# Boat Landing Problem - Subcase2: v_r=v_w

 Sample Problem   A person on a boat in a lake is 4 km from the shore and must go to a point 8 km down the shoreline and 2 km directly inland from there in the shortest possible time. The person walks and the boat travels at the same rate of 6 km per hour. a) Sketch the problem situation. Find total time t . This is a function of what variable? b) Using calculus, find the point on the shore where the person should land to minimize the time. c) Draw a scale drawing of the problem situation. What do you notice about the line connecting the start and stop points? Why do you think it is a straight line? Thinking hints: Suppose the person walks and the boat travels at the same rate of 10 km per hour. What changes and what doesn't change in the above? What do you notice about the derivative function? What would change if the person walked and the boat traveled at different speeds? Does this look like a harder problem? d) Using only geometry (no calculus) and traveling on a straight line between the start and stop points, find the point on the shore where the person should land. (Your answer should be the same as in b.) Solution - with hints! (a): Sketch the problem situation. Find total time t . This is a function of what variable (according to your sketch)? (1) Sketch the problem situation. (2) Find total time t . Find the length of \overline {SR} and \overline {RQ} \overline {SR} = \sqrt {4^2 + x^2 } = \sqrt {x^2 + 16} \overline {RQ} = \sqrt {(8-x)^2 + 2^2 } = \sqrt {x^2 -16x + 68} Find the time for rowing and for walking. t_r = {{\overline {SR} } \over {v_1 }} = {{\sqrt {x^2 + 16} } \over 6} t_w = {{\overline {RQ} } \over {v_2 }} = {{\sqrt {x^2 -16x + 68} } \over 6} Find the total time. t=t_r+t_w = {{\sqrt {x^2 + 16} } \over 6} + {{\sqrt {x^2 -16x + 68} } \over 6} (3) This is a function of what variable (according to your sketch)? According to our sketch, t is a function of x , that is: t = t(x) .   (b):Using calculus, find the point on the shore where the person should land to minimize the time. Extreme values are found where the derivative is zero. (1) Find the derivative: {{{\rm{d}}t} \over {{\rm{d}}x}} . {{{\rm{d}}t} \over {{\rm{d}}x}} = {x \over {6\sqrt {x^2 + 16} }} + {{x - 8} \over {6\sqrt {x^2 - 16x + 68} }} (2) Set the derivative equal to zero and simplify. \displaylines{ 0 = {x \over {6\sqrt {x^2 + 16} }} + {{x - 8} \over {6\sqrt {x^2 - 16x + 68} }} \cr {{ - x} \over {6\sqrt {x^2 + 16} }} = {{x - 8} \over {6\sqrt {x^2 - 16x + 68} }} \cr {{ - x} \over {\sqrt {x^2 + 16} }} = {{x - 8} \over {\sqrt {x^2 - 16x + 68} }} \cr - x\sqrt {x^2 - 16x + 68} = (x - 8)\sqrt {x^2 + 16} \cr x^2 (x^2 - 16x + 68) = (x - 8)^2 (x^2 + 16) \cr x^4 - 16x^3 + 68x^2 = x^4 - 16x^3 + 80x^2 - 256x + 1024 \cr 12x^2 - 256x + 1024 = 0 \cr 3x^2 - 64x + 256 = 0 \cr} (3) Solve for x . \eqalign{ & x_1 ,x_2 = {{64 \pm \sqrt {64^2 - 4 \cdot 3 \cdot 256} } \over {2 \cdot 3}} = {{64 \pm \sqrt {1024} } \over 6} = {{64 \pm 32} \over 6} \cr & x_1 = {{64 + 32} \over 6} = {{96} \over 6} = 16 \cr & x_2 = {{64 - 32} \over 6} = {{32} \over 6} = 5.3 \cr} (4) Choose and check the reasonable result. From our sketch, x must be between 0 and 8, so x=5.3 Solution: The point R is 5.3 km along the shore from point P.   (c): Draw a scale drawing of the problem situation. What do you notice about the line connecting the start and stop points? Why do you think it is a straight line? (1) Draw a scale drawing of the problem situation. (2) What do you notice about the line connecting the start and stop points? The lines \overline {SR} and \overline {RQ} form a straight line. (3) Why do you think it is a straight line? It is a straight line because the speed is the same both over water and over land. This means that it takes the same amount of time to travel the same distance whether the line is over water or over land. We notice that the rate of 6 km per hour cancelled when setting the derivative =0. This means that as long as the rate over water and over land is the same they will cancel and the value of x when the derivative =0 is the same no matter the rate. This is true for 10 km per hour and any other speed as long as both rates are the same.   (d): Using only geometry (no calculus) and traveling on a straight line between the start and stop points, find the point on the shore where the person should land. (Your answer should be the same as in b.) (1) Look at the scale drawing of the problem situation. (2) What triangles are similar? The triangles: \Delta SPR and \Delta SAQ are similar, that is \Delta SPR \sim \Delta SAQ (3) Make a ratio to find the length of \overline {PR} . \eqalign{ &{{\overline {PR} } \over {\,\,\overline {AQ} \,\,}} = {{\overline {SP} } \over {\,\,\overline {SA} \,\,}} \cr &{{\,\overline {PR} \,} \over {\,\,8}} = {4 \over {4 + 2}} \cr &\overline {PR} = {{4 \cdot 8} \over 6} \cr &\overline {PR} = 5.3 \cr}