Vector Analysis:Differential Calculus

[Benzoate]

1. The problem statement, all variables and given/known data

The height of a certain hill(in feet) is given by

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

where y is the distance (in miles) north, x the distance east of South Hadley

a)Where is the top of the hill located

b) How high is the hill?




2. Relevant equations

grad T=dT/dx xhat+dT/dy yhat+ dT/dz zhat

3. The attempt at a solution

a) I need to find the distance in the x direction , so I would take the derivative of h(x,y) with respect to x

dh/dx=20*x-12=0=> x=3/5 feet

b) same algorithm, only I am now ask to calculate how high the hill is and so I would take the derivative of h(x,y) with respect to y:

dh/dy=y=3*x+9=3*(.6)+9=10.8 feet

or maybe I should calculate h(x,y) in order to determine the height of the hill. Therefore , I'd plugged x and y into h(x,y) right?

[tiny-tim]

Hi Benzoate! :smile:

(have a curly d: ∂ :smile:)
The height of a certain hill(in feet) is given by

h(x,y)=10(2xy-3x^2-4y^2-18x+28y+12)
…
a) I need to find the distance in the x direction , so I would take the derivative of h(x,y) with respect to x

dh/dx=20*x-12=0=> x=3/5 feet

b) same algorithm, only I am now ask to calculate how high the hill is and so I would take the derivative of h(x,y) with respect to y:

dh/dy=y=3*x+9=3*(.6)+9=10.8 feet

Nooo … :cry:

Your ∂h/∂x and ∂h/∂x are completely wrong …

for example, ∂h/∂x should start with 20*y, not 20*x

and what happened to all the other terms (and all the other 10s)?

You need to go back to your book and look again at how to do partial derivatives … :smile:

[Benzoate]

1. The problem statement, all variables and given/known data

The height of a certain hill(in feet) is given by

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

where y is the distance (in miles) north, x the distance east of South Hadley

a)Where is the top of the hill located

b) How high is the hill?




2. Relevant equations

grad T=dT/dx xhat+dT/dy yhat+ dT/dz zhat

3. The attempt at a solution

a) I need to find the distance in the x direction , so I would take the derivative of h(x,y) with respect to x

dh/dx=20*x-12=0=> x=3/5 feet

b) same algorithm, only I am now ask to calculate how high the hill is and so I would take the derivative of h(x,y) with respect to y:

dh/dy=y=3*x+9=3*(.6)+9=10.8 feet

or maybe I should calculate h(x,y) in order to determine the height of the hill. Therefore , I'd plugged x and y into h(x,y) right?

wow I messed up big time with calculating my partial derivatives.

Anyway, dh/dx= 20y-90x-180= and dh/dy= 20x-80y+280 =0. once I calculate my values for x and y , I would be able to calculate the height which is h(x,y), correct?

[tiny-tim]

Anyway, dh/dx= 20y-90x-180= and dh/dy= 20x-80y+280 =0.

(what happened to that ∂ I gave you? :smile:)

erm … 90x is wrong :rolleyes:

and the equations would be a lot more manageable if you'd divided them by 20 :wink:
once I calculate my values for x and y , I would be able to calculate the height which is h(x,y), correct?

That's right! :smile:

[HallsofIvy]

Why is this listed under "physics" rather than "mathematics"?

[Benzoate]

Why is this listed under "physics" rather than "mathematics"?

well because the problem came from my intro to Electrodynamics textbook

[tiny-tim]

well because the problem came from my intro to Electrodynamics textbook
The height of a certain hill …

:biggrin: The hill was part of a field! :biggrin: