# Thread: Construct involute

1. ## Construct involute

Construct the involute for $y = \frac {x^2}{2}$ starting from (0,0).

My solution so far:

Let $\alpha (t) = (t, \frac {t^2}{2})$

I use the formula $\alpha (t) - t(t) s(t)$ where $s(t) = \int ^{t} _{t_{0}} V(u)du$

I find that $t(t) = \frac { \alpha ^{'} } { | \alpha ^{'} | } = \frac { (1,t)} { \sqrt {1+t^2} }$

But in solving s(t), I have: $s(t) = \int ^{t} _{0} \sqrt {1+t^2}dt = \frac {t}{2} \sqrt {1+t^2} + \frac {1}{2} ln(t+ \sqrt {1+t^2} )$, but I can't see a way to simple this, am I doing this right?

Thank you.

2. Originally Posted by tttcomrader
Construct the involute for $y = \frac {x^2}{2}$ starting from (0,0).

My solution so far:

Let $\alpha (t) = (t, \frac {t^2}{2})$

I use the formula $\alpha (t) - t(t) s(t)$ where $s(t) = \int ^{t} _{t_{0}} V(u)du$

I find that $t(t) = \frac { \alpha ^{'} } { | \alpha ^{'} | } = \frac { (1,t)} { \sqrt {1+t^2} }$

But in solving s(t), I have: $s(t) = \int ^{t} _{0} \sqrt {1+t^2}dt = \frac {t}{2} \sqrt {1+t^2} + \frac {1}{2} ln(t+ \sqrt {1+t^2} )$, but I can't see a way to simple this, am I doing this right?

Thank you.
That's the arc length you're stuck with, I'm afraid. But did you really expect the equations defining the involute to be simple?

Things aren't that bad .... work with what you've got.

By the way, the integral for the arc length can be expressed in terms of the inverse hyperbolic sine function. Because I'm lazy, I'll just suggest you fire up the ol' Wolfram Integrator.

To put your mind at rest, you might want to look at this.

3. Originally Posted by tttcomrader
Construct the involute for $y = \frac {x^2}{2}$ starting from (0,0).

...
If I understand and translate the expression "involute" correctly this curve is defined (in your question) as the locus of all intersection points of the tangents to the parabola and the normal to the tangents passing through the origin.

Let $T(t, \frac{t^2}{2})$ the tangent point of the parabola.

Then the equation of the tangent to the parabola in T is:

$y = tx-\frac12 t^2$ . The equation of the normal to this tangent through the origin is:

$y = -\frac1t x$

Then the intersection point has the coordinates:

$x_S=\frac{t^3}{2(t^2+1)}$ and $y_S=-\frac{t^2}{2(t^2+1)}$

This is simultaneously the parametrized equation of the involute. There exists a asymptote because $\lim_{t \rightarrow \infty} y = -\frac12$

After a few steps of transformation (elimination of the parameter t, etc) you'll get the equation:

$2y^3 + 2x^2y + x^2 = 0$ which describes the involute.