Estimating partial derivatives

In summary, the problem involves a metal plate with temperature T(x,y) measured in degrees Celsius. Part a asks for the partial derivatives of T with respect to x and y at point (6,4) and the units are degrees Celsius per meter. Part b involves calculating T(u), where u = (i + j)/√2, and interpreting the result. Part c asks for the estimate of the second derivative Txy(6,4) by taking the partial derivative of Tx with respect to y at point (6,4). The final answer is 0.25.
  • #1
joemama69
399
0

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 celcius.
note the attached table

a- estimate the values of teh partial derivatives of Tx(6,4) and Ty(6,4). What are the units of measure
b - Estimate the values of Tu(6,4) where u = (i + j)/(2^1/2). Interpret your result
c - Estimate the value of Txy(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 celcius

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
 

Attachments

  • 10.pdf
    4.9 KB · Views: 370
Physics news on Phys.org
  • #2
joemama69 said:

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 celcius.
note the attached table

a- estimate the values of teh partial derivatives of Tx(6,4) and Ty(6,4). What are the units of measure
b - Estimate the values of Tu(6,4) where u = (i + j)/(2^1/2). Interpret your result
c - Estimate the value of Txy(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 Tx(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 "[itex]\Delta x[/itex]".

Finally, the units are not "degrees celcius". Because this is a rate of change, the units are "degrees celcius per meter"- again, you did not divide by the "[itex]\Delta x[/itex]".

b - for b i pluged my x and y partials from part a into the i & j component and got u = 7.8 degrees celcius
Did you divide by [itex]\sqrt{2}[/itex]?

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.
 
Last edited by a moderator:
  • #3
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)/21/2 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.
 
  • #4
Ok I think i got Txy(6,4)

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

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

Can i get a ahmen
 

What are partial derivatives?

Partial derivatives are a type of derivative used in multivariable calculus to measure the rate of change of a function with respect to a specific variable, while holding all other variables constant.

Why are partial derivatives important?

Partial derivatives are important because they allow us to analyze how a function changes in multiple directions. This is essential in fields such as physics and economics, where variables are often interdependent.

How do you estimate partial derivatives?

To estimate partial derivatives, we use the limit definition of a derivative. This involves taking the limit as the change in the variable approaches zero, and calculating the slope of the tangent line at that point.

What is the difference between partial derivatives and ordinary derivatives?

The main difference between partial derivatives and ordinary derivatives is that ordinary derivatives measure the instantaneous rate of change of a function with respect to one variable, while partial derivatives measure the rate of change with respect to a specific variable while holding all others constant.

What are some real-life applications of partial derivatives?

Partial derivatives have many real-life applications, such as in physics for calculating the acceleration of objects in multiple dimensions, in economics for analyzing the impact of multiple variables on a market, and in engineering for optimizing complex systems with multiple variables.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
489
Replies
9
Views
646
Replies
4
Views
594
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
705
  • Calculus and Beyond Homework Help
Replies
1
Views
945
  • Calculus and Beyond Homework Help
Replies
1
Views
535
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
28
Views
2K
Replies
6
Views
1K
Back
Top