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 $\frac {dx}{dt} = \frac {1}{3}$ unit/sec
I derive the formula: $\frac {dy}{dt} = 4 \frac {dx}{dt} - 3x^2 \frac {dx}{dt}$
I substitute to solve for $dy \over dt$.
$\frac {dy}{dt} = 4(\frac {1}{3}) - 3(2^2)(\frac {1}{3}) = \frac {-8}{3}$
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