Partial derivative problem.... why is my answer wrong?

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Homework Help Overview

The discussion revolves around a problem involving partial derivatives, specifically related to the rates of change of two variables, x and y, over time. Participants are attempting to reconcile their calculated results with those expected by a homework program.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the calculation of the partial derivative and questioning the sign of the result. There is a focus on the velocities of x and y, with some participants noting discrepancies in the expected answer from the homework program.

Discussion Status

Multiple participants have arrived at the same numerical result but are uncertain about the sign and the acceptance of their answers by the homework program. Some suggest that the issue may relate to significant figures, and there is a shared frustration regarding the program's requirements.

Contextual Notes

Participants mention that the distances x and y are defined in terms of time, leading to negative rates of change. There is a concern about the homework program's handling of significant figures, with suggestions that it may not align with the expected precision of the answers.

Daniel Sellers
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Homework Statement


The entire problem is in the attached picture. I have been checking and double checking for about an hour, found solutions online which agree with my solution, but I cannot find any answer beside -3.697 m/s which is marked wrong by the computer program.

Homework Equations


Is the homework program wrong or am I somehow missing something?

df/dt = (2x-y)/(2(x^2+y^2-xy)^(1/2))(dx/dt) + (2y-x)/(2*(x^2+y^2 -xy)^(1/2))(dy/dt)

The Attempt at a Solution


All of this works out to -3.697. I have tried rounding, leaving the answer as positive, nothing seems to work. Very frustrated.
upload_2017-11-28_22-25-24.png
 

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I also get 3.697... but where did u get the minus sign from, you input the velocities as negatives?
 
Delta² said:
I also get 3.697... but where did u get the minus sign from, you input the velocities as negatives?

The minus sign comes from the fact that the distances x and y at time t will be given by x = 21 - 5t and y = 25 - 3t hence dx/dt = -5 and dy/dt = -3. But the computer will not accept it either way. Unless anyone else can see some mistake we've both made, I will now go complain to my teacher!

Thanks
 
Daniel Sellers said:
The minus sign comes from the fact that the distances x and y at time t will be given by x = 21 - 5t and y = 25 - 3t hence dx/dt = -5 and dy/dt = -3. But the computer will not accept it either way. Unless anyone else can see some mistake we've both made, I will now go complain to my teacher!

Thanks

I agree with your answer.
 
Daniel Sellers said:
The minus sign comes from the fact that the distances x and y at time t will be given by x = 21 - 5t and y = 25 - 3t hence dx/dt = -5 and dy/dt = -3. But the computer will not accept it either way. Unless anyone else can see some mistake we've both made, I will now go complain to my teacher!

Thanks

More simply: the two people get closer together as ##t## increases.
 
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Have you tried fewer decimals? Four significant digits is a bit much considering the numbers given.
 
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It turns out that whoever wrote my homework program has never in their life heard of significant figures. Given the values involved the answer should contain only 1 sigfig, but the solution was to input SIX MORE DECIMAL PLACES from my calculator.
 
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Daniel Sellers said:
It turns out that whoever wrote my homework program has never in their life heard of significant figures. Given the values involved the answer should contain only 1 sigfig, but the solution was to input SIX MORE DECIMAL PLACES from my calculator.
And I suggested you use fewer :rolleyes:
 

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