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

## 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 postive, nothing seems to work. Very frustrated. #### Attachments

• 35.3 KB Views: 218

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Delta2
Homework Helper
Gold Member
I also get 3.697... but where did u get the minus sign from, you input the velocities as negatives?

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 any one else can see some mistake we've both made, I will now go complain to my teacher!

Thanks

PeroK
Homework Helper
Gold Member
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 any one else can see some mistake we've both made, I will now go complain to my teacher!

Thanks

Ray Vickson
Homework Helper
Dearly Missed
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 any one 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.

• WWGD
DrClaude
Mentor
Have you tried fewer decimals? Four significant digits is a bit much considering the numbers given.

• Delta2
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.

• DrClaude
DrClaude
Mentor
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 