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For the function f(xy) =1-x3ry on the disc x +y < [(a) Find all critical points of fAre the critical points of (a) degenerate? If not; what kind of critical poin...

Question

For the function f(xy) =1-x3ry on the disc x +y < [(a) Find all critical points of fAre the critical points of (a) degenerate? If not; what kind of critical points are they?(c) Find the absolute maximum of f on the disc_(d)Find the absolute minimum of on the disc.(e) Give rough sketch of the gradient vector field on the disc:(f) Give rough sketch of the level curves of the function on the disc .

For the function f(xy) =1-x 3ry on the disc x +y < [ (a) Find all critical points of f Are the critical points of (a) degenerate? If not; what kind of critical points are they? (c) Find the absolute maximum of f on the disc_ (d)Find the absolute minimum of on the disc. (e) Give rough sketch of the gradient vector field on the disc: (f) Give rough sketch of the level curves of the function on the disc .



Answers

Determine the critical points for each function. Determine whether the critical point is a local maximum or minimum, and whether or not the tangent is parallel to the horizontal axis. a. $h(x)=-6 x^{3}+18 x^{2}+3 \quad$ c. $y=(x-5)^{\frac{1}{5}}$ b. $g(t)=t^{5}+t^{3}$ d. $f(x)=\left(x^{2}-1\right)^{\frac{1}{3}}$

For this problem, we are asked to find all critical points of the function F of x Y equals x squared plus a squared minus to a x. Coast of y. Then we want to determine whether each point is a local max, local men or a saddle point. To begin, we take the partial derivatives with respect to X and Y. So the partial derivative with respect to X is going to be two x minus two. A coast of y partial derivative with respect why is going to be to a X sign of Y? And we want to solve for when both of these will equal zero. So we can see that for uh for the partial derivative with respect to why to equal zero, You have either x equals zero or why equals. Oh yes. Rather I should specify that we were given a domain for why? Wise between negative pi and pi. So uh for the partial derivative with respect to Y two equals zero. X could equal zero or like an equal zero. Then we can see that if Y equals zero then the partial derivative with respect to X is never going to equal zero. So we can exclude that as a possibility. So we have then that X needs to equal zero. Let's see daisy. Which then means that we need coast of Y. two equals 0. Which means then oh excuse me, I need to take that back. So if X equals zero then we would get that two x minus two. A coast wine needs to equal zero. Which would then mean that to a coast. Why would need to equal zero? Which would then say that coast of Why would need to be an integer multiple of pi But that's not going to be a valid option here. So in fact we get from that partial derivative with respect to why we should get actually that. Why must equal zero? Which in turn means that we have two X -2. A coast y queens, two X minus two. A coast of zero means just two X minus two. Amy's equals zero. So we get that X has to equal a whatever A. Is whatever A. Is there. So our candidate point then it's going to be the point A zero. Having that. We next need to start taking our second partial derivatives to apply the second partial derivative test. So the second partial derivative with respect to X is going to be too the mixed derivative Xy is going to be to a sign of Y. The second partial derivative with respect to why is going to equal to a X. Coast of Y. Which then means that our d function D of X. Y is going to equal uh for a sign of wine. It's not signed by. Excuse me, would be for a X. Coast of y. Yeah, minus for a squared sine squared Y. So evaluating at our .80, then We get for a square times coast of Y, which is just going to be one uh minus for a squared sine squared zero, which is going to just be zero. So our point then is going to be for a square which has to be greater than zero. And our second partial derivative with respect to X is greater than zero, which by theorem C then tells us that this point is going to have to be a I believe that's going to have to be a maximum. I'm going to double check as I speak here. Yes. No, actually. Rather that's going to be a minimum because both the second partial derivative with respect to X and D are greater than zero.

