Skip to content

Course home

Differentiation

Differentiation

EasyMediumHard
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599600601602603604605606607608609610611612613614615616617618619620621622
Question 619

The vertical displacement, y y\,y millimetres, of a vibrating plate in a laboratory experiment is modelled by the function y=f(t)y = f(t)y=f(t), where

f(t)=(t−4)(2t+1)2 f(t) = (t - 4)(2t + 1)^2 f(t)=(t−4)(2t+1)2

for t≥−1t \ge -1t≥−1, where t t\,t is the time in seconds.

The graph of y=f(t)y = f(t)y=f(t) touches the ttt-axis at the point P P\,P and crosses the ttt-axis at the point QQQ.

a.

State the coordinates of the point PPP.

[2]
b.

Find f′(t)f'(t)f′(t).

[3]
c.

Hence show that the equation of the tangent to the curve at the point where t=2.5t = 2.5t=2.5 can be expressed in the form y=ky = ky=k, where k k\,k is a constant to be found.

[3]
d.

The displacement is modified to y=f(t+b)y = f(t + b)y=f(t+b), where b b\,b is a constant. This new curve passes through the origin (0,0)(0, 0)(0,0).

State the possible values of bbb.

[2]
Markscheme

Differentiation Questions

  1. A Level
  2. /Maths
  3. /Differentiation

717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.

Question bank