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20. The contour plot for f (xy) is shown at the right. Determine if each quantity is positive, negative, Or Zero_ (a) f (L,1) (b) f.(-11) (c) f,6-2,-2) (d) f,6-2,-2...

Question

20. The contour plot for f (xy) is shown at the right. Determine if each quantity is positive, negative, Or Zero_ (a) f (L,1) (b) f.(-11) (c) f,6-2,-2) (d) f,6-2,-2) (e) f,(1,1) (f) f.(,-1) where 1 = 1i Fj52-3 -3 52 -1 221. Let W(x;y) = 3xcos(ty)_ Find %w] and & if x=u +V and y=~ Tu k, 12) Tv k,12) dwl (b) Find if r=e and y= Int _ dt I=l

20. The contour plot for f (xy) is shown at the right. Determine if each quantity is positive, negative, Or Zero_ (a) f (L,1) (b) f.(-11) (c) f,6-2,-2) (d) f,6-2,-2) (e) f,(1,1) (f) f.(,-1) where 1 = 1i Fj 52 -3 -3 52 - 1 2 21. Let W(x;y) = 3xcos(ty)_ Find %w] and & if x=u +V and y=~ Tu k, 12) Tv k,12) dwl (b) Find if r=e and y= Int _ dt I=l



Answers

Using the contour diagram for $f(x, y)$ in Figure $8.36,$ decide whether each of these partial derivatives is positive, negative, or approximately zero. (a) $\quad f_{x}(4,1)$ (b) $f_{y}(4,1)$ (c) $f_{x}(5,2)$ (d) $\quad f_{y}(5,2)$

Hey, it's clear. So when you right here. So here we have our function. Her part Amy have to differentiate. So we get a kind is equal to six times two X minus one on X plus one. And for it to be increasing, the derivative has to be bigger or equal to zero. So when we make a bigger equal to zero, we got it. Thanks. Going negative. Infinity. Good. And for a function to be decreasing, the prime must be smaller or equal to zero. So we're gonna do this equation. But this time, scorn only smaller are equal to zero. So we get that's they could have won calmer one, huh? Her part b. We got our differential equation and we're gonna different you again to find the second derivative, we got 24 x plus six to get our local minima and maximally up to set the prying zero. So when we set this equal to zero, we got access equal to, and I could have one Can no one have Now we're gonna plug these X values into our second derivative. Kerr won. It's equal to negative 18. Then when it's half is equal to just regular 18. And since the second derivative, when it's negative, one is smaller than zero and, um, left double prime of half this bigger, bigger than zero. The function has a local Maxa xs equal to negative one in a local minimum access equal to have so f a negative one. You go to six, enough of half that's equal to negative. 34 So, Max, um, and for part C for points of inflection, um, we're gonna first start with the Khan cavity tests. So if f double promise bigger than zero, we know that it's calm. Keep up. So even this equation 24 x plus six we're gonna make it bigger than certain zero when we get access bigger than negative one for so we know it's Kong cake up when it's negative, one forthe common infinity. And then if it's smaller than zero, we know that Khan Cake down 24 X plus six Texas Smaller than negative one ploy. We know it's called gave down and I could have infinity common negative 14 where the points of a inflection it has two F double crime has to change signs, so we see that we'll make it equal to zero when we got X is equal to negative 1/4 when we plug in the good of 1/4 you got 21 overeat. So we know our point of inflection as negative 1/4 comma 21 8 21 over means

Hey, it's clear. Someone. You right here. So here you ever function F of X is equal to x square Over at square plus three We differentiate it and we get six x over X square plus three square cute squared so far apart A for a function to be increasing F prime must be bigger than zero. So we're gonna make this equation one of prime is bigger than zero We get X this bigger than zero so we know that it's increasing in the interval zero comma infinity And for a function to be decreasing, it has to be less than thorough. This is one excess smaller than zero. This occurs will negative infinity home a zero part B We have our different you'd equation. So we have six x over x scripless three square. We're gonna differentiate it again. When we had 18 times x square plus three comes one minus x square all over That's square. Let's next square plus re cute to get local mats or men, we have to set the prime to be zero and soft reserve solve for X. So we're gonna make this a crime. Nicole zero. When we get a value of Xs equal to zero. You plug in X equals zero until our double time and then we got to. And since the double prints begin and zero, it has to be a local minimum. So it's a local minimum at, um, X equals zero for part C. You're going thio first. Um, our con cavity tests. So if f double crime this bigger than zero we know that it's calling cape up. So we're gonna put Got in and Teaneck Square was three comes linguine. It's X square over X square, minus plus three Q Bigger than zero on we get our inequality. Negative one and one. So the interval is calling cape up when it's negative one one and for it to be called gave down f double prime has to be smaller than zero. So we're just gonna switch this sign and we get ups. It's bigger than one. Were negative x smaller than negative one. So the interval is negative. Infinity common negative one one comma infinity for points of inflection. It's when the double prime changes, so we make us equal to zero. When the double primacy equal to zero access equal to negative one or one we're going to calculate by plugging them in. So an effort see half of negative one is equal to 1/4 and F one this equal to 1/4 as well. So the points of inflection or negative 1 1/4 and one comment on four.

For this exercise will assume F X equals one plus one over X squared and through the power of air. And hence according to a question, this limit equals the limit. As X approaches zero one plus X squared to the power of one over X. We let y echoes one plus X squared to the power of one over X and log Y. Because log one plus X squared over X. And the limit as X approaches zero of log one plus X squared over eggs. This limit is of the form 00 and we can apply the lobby to its roots. Saudi calls the limits of two X over one plus X squared. My ex approaches zero. This expression into zero mm. So the limit as X approaches zero or one, he calls the limits as X approaches zero of E to the power of log one, you know this vehicles each power of zero great vehicles. What

Moving parallel to the y axis from a .11 to the nearest level curve. And this point is approximately here and the partial derivative with respect to Y equals Georgia Z. over Georgia one 30 Z. He calls fine four minus four And Georgia Z. because 1.25 -1. Since the y coordinates of this point is approximately one point 25 And the answer is miners full. For part B. We move parallel to the X axis from a point Miners for 2 to the nearest level curve. So he tickles Darda Z over dirt to X. Start A Z. Because one miner's zero. The X coordinate of this point is approximately Miners 2.5 and has, They're dioxin pickles miners 2.5 miners. My nerves full. He has this to you girls. Two thirds. We'll proceed, remove parallel to the X axis from the point -5 -2 to the nearest level curve. This is also equal. Start A Z over dirt tracks. George has the he calls why miners there? And the x coordinate of this point is approximately -2.5. Hence dirtbag sickles, Miners 2.5 miners -5 and their sickles 2/5 for per deep. We move parallel to the Y axis from the point zero miners, too, to the nearest nah Volker, and the dirt has the echoes zero minus one, and the white coordinator of this point has approximately miners 19/8 and miners miners, too. And hence the answer is 8/3.


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