Distance between point and parametric equation

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I've just had a brain block... how do I work out the distance between a point (-5,10,13) and a parametric equation:

x(t) = 57- 4t
y(t) = 75 + 5t
z(t) = -t
 
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You write out an expression for the distance between (-5,10,13) and (x(t),y(t),z(t)) and minimize with respect to t.
 
he didn't ask for the minimum distance, he asked for the distance. hence the distance will be a function of t
 
ice109 said:
he didn't ask for the minimum distance, he asked for the distance. hence the distance will be a function of t

Good point. Don't minimize.
 
First, he asked for the distance from a point to a "parametric equation" which makes no sense! I assume he meant the distance from a point to the line represented by those parametric equations.

Dick was right to start with: the distance from a point to a line is the minimum distance from that point to any point on the line.

In my opinion the simplest way to find the distance from the point (-5,10,13) to the line represented by the parametric equations, x(t) = 57- 4t, y(t) = 75 + 5t, z(t) = -t, is to determine the plane perpendicular to the given line containing the given point and determine where the line intersects that plane. The distance from point to line is the distance from the given point to that point.

There is, of course, a standard formula for that problem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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