(I4pts) The surface z =y ~x is given nlong with the points A(-I, 0, 1) and B(2, 0,2) 0) If an ant is pluced at A what would its slope be ifit walked down the steep...

Find the slope of the surface at the given point in (a) the $x$ -direction and (b) the $y$ -direction. $$ \begin{array}{l}{z=x^{2}-y^{2}} \\ {(-2,1,3)}\end{array} $$

We're looking at the 0.1 to 2 mhm and we're looking at the function Z equals X times. Why, for part A were asked to calculate well, basically rephrasing the question were just asked to calculate the partial derivative of Z at this particular point with with respect to X. So this is just equal. Why and then if we evaluate that 0.12 to 1 to two, that is where X is one, and why is, too we get to now when we calculate the partial derivative with respect, why we get X. And when we evaluate this guy at the given points, we get one and that's it.

So for this problem, we have f of X y being equal to 1000 minus 0.5 x squared minus 0.1 y squared and were given the point equal to 60 40. And we want to go. Um, we're trying to find the directional derivative at a point p along V. So we have our directional derivative equation. We know that what we ultimately want to find them is our unit factor. So since we're moving due south, we know that our unit vector is just going to be zero negative one. Because if we think about a compass where this is north, this is South due South is this directional derivative of zero negative one and that's a unit vector already. So then our gradient of F X wine is going to be equal to negative 0.1 x negative 0.2 Why? And then we want to evaluate it at the 0.60 40 so that grading of F at 60 40 equals negative zero point 06 negative 0.8. Then from the equation nine, we can write our directional derivative as being equal to this dot our unit vector. So what we end up getting is just 0.8 as our directional derivative. Um, right here And since the directional derivative is greater than zero, we know that we shall move uphill as we're moving south. Then for part B, um, we have the same f of X y we have all that. Um The only thing that's different now is we are going to have a different unit vector now, So V in this case is equal to negative 11 So now our unit vector is actually going to be one over root two a negative one over it, too. Positive, one over root two. So when we take our, um, are directional our gradient again. So that was zero negative. 06 negative 08 Then when we do this, we end up getting, uh, when we got that with you, we end up getting a negative 1/5 route to. So since in this case it's less than zero. We see that we're moving downhill as we move northwest. And then lastly for part C, we want to determine in which direction slope is the largest, and we know that's going to be the case when the directional derivative we have 60 40 take the magnitude of that and we end up getting one. Um, and that's the maximum rate of change. So, max rate of change and we know that's going to occur, um, in the direction of the gradient of F, which, obviously, as we already have suggested, it is a long the vector negative. 0.6 negative, 0.8

In this problem we will cover computing partial derivatives. So the first part of this problem requires us to find the partial derivative of this function Z. With respect to X. And that means we are going to treat the variable. Why fixed? So we're going to treat it like a constant. So we want to find the derivative of our first X term which is three X squared. And We see that our 2nd charm for wise weird is just going to have a derivative of zero since we're treating why like the constant. And our last term is going to be a X. Y. And again, since we're treating why like a constant, the derivative of that is going to be just a white. So our final answer is going to be six X -AY. Now to find the partial derivative with respect to X at the 0.12 we're just going to plug in one for X into our function And two for why? And we get an answer Of 6 -2. 8. Now to keep the surface represented by the function Z sloping upward in the X direction. We want our partial derivative with respect to X to remain positive. and that means keeping 6 -2 a greater than zero. And that means our values of a have to be less than three because if A is equal to three, that means are partial derivative is just zero. And if it's greater than three we will get a negative partial derivative, which means the surface won't be sloping upwards anymore, it will be sloping downwards. So now moving on to the second part of our problem. This requires us to find the partial derivative with respect to why? So now instead of holding why fixed, we're going to hold xbox and treat it like a constant. So we know that the first X term three X square is just going to have a derivative of zero. And for our second term, we see that it has A. Y. So we're going to take the derivative of that. And also lastly because our last term has a Why in it as well, we can take the derivative of it and we're going to treat X constant. So our final answer is going to be eight wide minus A. X. So now we want to find the partial derivative respects why? At the .12? And we're just going to plug in one for X. And we're going to plug in two for why. And we see that our answer is 16 -A. Yeah. Now, in order to keep this partial derivative positive, we want a to be Less than 16 because of a equal 16. The partial derivative zero. And if a is greater than 16 are partial derivative, it's going to be negative, which makes our surface slope downwards instead of upwards. And we see that the restrictions placed on A from the first part of our problem does in fact Coincide with our restrictions from our 2nd part, so we know that the surface represented by the functions E. Is going to slope upwards with our values of A. From the first part of the problem.

Here we have Z is equal to four minus X squared minus y squared Part A is if we rephrase it, asking us to calculate the partial derivative with respect to X and then evaluated at the point that's given. So this is equal to negative two X and then if we evaluate at that given point where X is one and wise one, we get negative, too. Now, Part B is asking essentially the same thing. We need to calculate the partial derivative with respect to why so get negative two y and then we want to evaluate at a given point when X is one and wise one. So that's going to give us negative, too. That's it.