# Find the closest point on a curve?

• Apr 30th 2006, 03:13 PM
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?
• Apr 30th 2006, 03:36 PM
Jameson
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)! :)
• Apr 30th 2006, 03:40 PM
CaptainBlack
Quote:

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
• May 1st 2006, 08:36 AM
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?
• May 1st 2006, 02:19 PM
ThePerfectHacker
Quote: