Question

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

a.) how far is the particle closet to the point (6,0)
b.) How far is it from (6,0) at the moment?

Answer 1

You could write the distance of a point P(p, sqrt(p+4)) to (6,0) as a function, and try to calculate the minimum of that function using the derivative.

Here is hou it would look: \[d(p)= \sqrt{(p-6)²+(\sqrt{p-4}-0)^2}=\sqrt{(p-6)^2-(p-4)}\]However, the root makes it more difficult.
But if you realize d(p) is minimal if the number under the root sign has its minimum, you can focus on just that one.

So define the following function:

f(p)=(p-6)²-(p-4)=p²-12p+36-p+4

So

f(p)=p²-13p+40. Where f has its minumum, just calculate the root of that p to find the minimal distance.

Answer 2

f'(p)=2p-13=0, so p=6½. The minimum distance would then be \[\sqrt{\frac{ 13 }{ 2 }}=\frac{ \sqrt{13} }{ \sqrt{2} }=...=\frac{ 1 }{ 2 }\sqrt{26}\]

Answer 3

doesnt even say thank you..........

TThank You ZeHanz