Rotating a Graph?

Calculus Level 5

The above shows twos functions on a same Cartesian Plane: $f(x)$ and $f^\theta (x)$ colored blue and red respectively.

Let $f(x)=x^3-6x^2+6x+1$

Define $f^{\theta}(x)$ , where $\theta \geq 0$ , as the new graph obtained by rotating the graph of $f(x)$ $\theta$ degrees about the origin ( $0,0$ ) in the clockwise direction.

Find the maximum value of $\theta_M$ such that for all values of $0<\theta<\theta_M$ , $f^{\theta}(x)$ has every value of $x$ paired with only one $y$ (more simply, no two points can have the same $x$ -coordinate).

Input your answer as $\lfloor 100\cdot\theta_m \rfloor$ . Where $\lfloor x \rfloor$ is the floor function which means the greatest integer less than or equal to $x$ .

This is part of the set Trevor's Ten

The answer is 946.

1 solution

Trevor Arashiro
Mar 15, 2015

We want to find the maximum NEGATIVE slope of the function since to be a non surjective function, there must be at least one point on the graph that has an instantaneous slope of $\pm \infty$ (note that a slope of $\infty$ is the same as a slope of $-\infty$ ). Imagine the line tangent to $f(x)$ with the most negative slope. We need rotate that till it's perpendicular with the x-axis.

We normally take the first derivative for slope. But here, we need to take the second derivative and set it equal to 0 since we need to find the global minimum of the first derivative. Hence

$f(x)=x^3-6x^2+6x+1$

$f''(x)=6x-12$

Setting this equal to 0 to find the minimum of $f'(x)$

$0=6x-12$

$2=x$

Plugging this into $f'(x)$

$3x^2-12x+6=slope$

$3(2)^2-12(2)+6=slope$

$-6=slope$

If anyone needs an explanation of this next concept, please comment below.

Now, we take the arctan of this measurement to get the angle of this slope WRT to the x-axis ( $\arg(x)$ ).

$\tan^{-1}(|-6|)\approx 80.537$

Now, we need a right angle

$90=\theta_m+80.5377$

$\boxed{9.462=\theta_m}$

If we rotate the graph anymore than this clockwise, there will be two points with the same abscissa and different ordinate . Thus we have created a function with two y's per x

When the tangent at the point of inflection is vertical we get the limit. So turn the tangent at the point of inflection through $\theta$ to make it vertical. As done above, the point of inflection is when x=2. So the tangent at that point is turned through 9.462 degrees to make it vertical. It would have been clear if it was mentioned that $\theta$ is in degrees.

Niranjan Khanderia - 6 years, 2 months ago

I like how this problem employs the third derivative (I always check for $g''(x) > 0$ for minima, just to be sure).

Jake Lai - 6 years, 2 months ago

@Daniel Liu were you the one who changed the image? If so, did you use Geogebra?

Trevor Arashiro - 6 years, 2 months ago

Why since to be a non surjective function, there must be at least one point on the graph that has an instantaneous slope of +-inf?

Khánh Hưng Vũ - 2 years, 2 months ago

Rotating a Graph?

The answer is 946.

1 solution

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