Question: The Water Skier: Where’s the Dock?
Water-skiing is a popular summertime activity across Canada. A skier, holding onto a towrope, is pulled across the water by a motorboat.
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.
Graph: a U-shaped parabola opening upward is shown in the center-right of the page; the skiers/boat are labeled near the upper left branch with point A, point B is lower on the left branch, and point C is at the vertex; the dock icon is to the right of the parabola. position_hint = centersolve using power and tangent of the line
1. **Problem Statement:**
Linda is water-skiing and the boat follows a parabolic path with vertex at the origin $(0,0)$. The dock is at $(30,30)$ and the boat starts at $(-30,60)$. Points $A$, $B$, and $C$ lie on the parabola. We analyze Linda's trajectory if she releases the towrope at these points.
2. **Parabola Equation:**
Given vertex at origin and points $(-30,60)$ and $(30,30)$, assume parabola opens upward with equation $$y = ax^2 + bx + c.$$ Since vertex is at origin, $c=0$ and $b=0$ (vertex form $y = ax^2$).
Use point $(-30,60)$:
$$60 = a(-30)^2 = 900a \implies a = \frac{60}{900} = \frac{1}{15}.$$
So parabola equation is:
$$y = \frac{1}{15}x^2.$$
3. **Coordinates of points:**
- Point $A$: near upper left branch, approximate $x=-20$, then $y=\frac{1}{15}(-20)^2=\frac{400}{15}=26.67$.
- Point $B$: lower on left branch, approximate $x=-10$, then $y=\frac{1}{15}100=6.67$.
- Point $C$: vertex at origin $(0,0)$.
4. **Trajectory after release:**
Linda travels in a straight line tangent to the parabola at the release point.
5. **Find tangent line at point $(x_0,y_0)$:**
Derivative:
$$y' = \frac{d}{dx}\left(\frac{1}{15}x^2\right) = \frac{2}{15}x.$$
Slope at $x_0$:
$$m = y'(x_0) = \frac{2}{15}x_0.$$
Equation of tangent line:
$$y - y_0 = m(x - x_0).$$
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**a) Release at point A $(-20,26.67)$:**
Slope:
$$m = \frac{2}{15}(-20) = -\frac{40}{15} = -\frac{8}{3}.$$
Tangent line:
$$y - 26.67 = -\frac{8}{3}(x + 20).$$
Simplify:
$$y = -\frac{8}{3}x - \frac{160}{3} + 26.67 = -\frac{8}{3}x - 53.33 + 26.67 = -\frac{8}{3}x - 26.66.$$
Find where this line crosses dock at $(30,30)$:
Plug $x=30$:
$$y = -\frac{8}{3} \times 30 - 26.66 = -80 - 26.66 = -106.66,$$
which is far south of dock's $y=30$. So Linda will miss dock, headed far south.
Distance from dock to line:
Use point-line distance formula:
$$d = \frac{|m x_1 - y_1 + b|}{\sqrt{m^2 + 1}}$$
where line $y = mx + b$, $m = -\frac{8}{3}$, $b = -26.66$, dock $(x_1,y_1) = (30,30)$.
Calculate numerator:
$$|(-\frac{8}{3})(30) - 30 - 26.66| = |-80 - 30 - 26.66| = |-136.66| = 136.66.$$
Denominator:
$$\sqrt{(-\frac{8}{3})^2 + 1} = \sqrt{\frac{64}{9} + 1} = \sqrt{\frac{73}{9}} = \frac{\sqrt{73}}{3}.$$
Distance:
$$d = \frac{136.66}{\frac{\sqrt{73}}{3}} = \frac{136.66 \times 3}{\sqrt{73}} \approx \frac{409.98}{8.54} \approx 48.0 \text{ m}.$$
Linda will miss dock by about 48 m south.
