Problem

What is the minimum vertical distance between the…

04:20

Problem 5 Easy Difficulty

What is the maximum vertical distance between the line $ y = x + 2 $ and the parabola $ y = x^2 $ for $ -1 \leqslant x \leqslant 2 $?

Name: what is the maximum vertical distance between the line y x 2 and the parabola y x2 for 1 leqslant x
Uploaded: 2021-04-02
Duration: 6 min 47 s
Channel: Numerade
Description: What is the maximum vertical distance between the line $ y = x + 2 $ and the parabola $ y = x^2 $ for $ -1 \leqslant x \leqslant 2 $?

Answer

2.25

i seriously hope that no one is paying amrita for this--horrible explanation. Lazy and lousy teaching.

Maya R.

December 14, 2021

Amrita is the worst tutor of all the worst ones they have on Numerade, can't stand her. Get rif of her please before you start losing subscribes because of her. Lazy tutor.

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Video Transcript

maximum vertical distance between the line white was X plus two and the parameter of y equals X squared for X between negative one and positive too. Well, to do this first, let's find what the vertical distances between these two graphs recall. This distance D legal. There's a function of X. Well, this is going to be the square root of the difference of why coordinates. This is X squared minus X plus two squared. Now we know the vertical distance MM is maximized when he squared is maximized as well. So do you swear, Becks? I call this function f a bex. Well, this is X squared minus expose to squared. Now, to find the maximum vertical distance forward. To find the maximum of this function f of X, we're going to find its derivative. So at the prime of X is by the chain ruled two times X squared minus X minus two times two X minus one and we set this equal to zero. Well, this is only equal to zero when X squared minus X minus two equals zero or two x minus one equals zero. In the first case, the discriminate Delta E squared which is one minus four times a and see well, it's positive. So it has solutions. In fact, we can factor this as X minus two times X plus one equals zero. And so we get X equals two, where X equals negative one. Her second equation. This gives us X equals one half. However, we're only looking for excess between negative one and two. But all of these satisfy that. So we have three critical points native one one half hand, too. Now, we're going to compare the value of F I usually these critical points so f of negative one to plug this in. This is negative one squared, which is one minus one minus two squared negative two squared or four. I'm sorry. This should be one plus one minus two squared, which is zero f of one half. This is one half squared or four minus one half minus two. This is negative. 1/4 minus two. All this squared now. This is negative. 11 4th squared, no negative. 7/4 squared, no negative 9/4 square. My mistake. So this is 81 16th. Likewise. F two plug this in. This is two square which is four minus two minus two squared. This is zero squared or zero. So it's clear that yeah, the maximum value occurs at one of the critical values were at the endpoints, which happened to be the same as two of the critical values. And so f has absolute maximum on the interval. Negative 12 at X equals one half. Therefore, the maximum vertical distance the of one half. Well, this is the square root of one half squared, Just 1/4 minus one half minus two squared, which is the same as the absolute value. Uh huh. Negative 9/4 which is just positive. 9/4. So this is our answer.

Chris T.

Ohio State University

Related Courses

Calculus 1 / AB

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Optimization Problems