(Optimization) - Finding the Minimal Distance between a Point and a Parabola

In this video, we solve an optimization problem of finding a point on a parabola that is the closest to a fixed point off of the curve.
This is lecture 38 (part 3/4) of the lecture series offered by Dr. Andrew Misseldine for the course Math 1210 - Calculus I at Southern Utah University. A transcript of this lecture can be found at Dr. Misseldine's website or through his Google Drive at: drive.google.c...
This lecture is based upon Section 4.7 of Calculus by James Stewart. Please post any questions you might have below in the comment field and Dr. Misseldine (or other commenters) can answer them for you. Please also subscribe for further updates.

[@user-xd2dp1fy4g]

This is a phenomenal video, and the explanation make sense. Don't know why this doesn't have more views

[@Misseldine]

I'm glad to hear that. It's still a growing channel.

[@scottekoontz]

This can also be solved using a similar method to the "distance from point to line" problems. You have a point, get the slope. The slope of the parabola changes with values of x (or y) but you can still get the formula for the lines that can be made from the point to the parabola. The only calc needed is the first derivative for the slope of a parabola.
E.g., x = (y^2)/2 means x' = y, so perp line will have slope dx/dy of -1/y.
Point-slope formula: (x - 1) = -1/y (y - 4), thus x = 4/y (multiple points)
We have two equations, set the x values equal to each other: 4/y = (y^2)/2, thus y^3 = 8, etc.

[@Misseldine]

Interesting alternative approach. As you mentioned above, it is based upon the shortest path between a point and line is the perpendicular/orthogonal path (The Best Approximation Theorem). We don't immediately have the same condition, without calculus, for a parabola and a point, but we can instead find the shortest paths between the given points and all of the tangent lines of the parabola, so calculus is still required. Now, such a path is only valid if the point on the tangent is likewise on the parabola. Hence why you need to solve the system of two equations, you need points which are both on the parabola and the minimum point on its tangent line.
-How do we know such a point even exists? Consider a parabola with a removed vertex versus its removed point. In such a setting the min distance is unobtainable as the removed point is arbitrarily close to the parabola. Likewise, there is no path from the vertex to some other point which obtains a right angle with the tangent line and lands on the parabola.
-What if there were multiple such points? Consider a noncircular eclipse versus one of its foci. In this case every path from the focus to a point on the ellipse will obtain a right angle with the tangent line AND land in the ellipse. We haven't limited down any of the possibilities.
-If such a point exists, how do we know it obtains the minimum distance between the point and parabola? Considering the previous counterexample, an orthogonal path onto the curve does not necessarily produce the shortest path to the curve.
There are additional unstated assumptions in this approach which are needed to reliably find the solution that are trivial in the case presented. If the point is replaced with the vertex of the parabola, this alternative approach would fail to find the minimum distance (which would happen at the vertex BTW). On the other hand, the approach in this video of minimizing distance is versatile enough to handle all of these odd cases as well as not involving any more calculus than the alternative.

[@patheticotaku7112]

Wow!!! Understood the concept thanks professor 😍

[@Misseldine]

You're very welcome.

[@abhiacharya7377]

Thanks a lot sir, It's now cristal clear

[@Misseldine]

Always welcome

[@TheFnut]

This is great! Would love to see a version of this in code, can't find that anywhere.

[@Misseldine]

Thanks. Which code are you looking for exactly?

[@TheFnut]

Thank you for the reply! Well I'm working on a game in unity, and I'm using c#. Python is also an option, but would be nice to avoid numpy as I need to code the math in c#. Thank you!

[@diegotorres6319]

Thanks man

[@Misseldine]

Any time

[@berrybby]

i got finals in 2 hours and its 6am thank you so much

[@Misseldine]

Best of luck!

[@berrybby]

@@Misseldine THANK YOU I PASSEDDDDD❤️

[@Misseldine]

Congratulations! I am glad I was able to help.

[@shantanupoddar1831]

Thanks sir

[@Misseldine]

You're very welcome.

[@visible_radiation]

ধন্যবাদ sir..

[@Misseldine]

You're welcome.