Question #6cb9f

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Question #6cb9f

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Answer 1 · 2016-08-20T20:42:06

The length of subtangent is $a (1 - cos t) \cdot \frac{sin t}{1 + cos t}$ $a(1-cos t)*sin t/(1+cos t)$

Explanation:

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The subtangent is a segment $S T$ $ST$ on the X-axis from point $S$ $S$ , an abscissa of some point $P$ $P$ on a curve, to point $T$ $T$ , intersection of a tangent to a curve at point $P$ $P$ with the X-axis.

Knowing parametric expressions for $x = f (t)$ $x=f(t)$ and $y = g (t)$ $y=g(t)$ , for any parameter $t$ $t$ we know abscissa and ordinate of a point $P (t)$ $P(t)$ on a curve.
In this case we will use ordinate $y = a (1 - cos t)$ $y=a(1-cos t)$ and will determine $tan (Ψ)$ $tan(Psi)$ . That will be sufficient to find $S T$ $ST$ .

To calculate $tan (Ψ) = \frac{d y}{d x}$ $tan(Psi)=dy/dx$ we will use the property of parametric curve:
$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $dy/dx = (dy/dt)/(dx/dt)$

Using this approach,
$tan (Ψ) = \frac{d y}{d x} = \frac{a (sin t)}{a (1 + cos t)} = \frac{sin t}{1 + cos t}$ $tan(Psi) = dy/dx = [a(sin t)]/[a(1+cos t)] = sin t/(1+cos t)$

Now we can calculate $S T$ $ST$ :
$S T = P S \cdot tan (Ψ) = y \cdot \frac{d y}{d x} = a (1 - cos t) \cdot \frac{sin t}{1 + cos t}$ $ST = PS*tan(Psi) = y*dy/dx = a(1-cos t)*sin t/(1+cos t)$

This does not resemble any of the answers provided.
However, if $x = a (t - sin t)$ $x=a(t-sin t)$ then $\frac{d x}{d t} = 1 - cos t$ $dx/dt = 1-cos t$ and answer $a sin t$ $asin t$ would be correct.

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Question #6cb9f

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