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Find the point on the line $3 x+4 y+7=0$ closest to the point $(1,-2)$.

$$(61 / 25,-2 / 25)$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 4

Applications I - Geometric Optimization Problems

Derivatives

Campbell University

Oregon State University

University of Nottingham

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

01:30

Find the distance between …

00:39

Find the point on the line…

01:59

00:58

We're asked to find the minimum distance between the line defined by this equation and the .1 -2. So here's our point of interest, here's our line. So the distance, you know, is given by the square root, this thing here and then this is our constraints. Yeah. So we can solve this for X in terms of why and plug that into here and we get this for D. And then we can take the derivative of this respect to why is that why you want one? Are optimal value and we get this. Now, let's see here, we have we have, what do I want to say? We have. Let's see. We wind up with not making this up this equal to zero. And we get uh let's see here, we just get one solution. Obviously this denominator is irrelevant. So we get this. Um which is that make this zebra zero. And that means why one equals minus 58/25. Now it could have been a little easier. And they suggest this in the next problem to actually minimize the squared If we minimize the squared were also minimizing. So we've got rid of the square root here, we've made our life a little easier. But again, this isn't too bad. We take the derivative that So that's why one if why the actual value of Y is 58/25, then the actual value for X is 19/25. And that is this point right here. And we can actually show that that's perpendicular. Yeah. Um Go to point this this line here is perpendicular to exist, although it doesn't quite look like it because I don't have this, I should have plotted this on the same scale. I think this is this is the spanish cube. These are the scales here actually plotted it so that the distance is here and the distances here were the same. So you would have seen that this is actually a doesn't look part when they go here, but it really is. So you can always find the minimum. And you can show that the minimum distance between a line and a point is always um along the line perpendicular to the to the line line you have.

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