A particle travels along the curve y=sqr(x+4)

When is the particle closest to the point (6,0)?

Question

A particle travels along the curve y=sqr(x+4) 

When is the particle closest to the point (6,0)?

anonymous · Answer

when, or where?

luigi0210 · Answer

When

luigi0210 · Answer

Nobodies knows this?

luigi0210 · Answer

anybody? :/ I really need help on this..

anonymous · Answer

OK, assuming I have the sense of the question then :|dw:1360305542726:dw|we have some point (x0, y0) on the curve where the tangent to the curve at the point is orthogonal to the line from (6,0)?$$\frac{dy}{dx}=(1/2)(x+4)^{-1/2}$$$$\frac{dy}{dx}(x_{0})=\frac{1}{2\sqrt{x_{0}+4}}$$the product of the gradient of perpendicular lines is negative 1, so the slope from the point on the curve to (6,0) is$$-2\sqrt{x_{0}+4}$$equation of that line is: $$\frac{y_{0}-0}{x_{0}-6}= -2\sqrt{x_{0}+4}$$or knowing y0 from x0 :$$\frac{\sqrt{x_{0}+4}}{x_{0}-6}= -2\sqrt{x_{0}+4}$$which you can solve for x0.

luigi0210 · Answer

Thank you very much! Would this also be used to find how far it is from (6,) at the moment?

anonymous · Answer

A Mathematica solution including a plot is attached.

anonymous · Answer

$$y=\sqrt{x+4}$$or$$y=(x+4)^{2}$$

anonymous · Answer

As for 'at the moment' then we'd need a time dependence for that.

anonymous · Answer

@hewsmike
There is no sense of time expressed in the problem statement that I can see.

anonymous · Answer

@robtobey I'm not wanting to be rude but you've solved a different problem. :-)

anonymous · Answer

Second soluition based on$$y=\sqrt{(x+4)} $$