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 $(-30,60)$. Linda lets go of the towrope at points A, B, or C on the parabola and travels in a straight line from that point. We need to find where she is headed relative to the dock using linear equations of the form $y=mx+b$. 2. **Given points:** - Dock: $(30,30)$ - Boat start: $(-30,60)$ - Parabola vertex: $(0,0)$ - Points A, B, C lie on the parabola; assume coordinates for these points based on the description: - Point A near top-left: approximately $(-20,40)$ - Point B near vertex: approximately $(0,0)$ - Point C near bottom-right: approximately $(20,10)$ 3. **Approach:** When Linda lets go at a point, she travels in a straight line tangent to the parabola at that point. We find the slope $m$ of the tangent line at that point and write the line equation $y=mx+b$. 4. **Find the parabola equation:** The parabola passes through $(-30,60)$, $(0,0)$, and $(30,30)$. Assume form $y=ax^2+bx+c$. Using points: $$ \begin{cases} 60 = a(-30)^2 + b(-30) + c \\ 0 = a(0)^2 + b(0) + c \\ 30 = a(30)^2 + b(30) + c \end{cases} $$ From second equation: $c=0$. First: $60 = 900a -30b$ Third: $30 = 900a + 30b$ Add first and third: $$60 + 30 = 900a -30b + 900a + 30b = 1800a \implies 90 = 1800a \implies a = \frac{1}{20} = 0.05$$ Subtract third from first: $$60 - 30 = 900a -30b - (900a + 30b) = -60b \implies 30 = -60b \implies b = -\frac{1}{2} = -0.5$$ So parabola equation: $$y = 0.05x^2 - 0.5x$$ 5. **Find slope of tangent line at a point:** Derivative: $$y' = \frac{dy}{dx} = 2 \times 0.05 x - 0.5 = 0.1x - 0.5$$ 6. **Point A: $x=-20$** Slope: $$m = 0.1(-20) - 0.5 = -2 - 0.5 = -2.5$$ Point A coordinates: $$y = 0.05(-20)^2 - 0.5(-20) = 0.05(400) + 10 = 20 + 10 = 30$$ So $A=(-20,30)$. Equation of tangent line at A: $$y = m x + b$$ Use point-slope form: $$y - 30 = -2.5(x + 20)$$ $$y = -2.5x - 50 + 30 = -2.5x - 20$$ 7. **Where is Linda headed relative to the dock?** Check if line passes through dock $(30,30)$: $$y = -2.5(30) - 20 = -75 - 20 = -95 \neq 30$$ So Linda will not reach the dock if she lets go at A. 8. **Point B: $x=0$** Slope: $$m = 0.1(0) - 0.5 = -0.5$$ Point B coordinates: $$y = 0.05(0)^2 - 0.5(0) = 0$$ So $B=(0,0)$. Equation of tangent line at B: $$y - 0 = -0.5(x - 0) \implies y = -0.5x$$ Check if line passes through dock $(30,30)$: $$y = -0.5(30) = -15 \neq 30$$ Linda will not reach dock if she lets go at B. 9. **Point C: $x=20$** Slope: $$m = 0.1(20) - 0.5 = 2 - 0.5 = 1.5$$ Point C coordinates: $$y = 0.05(20)^2 - 0.5(20) = 0.05(400) - 10 = 20 - 10 = 10$$ So $C=(20,10)$. Equation of tangent line at C: $$y - 10 = 1.5(x - 20)$$ $$y = 1.5x - 30 + 10 = 1.5x - 20$$ Check if line passes through dock $(30,30)$: $$y = 1.5(30) - 20 = 45 - 20 = 25 \neq 30$$ Linda will not reach dock if she lets go at C. 10. **Summary:** - At A, Linda heads along $y = -2.5x - 20$, missing dock. - At B, Linda heads along $y = -0.5x$, missing dock. - At C, Linda heads along $y = 1.5x - 20$, missing dock. 11. **Find point on parabola where tangent line passes through dock $(30,30)$:** Let point be $(x_0,y_0)$ on parabola: $$y_0 = 0.05x_0^2 - 0.5x_0$$ Slope at $x_0$: $$m = 0.1x_0 - 0.5$$ Equation of tangent line: $$y - y_0 = m(x - x_0)$$ Since line passes through dock $(30,30)$: $$30 - y_0 = m(30 - x_0)$$ Substitute $y_0$ and $m$: $$30 - (0.05x_0^2 - 0.5x_0) = (0.1x_0 - 0.5)(30 - x_0)$$ Simplify left: $$30 - 0.05x_0^2 + 0.5x_0 = (0.1x_0 - 0.5)(30 - x_0)$$ Expand right: $$= 0.1x_0 imes 30 - 0.1x_0 imes x_0 - 0.5 imes 30 + 0.5 x_0 = 3x_0 - 0.1x_0^2 - 15 + 0.5x_0 = 3.5x_0 - 0.1x_0^2 - 15$$ Set equal: $$30 - 0.05x_0^2 + 0.5x_0 = 3.5x_0 - 0.1x_0^2 - 15$$ Bring all terms to one side: $$30 - 0.05x_0^2 + 0.5x_0 - 3.5x_0 + 0.1x_0^2 + 15 = 0$$ $$ ( -0.05x_0^2 + 0.1x_0^2 ) + (0.5x_0 - 3.5x_0) + (30 + 15) = 0$$ $$0.05x_0^2 - 3x_0 + 45 = 0$$ Multiply both sides by 20 to clear decimals: $$x_0^2 - 60x_0 + 900 = 0$$ 12. **Solve quadratic:** $$x_0 = \frac{60 \pm \sqrt{60^2 - 4 \times 1 \times 900}}{2} = \frac{60 \pm \sqrt{3600 - 3600}}{2} = \frac{60}{2} = 30$$ So $x_0 = 30$. Find $y_0$: $$y_0 = 0.05(30)^2 - 0.5(30) = 0.05(900) - 15 = 45 - 15 = 30$$ Point is $(30,30)$, the dock itself. 13. **Interpretation:** The tangent line at the dock is vertical to the parabola at that point, so Linda should release the rope exactly at the dock to head straight there. 14. **Answer to parts:** - a) At A $(-20,30)$, Linda heads along $y = -2.5x - 20$, missing dock. - b) At B $(0,0)$, Linda heads along $y = -0.5x$, missing dock. - c) At C $(20,10)$, Linda heads along $y = 1.5x - 20$, missing dock. - d) Linda should release the rope at the dock point $(30,30)$ to head straight there. 15. **Assumptions:** - Linda travels in a straight line tangent to the parabola at release. - No external forces like wind or water currents. - The parabola accurately models the boat path. 16. **Additional info for realism:** - Speed of boat and skier. - Water resistance. - Angle of release. Using this info, one could model trajectory more precisely.