Derivative with respect to third variable

[trainee engineer]

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 $$\frac{y^2+x^2}{y+2x}=1$$ . find $$\frac{dy}{dt}$$ given $$\frac{dx}{dt}=3$$ when y = 1 and x = 2.

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

My working is as follows:

$$\frac{y^2+x^2}{y+2x}=1$$
multiply out the denominator

$$y^2+x^2=y+2x$$
move the x and y to one side

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

Can anyone see where I have made the error?

Thank you in advance.

 

[topsquark]

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 $$\frac{y^2+x^2}{y+2x}=1$$ . find $$\frac{dy}{dt}$$ given $$\frac{dx}{dt}=3$$ when y = 1 and x = 2.

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

My working is as follows:

$$\frac{y^2+x^2}{y+2x}=1$$
multiply out the denominator

$$y^2+x^2=y+2x$$
move the x and y to one side

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

Can anyone see where I have made the error?

Thank you in advance.

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]

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

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.

 

[trainee engineer]

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

Thank you. That makes me feel much better.

 

[trainee engineer]

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.

Thank you. That was my thought too when I worked it out but double and triple checking, I had written it correctly.