Minimum distance Find the point P on the line y = 3x that...

Question

Minimum distance Find the point P on the line y = 3x that is closest to the point (50, 0). What is the least distance between P and (50, 0)?

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the point and the line Y equals -2X nearest to 150. Then calculate the minimum distance between them. A says the point is 36 and the minimum distance is 15 root 2 units. B says it's 36 and 6 route 5 units. C says it's -36 and 6 route 10 units, and D says it's -36 and 3 route 10 units. Now, what are we really trying to figure out here? Well, basically, what we're saying is that if we have the line Y equals 2 X, OK, or Y E equals negative to X rather. Then we're trying to find a point on the line or a point on this line that's going to be closest to 150. So let's say that 150 was here, OK? So now, if we call this point XY, OK, then we are trying to find this point and we're also trying to find the distance, the minimum distance between them, D. Now how are we going to do that? Well, here, notice that our distance will have a horizontal and a vertical component, OK? So in this case, We can use the formula for the distance on our grid to help us to figure out D or to write D in terms of X and Y, OK? And then from there we can minimize D to find the value of X, then use that for Y and then find D. So let's break that down a bit. First of all, let's try to determine what our distance would be using the distance formula. Recall That the distance formula. The distance far right now. Basically tells us, OK, that D is equal to the square root. Of our distance in X values X2 minus X1 squared. Plus Or difference in Y values, Y2 minus Y1 squared or coordinates, OK? So in this case, if we let 150 be X1 Y1, OK, and we let our point on the line. BX2Y2. Let me just rewrite this here. X2Y2, OK. Then no This would imply then. That our distance D is going to be equal to the square root of X minus 15. Squared Plus Y minus 0 squared, OK? I know. In this case, this is gonna be the square root of X minus, well, let me write it here, the square root of X minus 15 2 plus Y2. Now what do we know about Y? Well, remember, Y equals -2 X. And if we're going to minimize D, we'd want to express it purely in terms of one variable. In this case X. So let's substitute Y equals -2 X into our distance formula, and now it's gonna become X minus 15 2 + -2 X2. Where we know that negative 2 X2 will be equal to 4X, so this is going to really be the square root of X minus 15 squad. Plus 4 x 2. Now minimizing d is the equivalent of minimizing D2. So if we square both sides, that means dred will be equal to x minus 15 2 plus for X2. And no, we can minimize D2d by differentiating it with respect to X, setting it to 0, and solving for that value of X that makes it a minimum. So differentiating these squared. Notice here we have two terms, OK? And to differentiate, we can use the chain rule and the poor rule of differentiation. So now our derivative, the derivative of d squared with respect to X. Is going to be equal to 2 multiplied by x minus 15 multiplied by 2 to the power of 2 minus 1 because remember we subtract one from the poor, which is 1, and then we're going to multiply by the derivative of what's in our inner function and the derivative of X minus 15 is 1. There's gonna be 2 multiplied by X minus 15 multiplied by 1 as the derivative of X minus 15 squared. Plus the derivative of 4 X squad, which is 8 X. I know if we simplify that, this is going to be 2X minus 30 plus 8X and that is going to give us, give us 10 X minus 30. To know When we set our derivative to 0. OK. Then put this in red, we'll have 10 X minus 30 being equal to 0, which means that 10 X equals 30 and thus X is going to be equal to 3. And that makes sense because 10 multiplied by 3 is 30, 30 minus 30 equals 0. So now we know X minus X equals 3 is a critical point. Now we just need to confirm if it's a minimum or maximum. So confirming, confirming whether X equals 3. Is a minimum, yeah. We can use the 2nd derivative test. So we can use the second derivative test. I know what that test basically tells us, OK, is that if the second derivative is positive, that means the function is concave up, so it corresponds to a minimum, and if it's negative, it's concave down, it corresponds to a maximum, OK? Now, finding our second derivative here, that means we're gonna find the second derivative of the derivative of D squared with respect to X. Which means we're going to differentiate 10 x minus 30. Which is going to be equal to 10. 10 is greater than 0. So that means the function is concave up, which means that X is, sorry, X equals 3. Because it's positive, then X equals 3 is a minimum. So now we know that these critical values are minimal. OK, so we have the X value of the point. If we go back to our, our diagram here. We have the X value of our point that's going to make our function a minimum. Now we need the Y value and we know that our equation is Y equals negative to X. So all we need to do is to substitute this X value to get the Y value. So now, when X equals 3, let me put that in black here. When X equals 3. Then Y equals -2 multiplied by 3, which is -6. Therefore, This tells us that that sorry 36. Is the point on the line. Yeah, that's closest. To the 0.150. So we figured out the first part of our problem. No, we just need to find a minimum distance, and to find a minimum distance, all we need to do now. Is to is to substitute the value of or the value of our points here. So that's 3-6 into the distance formula, OK? So therefore, the minimum distance. OK. For 36 means then that we're gonna substitute. We're gonna take our formula. Remember it was D equals the square of X minus 15 + Y quad. So now that's gonna be. 3 minus the square root of 3 minus 15 square. Plus -6 Y 2, OK. Now if we think about it, this is gonna be the square root of -12 squared. Plus 36. Right. And negative 12 squared is 144. So this is gonna be the square root of 144 + 36, which is route 180, and the square of 180, we could rewrite this as 6 route 5, OK? Because here the skirt of 180 could be rewritten as the skirt of 36 multiplied by 5, and using the properties of roots, we could say this is Route 36 multiplied by route 5, which is how we got 6 route 5. Thus, 3 negatives, the 36 is 6 route 5 units away from 150. Now if we look back at our answer choices, that tells us then that B. Is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Minimum distance Find the point P on the line y = 3x that is closest to the point (50, 0). What is the least distance between P and (50, 0)?

Key Concepts

Distance Formula

Line Equation

Optimization

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