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**b) Release at point B $(-10,6.67)$:**
Slope:
$$m = \frac{2}{15}(-10) = -\frac{20}{15} = -\frac{4}{3}.$$
Tangent line:
$$y - 6.67 = -\frac{4}{3}(x + 10)$$
$$y = -\frac{4}{3}x - \frac{40}{3} + 6.67 = -\frac{4}{3}x - 13.33 + 6.67 = -\frac{4}{3}x - 6.66.$$
Check dock at $(30,30)$:
$$y = -\frac{4}{3} \times 30 - 6.66 = -40 - 6.66 = -46.66,$$
far south of dock.
Distance from dock:
Numerator:
$$|(-\frac{4}{3})(30) - 30 - 6.66| = |-40 - 30 - 6.66| = |-76.66| = 76.66.$$
Denominator:
$$\sqrt{(-\frac{4}{3})^2 + 1} = \sqrt{\frac{16}{9} + 1} = \sqrt{\frac{25}{9}} = \frac{5}{3}.$$
Distance:
$$d = \frac{76.66}{5/3} = 76.66 \times \frac{3}{5} = 45.99 \text{ m}.$$
Linda will miss dock by about 46 m south.
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**c) Release at point C $(0,0)$ (vertex):**
Slope:
$$m = \frac{2}{15} \times 0 = 0.$$
Tangent line:
$$y - 0 = 0(x - 0) \implies y=0.$$
Distance from dock $(30,30)$ to line $y=0$:
Vertical distance:
$$|30 - 0| = 30 \text{ m}.$$
Linda will pass 30 m south of dock.
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**d) Release point for heading straight to dock:**
We want tangent line passing through dock $(30,30)$ and release point $(x_0,y_0)$ on parabola $y_0 = \frac{1}{15}x_0^2$ with slope
$$m = \frac{2}{15}x_0.$$
Equation of tangent line:
$$y - y_0 = m(x - x_0).$$
Dock lies on line:
$$30 - y_0 = m(30 - x_0).$$
Substitute $y_0$ and $m$:
$$30 - \frac{1}{15}x_0^2 = \frac{2}{15}x_0 (30 - x_0).$$
Multiply both sides by 15:
$$450 - x_0^2 = 2 x_0 (30 - x_0) = 60 x_0 - 2 x_0^2.$$
Bring all terms to one side:
$$450 - x_0^2 - 60 x_0 + 2 x_0^2 = 0$$
$$450 + x_0^2 - 60 x_0 = 0.$$
Rewrite:
$$x_0^2 - 60 x_0 + 450 = 0.$$
Solve quadratic:
$$x_0 = \frac{60 \pm \sqrt{60^2 - 4 \times 1 \times 450}}{2} = \frac{60 \pm \sqrt{3600 - 1800}}{2} = \frac{60 \pm \sqrt{1800}}{2}.$$
$$\sqrt{1800} = \sqrt{900 \times 2} = 30 \sqrt{2} \approx 42.43.$$
Solutions:
$$x_0 = \frac{60 \pm 42.43}{2}.$$
1) $$x_0 = \frac{60 + 42.43}{2} = \frac{102.43}{2} = 51.215$$ (outside domain of parabola approach)
2) $$x_0 = \frac{60 - 42.43}{2} = \frac{17.57}{2} = 8.785.$$
Use $x_0 = 8.785$ (on left branch is negative $x$, so this is on right branch). Calculate $y_0$:
$$y_0 = \frac{1}{15} (8.785)^2 = \frac{1}{15} \times 77.17 = 5.145.$$
So Linda should release at approximately $(8.79, 5.15)$ to head straight to dock.
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**e) Assumptions:**
- The boat path is exactly parabolic.
- Linda travels in a straight line tangent to the parabola at release.
- No external forces (wind, current) affect trajectory.
- The dock is fixed at $(30,30)$.
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**f) Additional information for realism:**
- Speed of boat and skier at release.
- Water current and wind effects.
- Friction and drag forces.
Assuming constant speed and no external forces, the tangent line method applies.
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**Summary:**
- Release at $A$ or $B$ misses dock far south.
- Release at vertex $C$ misses dock by 30 m south.
- Release near $(8.79,5.15)$ on parabola heads straight to dock.