Father duck swim along $y=-x^2$,baby duck swim along $y=x^2$, Father duck’s movements parameterized by: $()=(−5,−(−5)^2)$, equation for baby?

Question

A father duck and a baby duck are swimming along opposing parabolic paths, with the father along $ = −^2$ and the baby along = $^2$. The father duck’s movements can be parameterized by: $_() = ( − 5, −( − 5)^2), 0 \le \le 5$ and he is always looking directly forward (along his path). The baby duck is trying to always stay exactly in front of her father. Find a parameterization of her path.

At the moment I don't even know what is the question is asking, I'm thinking like the baby duck is always ahead of the father duck as moving in the point along the tangent of the father duck at any time, any help will be appreciated.

Answer 1

The tangent line from the position of the father duck is

$ P(t, s) = ( t - 5, - (t - 5)^2 ) + s ( 1, - 2 (t - 5) ) $

This line intersects the parabola of the baby duck at

$ ( t - 5 + s)^2 = - (t - 5)^2 - 2 s (t - 5) $

Let $t' = t - 5 \le 0 $, then

$ (t' + s)^2 = - t'^2 - 2 s t' $

which gives us

$ t'^2 + 2 s t' + s^2 = - t'^2 - 2 s t' $

and this simplifies to

$ s^2 + 4 s t' + 2 t'^2 = 0 $

The solution of which is

$ s = \dfrac{ - 4 t' \pm \sqrt{ 8 t'^2 } } {2} = -2 t' \pm \sqrt{2} | t' | = t' ( - 2 \mp \sqrt{2} ) $

But $t' = t- 5 $ , therefore,

$ s = (5 - t) (2 \pm \sqrt{2} ) $

Substitute this into the equation $P(t,s)$

$ B(t) = ( t - 5, - (t - 5)^2 ) + (5 - t) (2 + \sqrt{2}) (1, -2 (t - 5) ) $

OR

$ B(t) = ( t - 5, - (t - 5)^2 ) + (5 - t) (2 - \sqrt{2}) (1, -2 (t - 5) ) $

These are the possible parametric equations for the paths for the baby duck.

Answer 2

The tangent line of the father duck's path $$r_1(t)=(t-5,-(t-5)^2)$$ at time $t$ has slope $$m=\left.(-x^2)'\right\vert_{x=t-5}=-2(t-5).$$ So, the tangent line is $$\frac{y-(-(t-5)^2)}{x-(t-5)}=-2(t-5)$$ that is $$y=-2(t-5)x+(t-5)^2.$$ Intersecting the tangent line with baby duck's swimming curve $y=x^2$ gives the quadratic equation $$x^2+2(t-5)x-(t-5)^2=0.$$ Solution of this equation $$x_{1,2}=(-1\pm\sqrt2)(t-5)$$ are the two possible $x$-coordinates of the baby duck's path. Hence the possible paths of the baby duck are $$\small r_2(t)=((-1+\sqrt2)(t-5), (3-2\sqrt2)(t-5)^2 )$$ and $$\small r_2(t)=((-1-\sqrt2)(t-5), (3+2\sqrt2)(t-5)^2 ).$$

Answer 3

Since the father duck is moving along the $y = -x^2$, we can look at the tangent line at $(x_0,-x_0^2)$: $$y-(-x_0^2)=y'(x_0)(x-x_0)\implies y = -2x_0x+x_0^2.$$ Its intersections with $y = x^2$ are given by the solutions of $x^2=-2x_0x+x_0^2$ which are $$x_{1,2} = (-1\pm\sqrt 2)x_0,\\ y_{1,2} = x_{1,2}^2 =(3\mp2\sqrt 2)x_0^2.$$ Finally, we know that $x_0(t) = t-5$, so the baby duck's path is given by parametrizations $$b_{1,2}(t)=((-1\pm\sqrt 2)(t-5),(3\mp2\sqrt 2)(t-5)^2),\ t\in[0,5].$$