# Trying to solve a related rates problem

I am trying to solve a related rates problem. The problem states:

If y = 4x -x^3 and the x-coordinate is increasing at the rate of 1/3 unit/sec. How fast is the slope of the graph changing at the instant when x = 2?

I have done this:

Let dx/dt = 1/3 unit/sec

I derive the formula: dy/dt = 4 dx/dt - 3x^2 dx/dt

I substitute to solve for dy/dt.

dy/dt = 4(1/3) - 3(2^2)(1/3) = -8/3

and then I solve for the slope as (dy/dt)/(dx/dt) = (-8/3)/3 which is -8 unit/sec

But the answer is supposed to be -4 units/sec

What am I doing wrong?

Ted

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Hint: How fast the slope is changing is $\frac{dw}{dt}$ where $w=\frac{dy}{dx}$. So you will be differentiating again.
Added: The question does not ask for the slope, it asks for the rate of change of the slope. We have $\frac{dy}{dx}=4-3x^2$, so the slope is $4-3x^2$.
For the rate of change of this, differentiate $4-3x^2$ with respect to $t$. We get that $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=-6x\frac{dx}{dt}.$$ Since $\frac{dx}{dt}=\frac{1}{3}$, when $x=2$ we have $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=(-6)(2)(1/3)=-4.$$
You need to differentiate $\frac{dy}{dx}$ with respect to $t$. This is because you are asked for the rate of change of the slope. – André Nicolas Oct 1 '12 at 3:08
@TedFlethuseo: It isn't what you wrote down. But by the the Chain Rule, it is what you wrote down, multiplied by $\frac{dx}{dt}$ – André Nicolas Oct 9 '12 at 23:51