1S Summer exam 2008 - Calculus Dr Paul May

 

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

 

Determine the following:

 

            (a) dy/dx   if    y = 16x36

 

(b) du/dg if    u = 1.7g3 + (3×106)g

 

(c) dΩ/dy   if    Ω = 3tan y

(d) dñ/d–   if    ñ = 2 exp(–3–) ‑ 5–
                                                                                                            (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 = 12e-2x sin x

 

(c)                                            (d) y =  5 ln
                                                            (8 marks)

 

3.      Answer all parts (a) to (c).

 

      Consider the quartic function:           y = (x – 1)(x + 1)( x – 3)( x + 3)

 

(a) Multiply out the brackets in this equation.

                                                                                                                        (2 marks)

(b) Differentiate either version of this equation and hence find the co-ordinates of the stationary points.

                                                                                                                        (4 marks)

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

                                                                                                                        (3 marks)

(d) Sketch this function between x = ‑4 and x = +4.

                                                                                                            (3 marks)

Answers

 

1)

a) dy/dx = 576x35                                   b) du/dg = 5.1g2 + 3×106

c) dΩ/dy = 3 / cos2y                             d) dñ/d– = ‑6 exp(‑3–) ‑ 5

2)

 

a) Rules for Indices:      y = ½x‑3 – x2/3 + 4x‑1/2              dy/dx = ‑(3/2)x-4 – (2/3)x-1/3 – 2x-3/2

                                                                                             = 

 

b) Product Rule:           12e-2x ( cos x)  +  (sin x) (-24e-2x)        =   12e-2x (cos x ‑ 2sin x)

 

c) Quotient Rule:          

 

d) Funct. of a Funct.:    dy/dx =    

Alternatively, by the Laws of Logs,  

5 ln is the same as  ‑5 ln(2x2),  so dy/dx =

 

3)  (a)   y = (x – 1)(x + 1)( x – 3)( x + 3)          =          x4 – 10x2 + 9

 

  (b) dy/dx = 4x3 ‑ 20x

 

  At the tps, dy/dx = 0,    so     4x3 ‑ 20x = 0

                                                x3 ‑ 5x = 0

                                                x (x2 – 5) = 0

 

Therefore, x = 0           or         (x2 – 5)=0, i.e.    x = ±Ö5

 

When x=0,   y = +9,

When x=+Ö5,   y = ‑ 16,

When x= ‑ Ö5,   y = ‑ 16.

 

So there are three turning points, at: (0, 9), (Ö5, ‑16), ( ‑Ö5, ‑16)

 

(c) d2y/dx2 = 12x2 ‑ 20

 

When x=0,   d2y/dx2 = ‑20 (i.e. negative),  so that the tp at  (0, 9) is a maximum.

When x=+Ö5,   d2y/dx2 = +40 (i.e. positive),  so that the tp at  (Ö5, ‑16) is a minimum.

When x= ‑Ö5,   d2y/dx2 = +40 (i.e. positive),  so that the tp at  (‑Ö5, ‑16) is a minimum.

 

(d) 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.