# Thread: Find the closest point on a curve?

1. ## Find the closest point on a curve?

Q: Find the point on the line -5 x + 3 y + 5 =0 which is closest to the point ( -3, 3 ).

Solution: I know I have to use the distance formula, and I have to plug in the x and y with x and f(x);

therefore I got to here d = (x+3)^2+[(5x-5)/3-3]^2;

and f'(x) = 2(x+3) + (10/3)(5x-14)/3;

then I have no idea what to do, as our prof didn't bother to explain it afterward...

Any idea how to solve it?

2. First thing is you're doing the right thing by differentiating the distance squared, because it will have the same critical points as the distance, but just as a note the distance between two points is $D=\sqrt{(x-x_0)^2+(y-y_0)^2$. You're working with $D^2$. But you've set up your problem correctly and differentiated correctly, so now set the derivative equal to zero and find your critical point(s)!

Q: Find the point on the line -5 x + 3 y + 5 =0 which is closest to the point ( -3, 3 ).

Solution: I know I have to use the distance formula, and I have to plug in the x and y with x and f(x);

therefore I got to here d = (x+3)^2+[(5x-5)/3-3]^2;

and f'(x) = 2(x+3) + (10/3)(5x-14)/3;

then I have no idea what to do, as our prof didn't bother to explain it afterward...

Any idea how to solve it?
If a function $f(x)$ is at a local minimum then $f'(x)=0$.

(Note you have to be careful as $f'(x)=0$ does not guarantee that $x$
is a minimum, it may be a maximum, or a point at which the tangent is horizontal
for an otherwise locally increasing or decreasing function.

Such problems don't arise with your $d$ function as it goes to
infinity as $x \to \pm \infty$, and has only a single root to the equation:

$d'(x)=0$

RonL

RonL

4. I find the critical point as 43/34. So the minimum distance is f(43/34)? But how do I find the coodinate after I have found the minimum distance?