1S Summer exam 2002 - Calculus Dr Paul May

 

1.      Answer all parts (a) to (d).

 

Determine the following:

 

            (a) dy/dx   if    y = 8x4       (1 mark)

 

(b) dy/dp if    y = 6p5 + 3p – 2        (1 mark)

 

(c) db/dq  if    b = -7cos q        (2 marks)

 

(d) dj/dk   if    j = 253e–27k       (2 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 possible:

 

(a) y = (7x + 2)(3 – 8x)                                    (b) y = -5x7 ln x

 

(c)                                               (d) y = sin (5x6 – 2x2)  
                                                            (8 marks)

 

3.      Answer all parts (a) to (c).

 

Consider the function   y(x) = (x + 2)3.

 

(a)  Differentiate this function (without multiplying out the brackets) and thence determine the co-ordinates (x,y) of the stationary point(s).
                                                                                                                        (4 marks)

 

(b) Do the stationary point(s) correspond to local maxima, minima, or point(s) of inflection?
                                                                                                                        (4 Marks)

 

(c) Hence sketch this function between x = -4 and x = +1.
                                                                                                                        (3 marks)

Answers

 

1) [1 mark for (a) and (b), 2 marks for the rest].

 

a) dy/dx = 32x3                                    b) dy/dp = 30p4 + 3

c) db/dq = +7sin q                               d) dj/dk = -6831 e-27k

 

2) [2 marks each].

 

a) Product Rule:            (7x + 2).(-8) + (3 - 8x).7         =          5 –112x

 

b) Product Rule:           -5x7(1/x) + (ln x).(-35x6)         =          -5x6(1 + 7ln x)

 

c) Quotient Rule:                             =         

 

d) Funct. of a Funct.:  cos (5x6 – 2x2).(30x5 – 4x)        =          (30x5 – 4x) cos (5x6 – 2x2)

 

3)

a) Using Func. Of Func. Rule, dy/dx = 3(x + 2)2.(1) = 3(x + 2)2         [2 marks]

 

At the t.p. dy/dx = 0, so 3(x + 2)2 = 0,   so there’s only 1 soln at x = -2.

Feeding x = -2 back into the original eqn, we get that the t.p is at: (-2, 0).       [2 marks]

 

b) d2y/dx2 = 6(x + 2), and at x = -2 this has a value of  0.  So the t.p. is a point of inflexion!

Checking the gradient either side of the t.p.:  at x = -3, dy/dx = +3, so it’s increasing.

At x = -1, dy/dx = +3, so it’s increasing too.         [3 marks]

 

c) When x is very large and +ve, y is v. large and +ve.  When x is large and –ve, y is large and –ve. When x = -2, y = 0, and this is the t.p.  When x = 0, the intercept y = 8.
(The curve must be the right shape, the axes must be labelled, and the t.p. and y intercept must also be labelled to get the full 3 marks.)