A rectangular metal block above has length y com and a square cross section of side x cm. When the metal block is heated, the area of cross-section A and the length of the metal block increase at a constant rate of $0.3 cm/s$ and $0.2 cm/s$ respectively. Find the rate in which the volume of the metal block, V is increasing when $A= 4cm^2$ and $y=5 cm$
Here’s my workings -
$\frac{dV}{dt} = \frac{\partial V}{\partial A} \cdot \frac{dA}{dt} + \frac{\partial V}{\partial L} \cdot \frac{dL}{dt} $
$ V = y \cdot x^2 $
$A = x^2$
$A= x^2 = 4$
$x=2$
$ \frac{\partial V}{\partial A} = 2xy $
$\frac{\partial V}{\partial L} = x^2 $
Therefore
$\frac{dV}{dt} = 2(2)(5) \cdot (0.3) + (2)^2 \cdot (0.2) $
However, my answer is wrong and my mistake was in $ 2(2)(5) \cdot (0.3) $
Why is this so ? Thanks..
$∂V/∂A=2xy$ should be $∂V/∂A=2y$, as ∂A in this case = 2x, saying $∂V/∂A=2xy$ double-counts for ∂A