1S Summer exam 2010 - Calculus Dr Paul May

1. Answer all parts (a) to (d). All parts carry equal marks.

Determine the following:

(a) dy/dx if y = 3x¹⁰¹

(b) du/dh if u = 3.7h⁴ + (3×10⁶)h²

(c) dΩ/dy if Ω = 5 cos y

(d) dB/d$ if $ = 2 exp(–7B) ‑ 5B²
(4 marks)

2. Answer all parts (a) to (d). All parts carry equal marks.

Differentiate the following functions with respect to x, and simplify the result where appropriate:

(a) (b) y = 7e^-2x sin x

(c) (d) y = ln
(8 marks)

3. Answer all parts (a) to (c).

Consider the function: y =

(a) Differentiate this equation and hence find the co-ordinates of the stationary point(s).

(4 marks)

(b) Find the second differential of this equation, and hence find whether the stationary point(s) is/are a local maximum, minima or points of inflection.

(4 marks)

(c) Sketch the function between x = –5 and x = +5.

(4 marks)

Answers

1)

a) dy/dx = 303x¹⁰⁰ b) du/dh = 14.8h³ + (6×10⁶)h

c) dΩ/dy = –5 sin y d) dB/d$ = ‑14 exp(‑7B) ‑ 10B

2)

a) Rules for Indices: y = – 6x^-5/3 dy/dx = + (30/3)x^-8/3

=

b) Product Rule: 7e^-2x ( cos x) + (sin x) ( – 14e^-2x) = 7e^–2x (cos x – 2sin x)

c) Quotient Rule: =

d) Funct. of a Funct.: dy/dx = × 2 × = 6 / x

alternatively, rewrite it as y = 2 ln

So that dy/dx = 2 ×

3) (a) y =

Use F-of-F Rule and then the Quotient Rule:

= 2

=

At t.p. =0, so either 4(x – 1) = 0 so that x = +1, and y = 0,

or , so that x = ¥ and y = –1. This is an unusual solution and indicates something odd happens at this point.

So there are only 2 t.p.s, at (1, 0) and (¥, ‑1)

(b) d²y/dx² =

(i) When x = 1, d²y/dx² = 1/2, i.e. +ve so this is a minimum.

(ii) When x = ¥, d²y/dx² is undefined….

(c) Need to sketch graph, get correct shape, label axes properly, and label the turning points and places where it crosses the axes to get full marks.

From original eqn:

When x = 0, y = +1.

When x = 1, y = 0 (the p.o.i)

When x = –1 y = ¥ ; so have an asymptote at x = –1

When x = large and +ve, y = tends to +1

When x = large and -ve, y = tends to +1