Estimating partial derivatives

[joemama69]

Homework Statement

a metal plate is situated in the xy plane and occupies the rectangle 0<x<10 and 0<y<8 where x a y are measure in meters. The temperature oat the pooint x,y on the plate it T(x,y), where T is measured in degrees celsius.
note the attached table

a- estimate the values of the partial derivatives of T_x(6,4) and T_y(6,4). What are the units of measure
b - Estimate the values of T_u(6,4) where u = (i + j)/(2^1/2). Interpret your result
c - Estimate the value of T_xy(6,4)

Homework Equations

The Attempt at a Solution

welp I am not sure if correct but hers what i did

a - t(x) = 6, because from point 6,4, moving down the x value the temp rises 6 degrees
and so t(y) = -5. the units are degrees celsius i belive

b - for b i pluged my x and y partials from part a into the i & j component and got u = 7.8 degrees celsius

c - t(x) = 6 just as in part a. and from there point 8,4, i moved the right for a change of -6 degrees

 

[HallsofIvy]

Homework Statement

a metal plate is situated in the xy plane and occupies the rectangle 0<x<10 and 0<y<8 where x a y are measure in meters. The temperature oat the pooint x,y on the plate it T(x,y), where T is measured in degrees celsius.
note the attached table

a- estimate the values of the partial derivatives of T_x(6,4) and T_y(6,4). What are the units of measure
b - Estimate the values of T_u(6,4) where u = (i + j)/(2^1/2). Interpret your result
c - Estimate the value of T_xy(6,4)

Homework Equations

The Attempt at a Solution

welp I am not sure if correct but hers what i did

a - t(x) = 6, because from point 6,4, moving down the x value the temp rises 6 degrees
and so t(y) = -5. the units are degrees celsius i belive

What is "t"? Is it the same as T? If so, what do you mean by T(x)? T is a function of both x and y. Do you mean T_x(x,y), the derivate with respect to x?

The temperature at (6,4) is 80 while the termperatures at (4, 4) and (8,4) are 78 and 96 respectively. you could estimate the rate of temperature there as (80- 78)/(6-4)= 1 or (96- 80)/(8- 6)= 3 or (96- 78)/(8- 4)= 2. Of those I would expect the last to be most accurate. You seem to have forgotten to divide by the "$\Delta x$".

Finally, the units are not "degrees celsius". Because this is a rate of change, the units are "degrees celsius per meter"- again, you did not divide by the "$\Delta x$".

b - for b i pluged my x and y partials from part a into the i & j component and got u = 7.8 degrees celsius

Did you divide by $\sqrt{2}$?

c - t(x) = 6 just as in part a. and from there point 8,4, i moved the right for a change of -6 degrees

This question is asking for a second derivative. Calculate the partial derivatives with respect to x at (6,4), (6,2), and (6,6), then do the same calculation for the partial derivative with respect to y.

 

[joemama69]

sorry i was being lazy t(x) was just the partial with respect to x

ok so here's what i did

T(x) = (86 - 80)/(8-6) = 3
T(y) = (75-80)/(6-4) = -2.5

T(u), where u = (i + j)/2^1/2 was i corect in saying that i = T(x) & j = t(y), so T(u) = .354

for T(xy) don't i have to take the partial T(x), and then partial that with respect to y.

how do i do that.

 

[joemama69]

Ok I think i got T_xy(6,4)

T_x(6,4) = 3
T_x(6,6) = 2.5

=(T_x)(6,6) - T_x(6,4))/delta T = (3-2.5)/2 = .25

Can i get a ahmen