Derivative WRT third variable

[trainee engineer]

Good afternoon everyone

I am working on my exam revision and stuck on an question involving x and y WRT t.

The question reads:

Quantities x and y change with time, t and are related by the formula [math]x+4yx=3[/math]. When x = 2 and [math]\frac{dx}{dt}=1[/math], find the value of [math]\frac{dy}{dt}[/math]
Firstly, I subbed x = 2 and solved for y, and [math]y=\frac{1}{8}[/math]
then I worked through the derivative WRT t

[math]\frac{d(x+4yx)}{dt}=\frac{d(3)}{dt}[/math][math]\frac{dx}{dt}+\frac{d(4yx)}{dt}=0[/math]
by subbing for [math]\frac{dx}{dt}[/math] and rearranging, I get

[math]\frac{d(yx)}{dt}=\frac{-1}{4}[/math]
Using chain rule I get

[math]y\frac{dx}{dt}+x\frac{dy}{dt}=\frac{-1}{4}[/math]
and subbing in known values I find that [math]\frac{dy}{dt}=-\frac{3}{16}[/math]
However my options are [math]\frac{1}{4}, -\frac{1}{4}, \frac{1}{2} or 0[/math]
Can you see my error?

 

[JeffM]

I do not see an error in your work.

 

[blamocur]

I do not see an error in your work.

Me neither.

 

[Deleted member 4993]

Good afternoon everyone

I am working on my exam revision and stuck on an question involving x and y WRT t.

The question reads:

Quantities x and y change with time, t and are related by the formula [math]x+4yx=3[/math]. When x = 2 and [math]\frac{dx}{dt}=1[/math], find the value of [math]\frac{dy}{dt}[/math]
Firstly, I subbed x = 2 and solved for y, and [math]y=\frac{1}{8}[/math]
then I worked through the derivative WRT t

[math]\frac{d(x+4yx)}{dt}=\frac{d(3)}{dt}[/math][math]\frac{dx}{dt}+\frac{d(4yx)}{dt}=0[/math]
by subbing for [math]\frac{dx}{dt}[/math] and rearranging, I get

[math]\frac{d(yx)}{dt}=\frac{-1}{4}[/math]
Using chain rule I get

[math]y\frac{dx}{dt}+x\frac{dy}{dt}=\frac{-1}{4}[/math]
and subbing in known values I find that [math]\frac{dy}{dt}=-\frac{3}{16}[/math]
However my options are [math]\frac{1}{4}, -\frac{1}{4}, \frac{1}{2} or 0[/math]
Can you see my error?

Taking a different tack:

x + 4yx = 3

y = (3-x)/(4x) = 3/(4x) - 1/4

dy/dt = -3/(4 * x^2) * dx/dt

at x = 2

dy/dt = - 3/(4 *4) * 1 = - 3/16 ........................... same as OP

 

[Steven G]

x+4xy =3
x(1+4y) = 3
(1dx/dt)(1+4y)+(x)(4dy/dt) = 0
4xdy/dt = -(dx/dt)(1+4y)
dy/dt = -(dx/dt)(1+4y)/(4x) = -(1)(1+1/2)/(4*2) = -3/16

 

[Dr.Peterson]

The question reads:

Quantities x and y change with time, t and are related by the formula [math]x+4yx=3[/math]. When x = 2 and [math]\frac{dx}{dt}=1[/math], find the value of [math]\frac{dy}{dt}[/math]
However my options are [math]\frac{1}{4}, -\frac{1}{4}, \frac{1}{2} or 0[/math]
Can you see my error?

Either their list of answers is wrong, or the problem is not what they intended.

Looking for a possible typo (and noticing the odd order of "4yx"), what if the problem is meant to be x + 4y/x = 3? Then one of the choices is correct!