# Change in x (Differentiation) - Calc III needed?

1. Oct 8, 2013

### Qube

1. The problem statement, all variables and given/known data

http://i.minus.com/jDtSwpCGlrMhP.jpg [Broken]

2. Relevant equations

Solving for dx/dy (change in x with respect to y) I get a solution that isn't one of the answer choices.

3. The attempt at a solution

3y^2 = 8x' + x'/x

Coordinate:

x = 1 (given)
y = 2 (solved for in original equation)

3(4) = 9x'
12/9 = dx/dy
4/3 = dx/dy (change of x with respect to y).

Last edited by a moderator: May 6, 2017
2. Oct 8, 2013

### Dick

You want to think about dy/dt and dx/dt not dy/dx. For example, d/dt(y^3)=3y^2*dy/dt. Try that again. There is a correct answer in there.

Last edited by a moderator: May 6, 2017
3. Oct 8, 2013

### brmath

No matter what derivatives you take, please make sure you apply the chain rule properly. I don't think you did that in your attempt.

4. Oct 9, 2013

### Staff: Mentor

The tipoff that they're looking for time rates of change (dx/dt and dy/dt) is the units in the answers - units/sec.

5. Oct 9, 2013

### Qube

I'm not seeing the issue when I take the derivative of the equation with respect to y.

6. Oct 9, 2013

### Qube

Thank you. That makes more sense. I found dx/dt and plugged in dy/dt which is -1/2.

3y^2(dy/dt) = 8(dx/dt) + dx/dt*(1/x)

y=2, as always.

3(4)(-1/2) = 9(dx/dt)

-6 = 9(dx/dt)
-2/3 = dx/dt

It appears the answer is D.

7. Oct 9, 2013

Yes, it is.