A man with a quadratically shaped head, with profile given by the function h(x)=4-3x^2 where both x and h(x) are measured in decimetres, puts on a conical hat whose profile has a right angle at the tip. How much taller does he appear with the hat on if he wears the hat symmetrically? What are the coordinates of the point at which the hat touches his head.

Answer 1

f'(x) = -6x
Assuming the hat is tangent to his head at the points of contact, if the hat touches his head at x = ±h, then we have
-6(-h) = -1/(-6h)
h = ±1/6
so y = 47/12
That makes the tangent lines y-47/12 = ±(x ± 1/6)
That means the tangent lines meet at (0,49/12)
The vertex of the parabola is at (0,4)
which makes the top of the hat 1/12 dm higher than his head

Answer 2

What a strange but interesting question

Make a sketch and you will see that you simply want two
tangents from a point on the y-axis that form a right angle.
Let the points of contact of the conical hat with his head be
P(x, 4-3x^2) and Q(-x, 4-3x^2) .... (they have the same height)
dy/dx = -6x

so slope of tangent at P is -6x and the slope at Q is +6x
they must form a 90° angle, so
-6x = -1/(6x)
36x^2 = 1
x = ± 1/6
let's find the equation of the tangent with slope 6x at P(1/6, 4-3(1/36))
P is (1/6, 47/12) , and the slope is -1
y = -x + b
47/12 = -1/6 + b
b = 49/12
y = x + 49/12
so the y-intercept is 49/12 or 4 1/12

so he increased his height by only 4 - 4 1/12 or 1/12 dm
The hat touches his head at (1/6, 47/12) and (-1/6, 47/12)

Answer 3

LOL, while I was working on this, oobleck had already posted
the answer. At least we got the same results.

Answer 4

Thanks so much for the help everyone! This practice problem was just a little beyond my grasp so it was nice to see it explained :)