Derivative with respect to third variable

T

trainee engineer

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May 16, 2022
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Good afternoon everyone

I am having a bit of trouble trying to differentiate an equation with x and y as the variables, but the derivative is taken with respect to a third variable, t.

My problem is:

A particle is moving and its motion can be described with the equation [math]\frac{y^2+x^2}{y+2x}=1[/math] . find [math]\frac{dy}{dt}[/math] given [math]\frac{dx}{dt}=3[/math] when y = 1 and x = 2.

The options given for the answer are -3, -1, 0 or 1.

My working is as follows:

[math]\frac{y^2+x^2}{y+2x}=1[/math]
multiply out the denominator

[math]y^2+x^2=y+2x[/math]
move the x and y to one side

[math]y^2-y=2x-x^2[/math]
differentiate both sides with respect to t
[math]\frac{d(y^2-y)}{dt}=\frac{d(2x-x^2)}{dt}[/math][math]2y\frac{dy}{dt}-1\frac{dy}{dt}=2\frac{dx}{dt}-2x\frac{dx}{dt}[/math][math](2y-1)\frac{dy}{dt}=(2-2x)\frac{dx}{dt}[/math][math]\frac{dy}{dt}=\frac{2-2x}{2y-1}\frac{dx}{dt}[/math]
then substitute and solve for [math]\frac{dy}{dt}[/math]
which gives me [math]\frac{dy}{dt}=-6[/math]
which is not one of the offered solutions.

Can anyone see where I have made the error?

Thank you in advance.
 
trainee engineer said:
Good afternoon everyone

I am having a bit of trouble trying to differentiate an equation with x and y as the variables, but the derivative is taken with respect to a third variable, t.

My problem is:

A particle is moving and its motion can be described with the equation [math]\frac{y^2+x^2}{y+2x}=1[/math] . find [math]\frac{dy}{dt}[/math] given [math]\frac{dx}{dt}=3[/math] when y = 1 and x = 2.

The options given for the answer are -3, -1, 0 or 1.

My working is as follows:

[math]\frac{y^2+x^2}{y+2x}=1[/math]
multiply out the denominator

[math]y^2+x^2=y+2x[/math]
move the x and y to one side

[math]y^2-y=2x-x^2[/math]
differentiate both sides with respect to t
[math]\frac{d(y^2-y)}{dt}=\frac{d(2x-x^2)}{dt}[/math][math]2y\frac{dy}{dt}-1\frac{dy}{dt}=2\frac{dx}{dt}-2x\frac{dx}{dt}[/math][math](2y-1)\frac{dy}{dt}=(2-2x)\frac{dx}{dt}[/math][math]\frac{dy}{dt}=\frac{2-2x}{2y-1}\frac{dx}{dt}[/math]
then substitute and solve for [math]\frac{dy}{dt}[/math]
which gives me [math]\frac{dy}{dt}=-6[/math]
which is not one of the offered solutions.

Can anyone see where I have made the error?

Thank you in advance.
Click to expand...
I don't see anything wrong with your work. According to my work (taking the derivative of the original expression directly) I get the same answer you do.

-Dan
 
blamocur said:
Me too. My guess is you might have a typo when writing down the problem. E.g., for x=1,y=2 you get 0, which is among multiple choices.
Click to expand...
Thank you. That was my thought too when I worked it out but double and triple checking, I had written it correctly.
 
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