A calculus problem by anushka sharma

Calculus Level 3

Consider the parabola $C: y=x^2$ . A circle $K$ , with center on the $y-$ axis and radius 1, is tangent to $C$ at 2 distinct points.

Find the area bounded by $C$ and $K$ (but not in $K$ ).

\frac{ 3 \sqrt{3}}{4} - \frac{\pi}{3}

\frac{ 3 \sqrt{3} - 4}{4}

\frac{ 3 \sqrt{3}}{4}

1

1 solution

Ron Gallagher
Jun 16, 2020

Let (a, a^2) be the positive point of tangency of the circle and parabola (a>0). By symmetry, the other point of tangency is (-a, a^2). We must have the derivatives of the two curves equal at these two points. For f(x) = x^2, we have f'(a) = 2*a. The circle has equation x^2 + (y-k)^2 = 1 for some constant k. Differentiating this implicitly yields dy/dx = -x/(y-k). Evaluating this derivative at x = a and y = a^2 gives -a = (a^2-k). Hence:

-a / (a^2-k) = 2*a, or

a^2 = k - 1/2

Substituting this into the equation of the circle yields:

a^2 + (-1/2)^2 = 1, or

k - 1/2 + 1/4 = 1, or

k = 5/4.

Hence, a^2 = 5/4 - 1/2, or a = sqrt(3)/2.

The area between the curves is now the integral from -a to a of the bottom half of the circle minus x^2. Doing the algebra, we see that the integrand is:

5/4 - sqrt(1-x^2) - x^2.

This integral can be evaluated by trigonometric substitution to get the required result.

A calculus problem by anushka sharma

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