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HI Dr Sasho,

here is my problem that i couldn't solve.

An ant walks along a cable hanging in the shape of a parabola y= x^2-2x+2, where x and y are in meters. If the x-coordinate of the ant is decreasing at 0.05 m/s when it is at the point (2,2), how fast is its distance from the point (3,0) changing at this point?

thanks for replying,

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The distance between (3,0) and any other point (x,y) is Square root of (3-x)^{2}+y^{2}. If the point (x,y) is on the parabola, then y=x^{2}-2x+2, so that the distance function becomes Square root of (3-x)^{2}+(x^{2}-2x+2)^{2}. Now differentiate this function with respect to time and use the given data (that dx/dt=-0.05 and that x=2 at the given moment).