ChronicQuantumAddict
Oct2-05, 10:27 PM
The question is as follows: The height of a hill (meters) is given by [z=(2xy)-(3x^2)-(4y^2)-(18x)+(28y)+12], where x is the distance east, y is the distance north of the origin. a). where is the top of the hil (x,y,z) and how high is it (z=?)? b). How steep is the hill at x=y=1, that is, what is the angle between a vector perpendicular to the hill and the z axis? c). In which compass direction is the slope at x=y=1 the steepest?
okay, i know part a)., which is to get the derivative of dz/dx and dz/dy to get your x and y maximum values, then plug them back into the equation for z, giving you the location and the height of the hill at the steepest point.
The part i am having trouble with is b). how am i supposed to draw a vector that is perpendicular to the hill and calculate hte angle it makes with the z-axis? i thought of plugging in the x and y values of x=y=1 into the z equation, and this would give you z at that point, then create a vector normal to that point, but how do u find the angle (theta) made with the z-axis.
I also figured out part c). which is in a south east direction.
Please help, homework is due soon, thank you. :surprised
okay, i know part a)., which is to get the derivative of dz/dx and dz/dy to get your x and y maximum values, then plug them back into the equation for z, giving you the location and the height of the hill at the steepest point.
The part i am having trouble with is b). how am i supposed to draw a vector that is perpendicular to the hill and calculate hte angle it makes with the z-axis? i thought of plugging in the x and y values of x=y=1 into the z equation, and this would give you z at that point, then create a vector normal to that point, but how do u find the angle (theta) made with the z-axis.
I also figured out part c). which is in a south east direction.
Please help, homework is due soon, thank you. :surprised