6 marks
1. At noon, ship A is 30 kilometres west of ship B. Ship A is sailing east at 20 kilometres per
hour and ship B is sailing north at 30 kilometres per hour. How fast is the distance between
the ships changing at 4:00 p.m. on the same day? (You should indicate in your answer if the
distance between the ships is increasing or decreasing.)
Solution:
1. Diagram
2. Variables: Define the distance between the two ships to be z. We want to find z 0 (t).
Define the distance between the ships and the origin to be a and b respectively. Set
the original location of ship B to be origin. After 4hrs, Ship A is 50km east of origin
and ship B is 120km north of origin.
3. Equations:
The Pythagorean theorem here will be useful. The distance between them
√
2
z = 50 + 1202 = 130km. Since A is going away the origin and B is also going away
from origin, we have a0 = 20 and b0 = 30.
4. Differentiating the distance formula: z 2 = a2 + b2 , we have: zz 0 = aa0 + bb0 , substitute
to get: 130z 0 = 50 · 20 + 120 · 30.
5. Rearrange to get z 0 = 460
13 ≈ 35.38km/hr. The distance between the ships are increasing at 460
≈
35.38km/hr.
13
3 marks
2. (a) Use a linear approximation to estimate ln(0.9).
Solution: For this linear approximation, we want to pick f (x) = ln(x) and a = 1.
Then
f (x) = ln(x)
1
f 0 (x) =
x
f (1) = 0
f 0 (1) = 1
So that means our linear approximation is:
L(x) = f (a) + f 0 (a)(x − a)
L(x) = 0 + 1(x − 1) = x − 1
Since we are interested in x = 0.9, we get: L(0.9) = (0.9 − 1) = −0.1. So ln(0.9) ≈
−0.1.
1 mark
(b) Is your answer in part (a) an overestimate, an underestimate, or exactly equal to, the
actual value of ln(0.9)? Justify your answer.
Solution: This function is an over approximation since the function L(x) is above
the function ln(x) at x = 0.9.