Question 10:
--- 10(a) ---
10(a) | dV
Obtain =40π−0.8πr or equivalent
dt | B1 | Need a complete correct statement seen or
implied.
dV dV dr
Obtain =4πr2 or equivalent e.g. =4r2
dr dt dt | B1 | Need a complete correct statement seen or
implied.
Use the chain rule to obtain given answer (including the derivative) | B1 | dr 50−r dr 40−0.8r
Allow if = follows =
dt 5r2 dt 4r2
without further explanation (π already cancelled)
and no incorrect statements seen.
3
--- 10(b) ---
10(b) | Commence division and reach quotient of the form
–5r ± 250
or 5r2 = (50 – r)(Ar + B) + C and reach A = –5 and B = ± 250 | M1 | Allow M1 if divide by r−50 to obtain
5r250.
Obtain quotient –5r – 250 | A1 | Do not need to state which is quotient and which
is remainder. However, if clearly muddled, then
M1A1A0 for both expressions correct.
Obtain remainder 12500 | A1 | Note: 12500 following division by r – 50 is
correct and scores this A1 ISW.
SC B1 only for correct use of remainder
theorem to obtain correct remainder.
3
Question | Answer | Marks | Guidance
--- 10(c) ---
10(c) | Prepare to integrate e.g. separate variables correctly
dt 5r2  12500
= =−( 5r+250 )+
Or express in the form  
dr 50−r 50−r  | B1FT |  5r2
 dr =1dt
50−r
Condone missing dr, dt or missing integral
signs, but not both.
Follow their division in (b) if substitute before
separating.
Obtain term t | DB1
A
Obtain terms r2 +Br−Cln(50−r)
2 | M1 | C
From their Ar + B + in (b) where
50−r
ABC ≠ 0.
Allow a single slip in the coefficients.
5
Obtain terms − r2 −250r−12500ln(50−r)
2 | A1FT | FT their (b), provided of the correct form.
Use t = 0, r = 0 to evaluate a constant or as limits in a solution containing terms of
the form r2, r, ln(50 – r) and t | M1
5
Obtain final answer t=− r2 −250r−12500ln(50−r)+12500ln50
2 | A1 | OE
Must be t = …..
Allow with 12500ln50=48900or better.
6
--- 10(d) ---
10(d) | Obtain t = 70.5 | B1 | May be more accurate (70.4605…).
1
Question | Answer | Marks | Guidance