1S Summer exam 2001 - Calculus Dr Paul May

1) Determine the following:

a) dy/dx if y = 3x³ b) dy/dm if y = 4m⁷ + 5m - 1

c) dy/dq if y = 9cos q d) dp/dq if p = 1252e^–56q (6 marks)

2) Differentiate the following functions with respect to x, and simplify the result where possible:

a) y = (5x + 2)(3 – 6x) b) y = 2x³ ln x

c) d) y = sin (5x⁴ – 9x) (8 marks)

3) The function :

y =

has stationary points at r = ± ¥

a) Differentiate this function and thence determine the co-ordinates (r,y) of the remaining stationary point. (3 marks)

b) The second differential of this function is:

Determine whether the stationary point you just found is a local maximum or minimum. (3 Marks)

c) Hence sketch this function between r = 0 and r = 8. (4 marks)

Answers

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

a) dy/dx = 9x² b) dy/dr = 28m⁶ + 5

c) dy/dq = -9sin q d) dp/dq = -70112e^-56q

2) [2 marks each].

a) Product Rule: (5x + 2).(-6) + (3 - 6x).5 = 3 - 60x

b) Product Rule: 2x³(1/x) + (ln x).6x² = 2x²(1 + 3ln x)

c) Quotient Rule: =

d) Funct. of a Funct.: cos (5x⁴ – 9x).(20x³ – 9) = (20x³ – 9) cos (5x⁴ – 9x)

3)

a)

Quotient Rule: dy/dr = (r.2e^r - 2e^r.1) / r² = 2e^r(r - 1) / r² [2 marks]

For turning point, 2e^r(r - 1) / r² = 0, so either: r² = ¥ Þ r = ± ¥,

or 2e^r = 0 Þ r = - ¥

or (r - 1) = 0, Þ r = 1

The last answer is the required one.

So the turning point is at (1, 5.44). [1 mark]

b) Determine the sign of the second differential, d²y/dr². Putting in the value of r = 1, we get d²y/dr² = 4e, which is +ve, so the t.p. is a local minimum. [3 marks]

c) Sketch: (must get correct shape, label axes, and indicate t.p., and infinities for full 4 marks).

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