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Suppose you are climbing a hill whose shape is given by the equation 1400 O05x? 0.01y" where X, Y, and 2 are measured in meters, and you are standing at a poin...

Question

Suppose you are climbing a hill whose shape is given by the equation 1400 O05x? 0.01y" where X, Y, and 2 are measured in meters, and you are standing at a point with coordinates (120, 80, 1264). The positive X-axis points east and the positive Y-axis points north: a) If you walk due south, will you start to ascend or descend?oscengdescendAt what rate? vertical meters per horizontal meter(b) If you walk northwest, will you start to ascend or descend? asccncdescendAt what rate? (Round your an

Suppose you are climbing a hill whose shape is given by the equation 1400 O05x? 0.01y" where X, Y, and 2 are measured in meters, and you are standing at a point with coordinates (120, 80, 1264). The positive X-axis points east and the positive Y-axis points north: a) If you walk due south, will you start to ascend or descend? osceng descend At what rate? vertical meters per horizontal meter (b) If you walk northwest, will you start to ascend or descend? asccnc descend At what rate? (Round your answer to two decimal places:) 0.28 vertical meters per horizontal meter In which direction is the slope largest? What is the rate of ascent in that direction? vertical meters per horizontal meter At what angle above the horizontal does the path in that direction begin? (Round your answer to two decimal places _



Answers

Suppose you are climbing a hill whose shape is given by the equation $ z = 1000 - 0.005x^2- 0.01y^2 $, where $ x $, $ y $, and $ z $ are measured in meters, and you are standing at a point with coordinates $ (60, 40, 966) $. The positive $ x $-axis points east and the positive $ y $-axis points north.
(a) If you walk due south, will you start to ascend or descend? At what rate?
(b) If you walk northwest, will you start to ascend or descend? At what rate?
(c) In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin?

Hey, it's clear, So inhumane here. So we're given us of X Comma Y is equal to 1000 minus 0.5 at square minus 0.1 Lie square and P is equal to 60. Calmer 40. So we have to find the directional derivative of the function at point P along V. So using equation nine, we know that the direction of their birth is this formula. We'll know why times were you represents a unit vector in the direction of fee. So we know that you is going to be equal to one over one zero comma, negative one and the Grady it of f is given bye, because when you are, it's come on. Well, not far. I at school and F X is equal to negative Thorough point Sarah one X and fly X comma y this equal to negative point to why, after re substitute the values of point keen, they get negatives. There have been six comma negative and from Equation nine we can right 60 40 is equal to Ciro Coin eight. This is after we simplify. So since it's bigger than zero, we're gonna move uphill as we move. So So our answer is a walk to south from 60 comma 40 common 966 is uphill at a rate of 0.8 meters 0.8 meters 0.8 meters in. So for a part B oh were given the same equation and we're gonna find the direction of durable again. So using, um, equation nine. We're finding you which is one over square root too. They could have one common one, the grainy it is still this formula work here. In this case, it's going to be also negative. 0.1 x and negative 0.2 Why? So it's clean too. I won't write it all but negative 0.1 x comma Negative point Thio And then after we substitute the values off point p, you get negative 0.6 negative point and using equation nine, we could find the direction derivative, which is negative one over five square root too. Is this this less than zero? We're gonna move downhill as we move northwest and for part c were given that theory, um 10 which states that if we have f as adrift differential function and a V or in the domain of F and the maximum value of the directional derivative. A B is absolute value of the greedy int of F A B. And it happens when you has the same direction as the greedy inspector. And this is given by this formula once again and were given the F x of X comma y is negative. Negative, um, 0.1 x and f y of X comma y is negative 0.0. Why so after substituting the values, we get negative 0.6 and negative 0.8. And according to the serum ton, maximum rate of change of is this formula is equal to one. So the maximum rate of change of F the grating of 60 karma 40 which is equal to one and it happens along negative 0.6 and they get common negative 0.8

In this problem. You're given these shown information and were asked to first pretend that we are walking in the southern direction due south, in fact, and to determine whether that means we are going up a hill or down a hill. And so to do that, we need first our Grady in Vector for the function. And so we need to take Z with respect to X and that is going to be equal to negative 0.1 x in the partial Ozy, with respect to why it's is going to be equal to again. This was the person of the with respect to X inside the pressure of the with respect to why is going to be equal to negative 0.2 Why now we're going to evaluate these that are given point. Now, we really only need the 60 and 40 in this case because Z is what or Z is a function of X and y So what? We're just gonna use a 60 40. So I'm going to Delhi way at this partial at the 400.60 40 and I get negative 2001 times 60 and that is equal to negative 600.6 in the night of the other way. The negative 0.2 walling at the 0.60 40. And so I got negative 0.2 times 40. And so I get negative 400.8. And so that is our Grady in vector negative 0.6 and negative 0.8. Now we need our direction. Vector. Since they were telling us that we're walking due south, that means that we're not going south or we're going straight down. So which means they're direction Vector is going to be zero in the X direction and negative one in the Y direction. And because we know this and we have to make the length of this doctor one, we're going to divide it by the square root of the some of each of its components squared. So zero squared. Those negative one squared zero squared 01 negative one. Screw this one. So we just get spirit of one, which is one which we are going to divide into each component and just get zero and negative one back. And now we are going to take the dot product of these two vectors, and when we do, we get negative 0.6 common negative 0.8. That product with zero negative one No, 2.6 times zero is zero negative 00.8 times out of one is 10.8. And so our answer is going to be 28 and all I asks us if we're is if we're starting to ascend or descend. And now we know that our label for this is going to be 0.8 meters her every meter we travel forward. Many. We go a 0.8 meters for every meter we go forward and therefore we know that we're ascending or we're going uphill. And so that is the answer to part A and our answer to part B. It asks us about walking in Northwest, and so if we go in Northwest, that's going to be in this direction that is going to have a directional doctor of negative one, because we're going to negative one in the X direction and one in the Y direction because that's how much we're going up. And we've already found our Grady and doctor up here, and so we want to take the DOT product of that with our new vector. Once we make its length of Loch And so I'm going to divide it by the square rule of the solemn each of its components squared. So negative one squared plus one squared negative once credits +11 sport is one. So the spirit of to And so I get negative 1/2 comma, one over route to and again I'm going to typically don't product of the spectre with the new vector So negative 0.6 negative 0.8 and I'm going to write those infractions or negative 0.6 is negative. 3/5 negative 0.8 is negative. 4/5. That product with one over you'd be negative. 1/2 and one over. Route to do the 3/5 sends out of 1/2 is going to be three over five route to and then negative 4/5 times. One or two is going to be my has four over five route too. And so we get negative one over five route to That means we go down 1/5 or two meters for every meter of travel that means again. Then we are in descending. If we walk Northwest's okay and now we want to find the answer Teoh Part C. And to do that, I asks us for the direction in which the slope is the largest. Define that direction. We actually just take our Grady inspector. And so we know our Grady, Inspector. It's right here, and it's also up here you can write in either way. I'm going to choose a torrid as a decimal. And so the direction again, this is the case for all that this happens in all cases for the direction in which a soap is the largest is the grating vector. And so that this direction is going to be negative. 26 negative 260.8. And to find the rate of change in that direction of the grating doctor, we're gonna take the magnitude of it. And so theory of ascent, which is what they want, is going to be equal to the magnitude of the doctor. Negative 0.6, common negative. 28. And when we do that, we got a negative. A 0.6 squared to me the square root of negative 0.6 square, which is 0.36 plus negative pointed squared, which is 0.64 0.36 point 64 is equal to one. We got the spirit of one which is one that's already of a sentence one. And so this is our direction. This is a rate of ascent and now we're asked to find the angle above the horizontal of the path begins. The way we do that is that we take or say that the tangent of the angle that we want is equal to our rate of ascent. So the tangent of some angles equal toe one, which means that the tangent in verse of one is equal to our angle. And we know that a 45 degrees the tangent is equal to one and therefore our angle is equal 2 45 degrees. Again, we got that by taking the tangent of some angle on sending equal to a rate of ascent which we got by taking the magnitude of our direction vector which we got from migrating in factor

