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 \(5\sqrt 5 \) minutes. When he runs he covers 1 km in 5 minutes.

(a)     If PQ = x km, \(0 \leqslant x \leqslant 2\) , 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.

(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]

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.