Alright for this problem, we want to find all critical points of the function F of X Y equals Cossacks plus coast Y plus cost of X plus Y for X between zero and pi by two and why between zero and pi by two. Then we want to find whether each critical point is a max min or saddle point. So to begin, we take the partial derivatives uh partial derivative with respect to X is going to be negative sine of x minus sine of X plus Y. Partial derivative with respect to why it's going to be negative sign of why minus sine of X plus Y. And now we want to solve for when both of these are going to equal zero. So what we can do here is for instance, uh we can get both. Okay, let's let's see here, we're going to get that sign of X plus Y needs to equal both negative sign of why? As well as it needs to equal negative sine of X. So what that indicates here, uh the only way that this is going to be possible Is the only way that this is going to be possible for X between zero and pi by two at one moment. What we see here is that the only way for this to be possibly true is if X equals Y and they both equal zero. The thing is that that's outside of the boundaries for where our function is actually defined what that tells us. I'll add that. This could also be true if we have both X and Y being integer multiples of pi. But again, that's outside of the boundaries of what is defined for us. So what that tells us is that there are actually no critical points within the domain.

Alright for this problem we are asked to find all critical points of the function F. Of x Y equals E. To the power of negative x squared plus y squared minus four Y. And then indicate whether each point is a max min or a saddle point. So we start by taking the partial derivatives with respect to X. Partial with respect to X. Is going to be negative two X times E. To the power of negative X squared plus y squared minus four Y. Ah By applying chain rule there. Uh Then partial with respect to why is going to be a negative two, Y plus four times E. To the power of negative X squared plus Y squared. Why squared minus four. Why? So we want both of these two equals zero, which tells us that we need X two equals zero and we need uh to Y two equal four, which means we need y two equal to. So our point is going to be 02 next to determine whether it's max min or saddle. We take the second partial derivative, the second partial derivative. This would be applying the product rule here uh on the first partial with respect to X. The second partial with respect to X is going to end up being four times X squared minus two times E. To the power of negative X squared plus y squared minus four Y 2nd partial with respect to why is going to be two times 2, y squared minus eight Y plus seven. Seven. There times E. To the power of negative X squared plus y squared minus four Y. Lastly we need not quite lastly, but next we need to take the mixed derivative which is going to come out to four x times y -2 times E. To the power of E. Or it's the power of negative X squared plus y squared minus four. Y. So d of 02 is going to be equal then to, let's see here will be D of X while right down first going to be four X squared minus two times. So week two times works were minus two times two, Y squared minus eight. Y plus seven times E. To the power of negative two, X squared plus y squared minus four Y minus are mixed partial squared. So that's going to be 16 X squared plus y minus two squared. And eat the power of negative two X squared plus y squared minus four. Why? So the last thing that we need to do is evaluate D of XY at the .02 Which will give us two times negative two times two times four minus eight times two plus seven. Yeah. Times E. To the power of negative two times for minus eight, then, because the X there is going to be zero, uh that last term is just gonna drop off and I'm just gonna calculate this off screen here. All right, we get a final answer of four E. To the power of eight. And the second derivative with respect to X is going to be negative there. So that tells us that since we have negative second derivative with respect to X positive D. That we have a max.

Alright for this problem, we want to find all the critical points and then classify the function at those points of the function F of x, Y equals x squared plus four, Y squared minus two, X plus eight wine minus one. So to find the critical points we want to find were both partial derivatives equal zero. The partial derivative with respect to X is going to be two X -2. Uh we said that equal to zero means that X is going to have to equal one. Then with respect to why you'll get eight. Why plus eight needs to equal zero? Which means that why is going to equal negative one? So our point is going to be one negative one. Then to determine the nature of the function at those points, we start taking the second partial derivatives. 2nd partial with respect to X is going to be too second partial. Why Is going to be 8? The mixed partial Is going to be zero. So our d function is going to be 16 0, or just 16 greater than zero. Our second partial derivative with respect to X, is greater than zero, so that tells us that this critical point is going to be a minimum.


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