Smooth transition

Calculus Level 3

Above are two quadratics, shifted and then rotated, which are related as follows: $\begin{aligned} \text{[Purple]}:\, -x\sin\alpha+y\cos\alpha &= (x\cos\alpha+y\sin\alpha-2)^2\\ \text{[Red]}:\, x\sin\alpha+y\cos\alpha &= (x\cos\alpha-y\sin\alpha+2)^2. \end{aligned}$ The quadratic is first shifted (i.e. 2 to the left/right), and then it is rotated $\alpha$ radians.

What is the first positive real value of $\alpha$ such that the quadratics, when graphed together, have a lower $y$ bound that is continuous and has a continuous slope (as in the graph above)? Round to the nearest 3 significant figures.

To the right is an example of an a value that doesn't work $(\alpha = 0.64)$ because, although the lower bound is continuous, its slope is not continuous $($ i.e. at $x = \pm1.78$ and $x = 0),$ and the change in slope is abrupt.

The answer is 1.03.

2 solutions

Arjen Vreugdenhil
Dec 14, 2017

I will write $s = \sin\alpha$ and $c = \cos\alpha$ . Things simplify if we use the unrotated coordinates $x',y'$ : $y' = (x'\mp 2)^2\ \ \ \ \begin{cases} x' = cx \pm sy \\ y' = cy \mp sx \end{cases}\ \ \ \ \begin{cases} dx = c\:dx' \mp s\:dy' \\ dy = c\:dy' \pm s\:dx'\end{cases}$ At the lowest point on the curve, the tangent is horizontal; we set $dy = 0$ . Also from $y' = (x'\mp 2)^2$ we have $dy' = 2(x'\mp 2)\:dx'$ so that $0 = dy = c\:dy'\pm s\:dx' = (2cx'\mp 4c\pm s)\:dx'.$ This gives the unrotated coordinates of the lowest points of the curves: $x'= \pm\frac{4c - s}{2c}\ \ \ \ \ \ y' = (x' \mp 2)^2 = \frac{s^2}{4c^2}.$

Rotating this gives $x = cx' \mp sy' = \pm\frac{4c - s}{2} \mp \frac{s^3}{4c^2} = \pm\frac{8c^3 - 2c^2s - s^3}{4c^2}.$ These are the rotated $x$ -coordinates of the lowest points of both curves. They must coincide, i.e. $x = 0$ . Thus we obtain the trigonometric equation $8c^3 - 2c^2s - s^3 = 0.$ Solve this numerically to find $\alpha \approx \boxed{1.03135}$ rad.

Incidentally, $y = cy' \pm sx' = \frac{s^2}{4c} + \frac{4cs - s^2}{2c} = \pm\frac{8cs - s^2}{4c} \approx 1.358.$

Hosam Hajjir
Jan 12, 2018

If we start with $y = (x + 2)^2$ (red parabola) ,

and rotate it clockwise by $\alpha$

then the rotated coordinates will

$x' = \cos(\alpha) x + \sin(\alpha) y$ $y' = -\sin(\alpha) x + \cos(\alpha) y$

Solving for $x$ and $y$ in terms of $x'$ and $y'$ results in

$x = \cos(\alpha) x' - \sin(\alpha) y'$ $y = \sin(\alpha) x' + \cos(\alpha) y'$

Substituting this results in the given equation for the red quadratic. At the final step, we may drop the primes on the primed variables.

At a point (x, y) on the original unrotated parabola, the slope is $2 (x + 2)$ , and since we want the tangent of the rotated parabola to be horizontal, then we require that $2( x + 2) = \tan (\alpha)$ , and that the rotated x-coordinate be 0. That is,

$0 = \cos(\alpha) x + \sin(\alpha) y$

Substituting for $x$ , and $y = (x + 2)^2$ , we obtain,

$0 = \cos(\alpha) (-2 + \tan(\alpha) / 2) + \sin(\alpha) ( tan(\alpha) / 2 )^2$

simplifying,

$0 = -2 \cos(\alpha) + 1/2 \sin(\alpha) + 1/4 \sin^3(\alpha) / \cos^2(\alpha)$

multiplying by $4 cos^2(\alpha)$

$0 = - 8 \cos^3(\alpha) + 2 \sin(\alpha) \cos^2(\alpha) + \sin^3(\alpha)$

Solving numerically, we find that $\alpha = 1.0313$

Smooth transition

The answer is 1.03.

2 solutions

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