Problem: The volume of a cone is increasing at a rate of 6 cubic units per second. When the volume of the cube is 81 cubic units, at what rate is the longest diagonal of the cube is increasing?
Here's what I tried: Let $D$ be the length of the longest diagonal and $s$ be the side length of the cube. Thus, we have $D=\sqrt3s.$ So, we can write $V=\big(\frac{D}{\sqrt3}\big)^3 = 3^{-3/2}D^3,$ and taking the derivative w.r.t. time of both sides gives $\frac{dV}{dt}=3^{-3/2}\cdot3D^2\cdot \frac{dD}{dt}.$ I solved for $D$ given that the volume is $81,$ and eventually I got $\frac{dD}{dt}=2\cdot3^{-11/6}.$
However, I did it another way as well and got a different answer: $V=s^3,$ so $\frac{dV}{dt}=3s^2\cdot\frac{ds}{dt}.$ Then solve for $ds/dt,$ and we know $\frac{dD}{dt}=\sqrt3 \frac{ds}{dt}.$ This gave me $\frac{dD}{dt}=2\cdot3^{-13/6}.$
Why did I get different answers? (I checked my work for each method several times, so I think it's something with the method. But I still could've made a calculation mistake.)
I get 13 not 11 for the first method. $D=3^{11/6}$ so $D^2/\sqrt3=3^{19/6}$. One factor of 3 cancels with the one in 6, leaving $3^{-13/6}$ on the left hand side.