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What is the shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve $ y = 3/x $ at some point?

shortest length is 2$\sqrt{6} \approx 4.899$

07:41

Wen Z.

01:51

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Campbell University

Oregon State University

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mhm, whereas to find the shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve y equals three over X at some point. So we have our function y equals three over X. We want to find the distance between the X and Y intercepts of the tangent line to the curve at some point. So the derivative Why prime, this is negative three over X squared, which is our slope of this line. Yeah, an equation of a line that passes through the point. A three over a with this slope is why minus three over a equals R slope. Negative three over a squared times X minus a. Okay, that's what I'm saying. Yeah, we can expand this to y equals negative three over a squared X plus six over a. Now, to find the Y intercept, you simply set X equal zero. So have a wide intercept of 06 over a on the X intercept. Well, this is setting. Why equal to zero? This gives us the point. Negative six over a divided by three over a squared which is just to a zero this is from the X intercept to the Y intercept as a function of a I'll call it D. This is the square root of the difference of X coordinates, which is to a squared plus the difference in why coordinates, which is six over a squared. Now we know that this distance is minimized when it's square is also minimized. Yeah, yeah, yes, that's it. So we have our function f of A which is for a squared plus 36 over a squared. Now I want to minimize f so will differentiate and find where it's zero f prime of a is eight a minus 72 over a cubed and we set this equal to zero solving for a we get that a is equal to eventually the square root of three. Notice that the second derivative of F this is eight plus four times 72 over. I'm sorry. This should be three times 72 over eight of the fourth, which is positive for all values of a including equals Route three. Therefore, it follows that our function f has a minimum, he said at a equals route three. Yeah, mhm, of course, by our previous discussion. This is also the minimum. D also has a minimum at equals Route three. And so our minimum distance is going to be D of the square to three, which is the square root of four times three schools. Plus with me 36 over three, which is the square root of 12 plus 12, or to route six. So the shortest length is to route six, which is approximately 4.899 better side. Yeah, yeah.

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