Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/208

 Find the equation of the evolute of the ellipse $$b^2 x^2 + a^2 y^2 = a^2 b^2$$.


 * [[Image:wag 119-3 evolute of ellipse.jpg|228px|right|Evolute of the ellipse]]


 * Solution.
 * style="text-align: right;"|$$\frac{dy}{dx}$$
 * $$= -\frac{b^2 x}{a^2 y},\ \frac{d^2 y}{dx^2} = -\frac{b^4}{a^2 y^3}.$$
 * First step.
 * style="text-align: right;"|$$\alpha\ =$$
 * $$\frac{(a^2 - b^2)x^3}{a^4},$$
 * style="text-align: right;"|$$\beta\ =$$
 * $$-\frac{(a^2 - b^2)y^3}{b^4}.$$
 * Second step.
 * style="text-align: right;"|$$x\ =$$
 * $$\left( \frac{a^4 \alpha}{a^2 - b^2} \right)^{\frac{1}{3}}$$
 * style="text-align: right;"|$$y\ =$$
 * $$-\left( \frac{b^4 \beta}{a^2 - b^2} \right)^{\frac{1}{3}}$$
 * }
 * $$\left( \frac{a^4 \alpha}{a^2 - b^2} \right)^{\frac{1}{3}}$$
 * style="text-align: right;"|$$y\ =$$
 * $$-\left( \frac{b^4 \beta}{a^2 - b^2} \right)^{\frac{1}{3}}$$
 * }
 * $$-\left( \frac{b^4 \beta}{a^2 - b^2} \right)^{\frac{1}{3}}$$
 * }


 * Third step. $$(a\alpha)^{\frac{2}{3}} + (b\beta)^{\frac{2}{3}} = (a^2 - b^2)^{\frac{2}{3}}$$, the equation of the evolute EHE'H '  of the ellipse ABA'B ' . E, E', H', H are the centers of curvature corresponding to the points A, A', B, B ' , on the curve, and C, C', C   correspond to the points P, P', P  .

When the equations of the curve are given in parametric form, we proceed to find $$\tfrac{dy}{dx}$$ and $$\tfrac{d^2 y}{dx^2}$$, as in §103, from

and then substitute the results in formulas (50), §117. This gives the parametric equations of the evolute in terms of the same parameter that occurs in the given equations.

The parametric equations of a curve are

Find the equation of the evolute in parametric form, plot the curve and the evolute, find the radius of curvature at the point where $$t = 1$$, and draw the corresponding circle of curvature.


 * {| style="width: 100%;"


 * Solution.
 * $$\frac{dx}{dt} = \frac{t}{2},\ \frac{d^2 x}{dt^2} = \frac{1}{2},$$
 * $$\frac{dy}{dt} = \frac{t^2}{2},\ \frac{d^2 y}{dt^2} = t.$$
 * }
 * $$\frac{dy}{dt} = \frac{t^2}{2},\ \frac{d^2 y}{dt^2} = t.$$
 * }


 * Substituting in above formulas (A) and then in (50), §117, gives