How steep is the hill at a point 1 mile north?

[noblegas]

Homework Statement

the height of a certain hill is given by

h(x,y)=10(2xy-3x^2-4y^2-18x+28y+12)

wher y is the distance north and x is the distance east of south hadley

a)Where is the top of the hill

b) how high is the hill

c) how steep is the hill at a point 1 mile north

Homework Equations

The Attempt at a Solution

dh/dx=y-3x-9=0

dh/dy=x-4y+14=0

using linear system of algebra I find that my coordinates on top of the hill are

(x=-26,y=-3)

b) just plug in the coordinates you got in part a, into the h function; answer: h(x=-26,y=-3)

c) the slope would just be the gradient of h , which is: \nablah=(dh/dx)x-hat+(dh/dy)y-hat=-7x-hat+11y-hat

|\nablah|=sqrt(49+196)

direction is just: cos(theta)=\nablah dot dl/(
|\nablah||dl|), you take the inverse of cos(theta)

 

[rl.bhat]

In the problem it is given that the x is the distance east of south.
Will it make any difference on the co-ordinates of the tip of the hill?

 

[gabbagabbahey]

dh/dx=y-3x-9=0

dh/dy=x-4y+14=0

using linear system of algebra I find that my coordinates on top of the hill are

(x=-26,y=-3)

Your equations look correct, but your result (-26,-3) is not...check your algebra.

b) just plug in the coordinates you got in part a, into the h function; answer: h(x=-26,y=-3)

Right idea

c) the slope would just be the gradient of h , which is: \nablah=(dh/dx)x-hat+(dh/dy)y-hat=-7x-hat+11y-hat

Check your math on that again...you are missing a factor of 20(!) and your x-component is incorrect.

 

[gabbagabbahey]

In the problem it is given that the x is the distance east of south.
Will it make any difference on the co-ordinates of the tip of the hill?

No, "x is the distance East of South Hadley"...South Hadley being the name of a town

 

[Nabeshin]

To be fair, this isn't a very good characterization of the topography of the region. =( On what interval is this function valid? I get the feeling they're aiming for Mount Holyoke, simply because it's close to where the peak of the function is, though.

 

[noblegas]

Check your math on that again...you are missing a factor of 20(!) and your x-component is incorrect.

I did check my math and I continue to come up with: \nablah=(y-3x-9)x-hat+(x-4*y+14)y-hat, with x=1,y=1

 

[gabbagabbahey]

I did check my math and I continue to come up with: \nablah=(y-3x-9)x-hat+(x-4*y+14)y-hat, with x=1,y=1

Look at your original expression for 'h'...you are missing the factor of 10 in front and also another factor of 2 that you seem to have divided by...in part (a) it didn't matter, because you were setting it equal to zero, but here it does matter.

Also, when you plug in (1,1) to your above expression you do not get (-7,11).