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Let $f(x)$ be a real valued diffrentiable function in its domain. Let $PQ$ be a line which is inclined at an angle of $k$ with the tangent of the graph $f(x)$ at $x_0$. The rate of change of the magnitude of the line $PQ$ be $m(x)$. $PQ$ moves(rotates from its mid point in one direction) on the graph $f(x)$ such that its one end always touches the graph while the other traces a curve. What would be the equation of the curve in terms of $x$?

PS: I know its horrible phrasing of sentences but its the best I could do,I would be grateful if someone edit it to a better form.

Hashir Omer
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1 Answers1

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I will assume that PQ is a segment of constant length 1. The end P is at the point $(t, f(t))$ on the graph of f(x) and the point Q is inclined at an angle k to the tangent line at $(t, f(t))$. Then the tangent is inclined at an angle $\theta$ to the horizontal where $tan(\theta) = f'(t)$ and ' denotes the derivative. Then
$\alpha = k + \tan ^{-1}(f'(t))$

is the angle that PQ makes with the horizontal. The end Q of the segment is at

$(t, f(t)) + (cos(\alpha), sin(\alpha)) = (t+cos(\alpha), f(t)+sin(\alpha))$

Here is an image of the locus of the point Q as we move along a point on the curve f(x) = sin(x).

enter image description here

Paul
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