Vector Analysis ifferential Calculus

[Benzoate]

Vector Analysis:Differential Calculus

Homework Statement

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?

Homework Equations

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

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!

(have a curly d: ∂ )

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 …

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 …

 

[Benzoate]

Homework Statement

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?

Homework Equations

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

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? )

erm … 90x is wrong

and the equations would be a lot more manageable if you'd divided them by 20

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!

 

[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]

it's field theory …

well because the problem came from my intro to Electrodynamics textbook

The height of a certain hill …​

The hill was part of a field! ​