User interface language: English | Español

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 5555 minutes. When he runs he covers 1 km in 5 minutes.

(a)     If PQ = x km, 0x2 , find an expression for the time T minutes taken by André to reach point Y.

(b)     Show that dTdx=55xx2+45.

(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=205(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=(2x) (km)     (A1)

T=55AQ+5QY     (M1)

 = 55(x2+4)+5(2x) (mins)     A1

[4 marks]

 

(b)     Attempting to use the chain rule on 55(x2+4)     (M1)

ddx(55(x2+4))=55×12(x2+4)12×2x     A1

(=55xx2+4)

ddx(5(2x))=5     A1

dTdx=55xx2+45     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=551+4+5(21)     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+4dudx=1 and dvdx=x(x2+4)1/2     (A1)

d2Tdx2=55[x2+412(x2+4)1/2×2x2(x2+4)]     A1

Attempt to simplify     (M1)

=55(x2+4)3/2[x2+4x2]or equivalent     A1

=205(x2+4)3/2     AG

When x=1 , 205(x2+4)3/2>0 and hence T = 30 is a minimum     R1     N0

Note: Allow FT on incorrect x value, 0x2.

 

METHOD 2

Attempting to use the product rule     M1

u=x , v=x2+4dudx=1 and dvdx=x(x2+4)1/2     (A1)

d2Tdx2=55(x2+4)1/255x2(x2+4)3/2×2x     A1

(=55(x2+4)1/255x2(x2+4)3/2)

Attempt to simplify     (M1)

=55(x2+4)55x2(x2+4)3/2(=55(x2+4x2)(x2+4)3/2)     A1

=205(x2+4)3/2     AG

When x=1 , 205(x2+4)3/2>0 and hence T = 30 is a minimum     R1     N0

Note: Allow FT on incorrect x value, 0x2.

 

[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.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Optimization problems.

View options