Date | May 2008 | Marks available | 18 | Reference code | 08M.1.hl.TZ2.13 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find, Show that, Solve, Determine, and Hence | Question number | 13 | Adapted from | N/A |
Question
André wants to get from point A located in the sea to point Y located on a straight stretch of beach. P is the point on the beach nearest to A such that AP = 2 km and PY = 2 km. He does this by swimming in a straight line to a point Q located on the beach and then running to Y.
When André swims he covers 1 km in 5√55√5 minutes. When he runs he covers 1 km in 5 minutes.
(a) If PQ = x km, 0⩽x⩽2 , find an expression for the time T minutes taken by André to reach point Y.
(b) Show that dTdx=5√5x√x2+4−5.
(c) (i) Solve dTdx=0.
(ii) Use the value of x found in part (c) (i) to determine the time, T minutes, taken for André to reach point Y.
(iii) Show that d2Tdx2=20√5(x2+4)32 and hence show that the time found in part (c) (ii) is a minimum.
Markscheme
(a) AQ=√x2+4 (km) (A1)
QY=(2−x) (km) (A1)
T=5√5AQ+5QY (M1)
= 5√5√(x2+4)+5(2−x) (mins) A1
[4 marks]
(b) Attempting to use the chain rule on 5√5√(x2+4) (M1)
ddx(5√5√(x2+4))=5√5×12(x2+4)−12×2x A1
(=5√5x√x2+4)
ddx(5(2−x))=−5 A1
dTdx=5√5x√x2+4−5 AG N0
[3 marks]
(c) (i) √5x=√x2+4 A1
Squaring both sides and rearranging to obtain 5x2=x2+4 M1
x = 1 A1 N1
Note: Do not award the final A1 for stating a negative solution in final answer.
(ii) T=5√5√1+4+5(2−1) M1
= 30 (mins) A1 N1
Note: Allow FT on incorrect x value.
(iii) METHOD 1
Attempting to use the quotient rule M1
u=x , v=√x2+4, dudx=1 and dvdx=x(x2+4)−1/2 (A1)
d2Tdx2=5√5[√x2+4−12(x2+4)−1/2×2x2(x2+4)] A1
Attempt to simplify (M1)
=5√5(x2+4)3/2[x2+4−x2]or equivalent A1
=20√5(x2+4)3/2 AG
When x=1 , 20√5(x2+4)3/2>0 and hence T = 30 is a minimum R1 N0
Note: Allow FT on incorrect x value, 0⩽x⩽2.
METHOD 2
Attempting to use the product rule M1
u=x , v=√x2+4, dudx=1 and dvdx=x(x2+4)−1/2 (A1)
d2Tdx2=5√5(x2+4)−1/2−5√5x2(x2+4)−3/2×2x A1
(=5√5(x2+4)1/2−5√5x2(x2+4)3/2)
Attempt to simplify (M1)
=5√5(x2+4)−5√5x2(x2+4)3/2(=5√5(x2+4−x2)(x2+4)3/2) A1
=20√5(x2+4)3/2 AG
When x=1 , 20√5(x2+4)3/2>0 and hence T = 30 is a minimum R1 N0
Note: Allow FT on incorrect x value, 0⩽x⩽2.
[11 marks]
Total [18 marks]
Examiners report
Most candidates scored well on this question. The question tested their competence at algebraic manipulation and differentiation. A few candidates failed to extract from the context the correct relationship between velocity, distance and time.