# Application of chain rule for partial derivatives

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..

• Does the side length A increase at .3cm/s or the area of square cross section by .3 square cm/s? – user583156 Aug 12 '18 at 17:23
• @ArthurDent area of the square cross section – user185692 Aug 12 '18 at 17:30
• ∂V/∂A=2xy is where the error is, think about your variable A and its relationship to x – user583156 Aug 12 '18 at 17:38
• Is the $x$ parameter of the area, $A=x^2$ increasing at a rate of $0.3 \frac{cm}{s}$ or it is the area itself, $A$, is increasing at a rate of $0.3 \frac{cm^2}{s}$? Note the units used. – Winter Soldier Aug 12 '18 at 17:53

$∂V/∂A=2xy$ should be $∂V/∂A=2y$, as ∂A in this case = 2x, saying $∂V/∂A=2xy$ double-counts for ∂A