Mhm. In this problem we have a border that is launched vertically upwards by a volcano. The initial speed of the world. Er He's 40 m/s. The border is in projectile shot rob, then asked for the Halloween first. What is the time when the velocity of the boulder equals to 20 m per second? Appoints will denote this us capacity so to drive an expression. For time we will use velocity equation for a projector shot. So B. Y. Equals two. Vignette, Y plus A Y. D. And we know that for a projected more soon the vertical acceleration or a Y. Is equal to negative G. So our equation becomes me, Y. Z. Equals two. We know why minus. Sure B. We are changing since this is what we're looking for. The antique is it has to bring up Y minus V. Y. Over G. Cirque. Indeed, living values Then 40 m/s -20 m/s. Over 9.8 m/s squared. The answer is 2.04 seconds. So this is the time When the velocity is 20 m/s, whites question B. What is the time if the velocity is drinking either for a second but don't work so well. They notice us negative. So using the same equation, we have the signals to 40 m per second minus negative. 20 m per second Over 9.8. Meet their per second. The answer is 6.12 seconds. So the time when the velocity of the boulder is negative, 20 m per second for 20 million% number is 6 2012 seconds. No, for a letter C. The problem is for the time when the displacement of the border sequence to syria or why minus the initial position, why not? It was 20. So we will use the This place vertical displacement equations to drive an expression 40. So we have right? Because to why not blast peanut right there plus one half A. Y. The script. We know that A right supposed to negative G. And why? Mine's why not? It was too serious. A. Regulation becomes negative one half G. T squared plus Y. T. Equals to zero. So isolating T. to the left side of patient and he is equals two to renew quite over. It's tricky given values into multiplied by the initial velocity. The acceleration due to gravity in the answer is 8.16 seconds. So this is the time when the vertical displacement, you know No for a letter D. What is that time when the velocity 0? Using the equation we have for A&B. Then T. Is equal to 40 m per second minus zero m per second. Over 9.8. Neither per second grade, and the answer is 4.08 2nd. So this is the time in the velocity becomes real. Straight with lunch letter D. Let her eat as 40 magnitude and direction of the vertical acceleration when the boulder is moving up while moving downward. And at the highest point we know that for a projectile motion that vertical acceleration has a constant value. And yes, always equal to negative G. As such vertical acceleration. He's constant and always done work at any point. Mhm. And it's about it's a value of negative 9.8 m/mi2. Yeah. Yeah. So let there f as as to graph the acceleration velocity and a vertical displacement with respect to time. So let's start with a Y versus grab suppose is zero. And since The acceleration is negative 9.8 since it is constant, then this is just a what is sometimes nine now for V way versus speak? Yeah, here, two points for do you need their per second? 20 m per second? Zero negative 20 m per second and -40 m/s. For the time we have to four 68 Initially At T. equals zero. Our velocity is 40 m/s. A 20 m/s. The time is 2.34 seconds At 4.38 seconds. The velocity east negative thinking with their per second. Noble as it is I mean At 20 m/s. The time is to consider for a second At zero velocity that time is for consider eight seconds. At the time of 6.1 seconds developed velocity is negative 20 m/s. So but good enough. Is Lena No. For the right words to speak, two million zero. And since the project, its graph is a curve inverted parabola.

We have a large boulders ejected vertically upwards from a volcano with a dispute of forty meters per second. Ignoring a resistance we want to figure out at what time after rejection is the boulder now moving at twenty meters per second? Let's find this. We want to use the velocity is a function of time equation. Final losses for the initial velocity equals acceleration times time we want to find when the final glasses honey no, it's this velocities forty and we know that acceleration is just gravity. Nine point eight seconds squared times time going for tea two point oh, four seconds R b you wanted to same thing. So now we want to find when the final velocity is negative, twenty second Or, in other words, once moving at twenty downwards. Putting up this equation again and solvent, your team is equal to six point one two seconds. Parte si we want to find with displacement of the boulder from its initial position. Zero. So when in fact, the starting point, well, if it's fun stuff with a velocity of forty meters per second and goes all the way up and that's Apex is now zero meters velocities zero meters per second when it when it comes all the way back down, when its velocity is exactly the negative of its initial velocity. That's when it will be at its starting point to students. Symmetry of the travel. So I see you want to find now in the final velocity is negative of the initial velocity. So negative forty It was positive. Forty. When it's nine point eight, see, this gives us a time of eight point one seconds. Our beef you want to find when the velocity the boulders. Zero So again. Zero disease with forty since nine point eight. Then he for final eight seconds. Party. A little bit of quick question because for all three points, yeah, acceleration is negative. Nine point eight meters per seconds. Squared, pointing downwards, sort of center of the earth as always as usual for all three cases. So when it's moving upward when it's moving downward and at the highest point finally, now you want to draw three plots. First of the acceleration. First time sounds of the acceleration is constant, and it's negative change with times. That's that's gonna be a straight line, its velocity. He's going to be a straight line. That slope. Something's changing, changing linearly lost time like something like this on DH. Finally, its position. There's a function of time. He's going to be a rebel. Uh, of course, of the rebels should be symmetric, but it's it's hard to draw.


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