Skip to content

Course home

Sign up

1 Pure Mathematics

EasyMediumHard
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497
Question 183
i.

The volume, VVV, of a spherical weather balloon is increasing at a constant rate of 120π cm3 s−1120\pi \text{ cm}^3\text{ s}^{-1}120π cm3 s−1. Find the rate of increase of the radius, rrr, of the balloon in cm s−1\text{cm s}^{-1}cm s−1 at the instant when the radius is 6 cm6 \text{ cm}6 cm. [The volume VVV of a sphere of radius rrr is given by the formula V=43πr3V = \frac{4}{3}\pi r^3V=34​πr3]

[4]
ii.

The height of a pile of sand, h metresh \text{ metres}h metres, under a conveyor belt is monitored over time. The rate of increase in the height of the pile is modeled by the differential equation

dhdt=kh2 \frac{\text{d}h}{\text{d}t} = \frac{k}{h^2} dtdh​=h2k​

where kkk is a positive constant and ttt hours is the time after the measurement began. Given that:

  • initially (at t=0t = 0t=0), the height of the pile was 3 metres3 \text{ metres}3 metres.
  • 444 hours after monitoring began, the height of the pile was 5 metres5 \text{ metres}5 metres.
  • TTT hours after monitoring began, the height of the pile was 8 metres8 \text{ metres}8 metres.

Solve the differential equation to find the value of TTT. Give your answer to one decimal place.

[6]
Markscheme

1 Pure Mathematics Questions

  1. A Level
  2. /Maths
  3. /1 Pure Mathematics

Question bank