=− x2 −1( ) x+1−1
1
∫ ⋅dx
area=− x−1( ) x+1( ) x+1−1
1
∫ ⋅dx
area= x−1( ) x+1( )32
1
−1
∫ ⋅dx
letu=x+1 and x−1=u−2terms: x=−1⇒ u=0
x=1⇒ u=2
area= u−2( ) u( )32
2
0
∫ ⋅du
area= u52 −2u
32
⎛
⎝⎜⎞
⎠⎟2
0
∫ ⋅du
area=27
u72 −
45
u52
⎡
⎣⎢
⎤
⎦⎥2
0
area= 0( )−27
2( )72 −
45
2( )52
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢⎤
⎦⎥
area=45
2( )52 −
27
2( )72 =
256 27
−128 2
5
area=32 235
Not A, C and D
B has the same co-efficient for x2 and y2. Asymptote y=x
Only E will have non-perpendicular asymptote.
Gradient is independent of t: NOT AGradient for positive V is negative: NOT B or DAs V increases Abs(V) increases NOT EMust be C
V =e−t
dVdt
=−e−t =−V
∴C
OR
F =ma∑ma=mgsinθ −μNma=mgsinθ −μmgcosθa=gsinθ −μgcosθa=9.8sin30− 0.077( ) 9.8( )cos30 =4.244ms−2
v=u+at
v=3×4.244 =12.7ms−1
y=Sin−1 2x2 −1( )
let u=2x2 −1⇒dudx
=4x
y=Sin−1 u( )⇒dydu
=1
1−u2
dydx
=dydu
dudx
=4x
1−u2
=4x
1− 2x2 −1( )2=
4x
4x2 1−x2( )
=2x
1−x2( )
and a=1
z3 =20012 cis
−3π4
⎛⎝⎜
⎞⎠⎟
z= 20012 cis
−3π4
⎛⎝⎜
⎞⎠⎟
⎛
⎝⎜⎞
⎠⎟
13
z=20016 cis
−π4
⎛⎝⎜
⎞⎠⎟
also 20016 cis
−π4
+2π3
⎛⎝⎜
⎞⎠⎟=200
16 cis
5π12
⎛⎝⎜
⎞⎠⎟
20016 cis
−π4
+4π3
⎛⎝⎜
⎞⎠⎟=200
16 cis
13π12
⎛⎝⎜
⎞⎠⎟=200
16 cis
−11π12
⎛⎝⎜
⎞⎠⎟
u =2 10cis −α( )
w= 10cis π / 2−α( )
⇒ u=2 10cis α( )
iw= 10cis π −α( ), rotatebyπ / 2
uiw
=2 10cis α( )10cis π −α( )
=2cis 2α −π( )