Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
the dock. The driver steers the boat toward the dock, turning in a parabolic path as it nears. If Linda lets go of the towrope at the right moment, she will glide to a stop near the dock. Let the vertex of the parabola travelled by the boat be the origin. The dock is located 30 m east and 30 m north of the origin. The boat begins its approach to the dock 30 m west and 60 m north of the origin.a) If Linda lets go of the towrope when she reaches point A, where is she headed relative to the dock? Use an equation to describe the path. (Hint: She will travel in a straight line.) b) If Linda lets go of the towrope when she reaches point B, where is Linda headed relative to the dock?c) If Linda waits until point C, how close will her trajectory be to the dock?d) At what point should Linda release the towrope to head straight for the dock?e) What assumptions have you made?f) To make this situation more realistic, what additional information would you need? Create this information, and solve part d) using what you have created. As you work through the task, be sure to explain your reasoning and use diagrams to support your answers.A
abc) The derivative of this function is 2x/15. This is the slope of a tangent at that point. If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point. Since that line has slope 2t/15, we can determine equation of line using point-slope formula: y = m(x-x0) + y0 y = 2t/15 * (x - t) + (1/15)t^2 Plug in the x-coordinate "t" that was given for any point.
d) We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30). So, use our equation for a general tangent picked at point (t, t^2 / 15): y = 2t/15 * (x - t) + (1/15)t^2 And plug in the condition that it must satisfy x=30, y=30. 30 = 2t/15 * (30 - t) + (1/15)t^2 t = 30 ± 2√15 = 8.79 or 51.21 The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point. So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
