$y=x^2+\frac{1}{4}$ is eventually intersected at two points by the line $y=tx,$ where $t$ is the time increasing linearly from $0$ (hence the line rotates from a horizontal towards a vertical position). Determine the speed at which the segment linking the two intersection points grows.
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Where do the two curves intersect?
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Find, in terms of
$t,$the coordinates of these points. -
… and the length of the segment joining the two points.
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How could you find the speed at which it grows from this length?
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We first find where the points of intersection are, in terms of
$t.$We equate the two curves to get$x^2-tx+\frac{1}{4} = 0.$Solving for$x$gives us their$x$-coordinates:$x_{A,B} = \frac{t\pm\sqrt{t^2-1}}{2}.$The two points
$A$and$B$lie on the line$y=tx$so we substitute in the above$x$-coordinates to get their$y$-coordinates. Thus the two points are:$A = \big(\frac{t-\sqrt{t^2-1}}{2},\frac{t^2-t\sqrt{t^2-1}}{2}\big)$and$B = \big(\frac{t+\sqrt{t^2-1}}{2},\frac{t^2+t\sqrt{t^2-1}}{2}\big)$The length of the segment joining
$A$and$B$is their Euclidean distance given by$D_{A,B}(t) = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}.$Substituting and simplifying yields$D_{A,B}(t) = \sqrt{t^4-1}.$To get the rate at which this distance increases, we differentiate with respect to
$t$to get the result$D'_{A,B}(t) = \frac{2t^3}{\sqrt{t^4-1}}.$