SCHOOL OF CHEMISTRY

CHEMISTRY 1S

Calculus I - Dr Paul May

Answers to Problems 1

1. (a)
xy = x3 - 2x + 1
-2-3
-1.50.625
-12
-0.51.875
01
0.50.125
10
1.51.375
25

For small values of x the leading term after the constant 1 is -2x. This has a negative multiplier, so the line passes through (x,y) = (0,1) with a negative slope. For large values of x the leading term is now x3 which is large and positive for positive large x; and is large and negative for large negative x. This explains the overall shape of the curve.

(b)
ω
7
0.5
9
1.25
10
2
11
2.5
12
2
13
1.25
15
0.5

Note: this curve is symmetrical about the line ω = 11. This is because the denominator can be written as [(ω-11)2 + 4], and is thus the sum of two squares.

2. (a) Since y is the product of three factors it will take the value zero when any one of the factors is itself zero. This will occur for x = 2, x = 3, and x = 4. Furthermore, if x 4 all three factors will be positive and thus y will be positive, and for x 2 all three factors will be negative and thus y will be negative. This is therefore another cubic equation with a shape very similar to that of 1(a).


(b) The numerator is zero for x = 0, so that the curve passes through the origin (0,0). The denominator is the sum of the square of x and a positive number, and is therefore always positive. Thus y will be positive for positive x, and negative for negative x. Finally, for large positive or negative x the denominator will become much larger than the numerator, so that y will tend to zero.


3. (a) 5x4 (b) 4x (c) 21x2

(d) 2x3 (e) 1½x5 (f) 3400x33

(g) 43.8x2 (h) 1.16x22.2 (i) 2104x3

4. Emphasise that we don't always use x and y, other symbols, even strange, unfamiliar ones may be used in Chemistry...

(a) 6p2 (b) 15 (c) 20h3

(d) 120z7 (e) 9x2 (f) 4½λ5

(g) 36ξ17 (h) 1½ζ2 (g) 63¥2