trainee engineer
New member
- Joined
- May 16, 2022
- Messages
- 11
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.
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.