Question 8:
8 | (a) | 3
( d I ) ( )
2 t − t 2 = 2 t − 2 − 2 2 t ( t − 1 ) I
d t
12
d I ( ) − 2 ( 1 ) t
 = − 2 t 2 t − I
d t 2 − 2 t t
12
d I − 2 ( t 1 )2 ( )
 + I = 2 − 2 t t
d t − 2 t t | *M1 | 3.3 | Rearranging to the form
d I
+ P ( t ) I = Q ( t )
d t
− 2 (t 1 )d − 2 2 t2
I F = e  2 − 2 t t t = e −  − 2 t t d t = e − ln (2 −t 2t ) = 1
2 t − t 2 | A1 | 2.2a | Finding correct integrating
factoras an expression not
involving exponentials and logs. | Ignore unnecessary constant
multiplier here
1
( 2t−t2)−1dI ( 2t−t2)−2 ( 2t−t2)−
 +2(t−1) I = 2
dt
1
d ( )−1  ( )−
 2t−t2 I= 2t−t2 2
dt  | *dep*M
1 | 1.1 | Multiplying both sides by IF
and recognising LHS as exact
derivative of ( 2 t − t 2 ) − 1 I | Can be implied by next M1
Can be awarded from slip in initial
rearrangement if their IF works for
their rearrangement
12
( 2 t − t 2 ) − 1 I =  ( 2 t − t 2 ) − d t =  1 d t
1 − ( t − 1 ) 2 | dep*M1 | 1.1 | Taking integral of both sides
and attempt to complete square
in order to express RHS in a
standard form
(or using the substitution
u = t – 1 including du = dt). | Condone1( t−1 )2 for M1
1
= du
1−u2
= s i n − 1 ( t − 1 ) + c | *A1 | 1.1 | For correctly integrating RHS to
a function of t. “+ c” not
necessary here.
t = 1 , I = 5  ( 2 − 1 − 1 ) 5 = s i n − 1 (1 − 1 ) + c
 c = 5
 ( 2 t − 2 t ) − 1 I = s i n − 1 ( t − 1 ) + 5
 I = ( 2 t − 2 t ) ( s i n − 1 − ( t 1 + ) 5 ) | dep*A1 | 3.3 | AG so use of relevant condition
must be explicit.
Verification of AG by
substitution is not sufficient;
value of c must be derived. | Ignore workings using other
condition (t = 0, I = 0)
[6]
(b) | I = 0  ( 2 t − t 2 ) ( s i n − 1 − ( t 1 ) + 5 ) = 0
 t ( 2 − t ) ( s i n − 1 ( t − 1 ) + 5 ) = 0
 t = 2
So length of surge is 2 – 0 = 2 (seconds) | B1 | 3.1a | If no other later comment or
statement to the contrary then
accept just t = 2
o r s i n − 1 ( t − 1 ) + 5 = 0
 s i n − 1 ( t − 1 ) = − 5
which is not possible (since
–½π sin–1(t – 1)  ½π.) | B1 | 2.4 | Do not accept incorrect
explanation eg
–1  sin–1(t – 1)  1 | If B0B0 thenSc1if length of surge
determined to be awrt 1.96 s from
sin(–5rads) + 1 after t = 2 found
[2]
(c) | ( I = − ( 2 t 2 t ) ( s − i n 1 ( t − ) + 1 ) 5 )   1 ( t − 2 ) 1
3
& ( 2 2 − t t d ) I ( = 2 t − ) 2 − 2 t 2 − ( t 1 ) I
d t
3
( 2 − t t 2 ) 2 − 2 ( − t 1 ) ( 2 t 2 − t ) ( s i n − 1 ( t − 1 ) + 5 )
d I
 =
d t − 2 t 2 t | *M1 | 3.4 | d I
Finding an expression for in
d t
terms of t by either eliminating
given I from given DE or
differentiating given I(t)using
product rule. | ( I = ( 2 t − t 2 ) ( s i n − 1 ( t − 1 ) + 5 )  d I ) =
d t
( 2 t − 2 t )
2 (1 − t ) ( s i n − 1 ( t − 1 ) + 5 ) +
− 1 ( t − 1 ) 2
12
 d I = ( 2 t − t 2 ) − 2 ( t − 1 ) ( s i n − 1 ( t − 1 ) + 5 )
d t | M1
dep * | 2.2a | dI
Simplifying their so that
dt
there is no denominator which
is zero at t = 0 | dI
 =
dt
(2t−t2)
2(1−t)(sin −1(t−1)+5)+
2t−t2
1
=(2−2t)(sin −1(t−1)+5)+(2t−t2)2
t 0  =
d I
0 2 ( 1 ) ( s in 1 ( 1 ) 5 )  = − − − − +
d t
1  
2 5 1 0 ( u n it s /s )   = − = −
2 | A1 | 3.4 | Must be in a simplified non-
trigonometric form.
Ignore units. | If M1M0 or M0M0, then Sc B1 for
correct answer
[3]