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√5 minutes. When he runs he covers 1 km in 5 minutes.
(a) If PQ = x km, 0⩽ , find an expression for the time T minutes taken by André to reach point Y.
(b) Show that \frac{{{\text{d}}T}}{{{\text{d}}x}} = \frac{{5\sqrt 5 x}}{{\sqrt {{x^2} + 4} }} - 5.
(c) (i) Solve \frac{{{\text{d}}T}}{{{\text{d}}x}} = 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 \frac{{{{\text{d}}^2}T}}{{{\text{d}}{x^2}}} = \frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{\frac{3}{2}}}}} and hence show that the time found in part (c) (ii) is a minimum.
Markscheme
(a) {\text{AQ}} = \sqrt {{x^2} + 4} {\text{ (km)}} (A1)
{\text{QY}} = (2 - x){\text{ (km)}} (A1)
T = 5\sqrt 5 {\text{AQ}} + 5{\text{QY}} (M1)
{\text{ = 5}}\sqrt 5 \sqrt {({x^2} + 4)} + 5(2 - x){\text{ (mins)}} A1
[4 marks]
(b) Attempting to use the chain rule on {\text{5}}\sqrt 5 \sqrt {({x^2} + 4)} (M1)
\frac{{\text{d}}}{{{\text{d}}x}}\left( {{\text{5}}\sqrt 5 \sqrt {({x^2} + 4)} } \right) = 5\sqrt 5 \times \frac{1}{2}{({x^2} + 4)^{ - \frac{1}{2}}} \times 2x A1
\left( { = \frac{{5\sqrt 5 x}}{{\sqrt {{x^2} + 4} }}} \right)
\frac{{\text{d}}}{{{\text{d}}x}}\left( {5(2 - x)} \right) = - 5 A1
\frac{{{\text{d}}T}}{{{\text{d}}x}} = \frac{{5\sqrt 5 x}}{{\sqrt {{x^2} + 4} }} - 5 AG N0
[3 marks]
(c) (i) \sqrt 5 x = \sqrt {{x^2} + 4} A1
Squaring both sides and rearranging to obtain 5{x^2} = {x^2} + 4 M1
x = 1 A1 N1
Note: Do not award the final A1 for stating a negative solution in final answer.
(ii) T = 5\sqrt 5 \sqrt {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{\text{ , }}v = \sqrt {{x^2} + 4} {\text{, }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1{\text{ and }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = x{({x^2} + 4)^{ - 1/2}} (A1)
\frac{{{{\text{d}}^2}T}}{{{\text{d}}{x^2}}} = 5\sqrt 5 \left[ {\frac{{\sqrt {{x^2} + 4} - \frac{1}{2}{{({x^2} + 4)}^{ - 1/2}} \times 2{x^2}}}{{({x^2} + 4)}}} \right] A1
Attempt to simplify (M1)
= \frac{{5\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}}[{x^2} + 4 - {x^2}]\,\,\,\,\,or equivalent A1
= \frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}} AG
When x = 1{\text{ , }}\frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}} > 0 and hence T = 30 is a minimum R1 N0
Note: Allow FT on incorrect x value, 0 \leqslant x \leqslant 2.
METHOD 2
Attempting to use the product rule M1
u = x{\text{ , }}v = \sqrt {{x^2} + 4} {\text{, }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1{\text{ and }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = x{({x^2} + 4)^{ - 1/2}} (A1)
\frac{{{{\text{d}}^2}T}}{{{\text{d}}{x^2}}} = 5\sqrt 5 {({x^2} + 4)^{ - 1/2}} - \frac{{5\sqrt 5 x}}{2}{({x^2} + 4)^{ - 3/2}} \times 2x A1
\left( { = \frac{{5\sqrt 5 }}{{{{({x^2} + 4)}^{1/2}}}} - \frac{{5\sqrt 5 {x^2}}}{{{{({x^2} + 4)}^{3/2}}}}} \right)
Attempt to simplify (M1)
= \frac{{5\sqrt 5 ({x^2} + 4) - 5\sqrt 5 {x^2}}}{{{{({x^2} + 4)}^{3/2}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { = \frac{{5\sqrt 5 ({x^2} + 4 - {x^2})}}{{{{({x^2} + 4)}^{3/2}}}}} \right) A1
= \frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}} AG
When x = 1{\text{ , }}\frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}} > 0 and hence T = 30 is a minimum R1 N0
Note: Allow FT on incorrect x value, 0 \leqslant x \leqslant 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.