# Science:Math Exam Resources/Courses/MATH200/December 2012/Question 02/Solution 1

We apply the Chain rule to calculate

{\displaystyle \displaystyle {\begin{aligned}{\frac {\partial G}{\partial t}}&={\frac {\partial F}{\partial z}}{\frac {\partial (At)}{\partial t}}={\frac {\partial F}{\partial z}}A\\{\frac {\partial G}{\partial \gamma }}&={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}={\frac {\partial F}{\partial x}}+{\frac {\partial F}{\partial y}}\\{\frac {\partial G}{\partial s}}&={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial s}}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial s}}={\frac {\partial F}{\partial x}}-{\frac {\partial F}{\partial y}}\end{aligned}}}

For the second derivatives, we apply the Chain rule again to the already calculated derivatives

{\displaystyle \displaystyle {\begin{aligned}{\frac {\partial ^{2}G}{\partial \gamma ^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial \gamma }}+{\frac {\partial }{\partial x}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y\partial x}}+{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}\\{\frac {\partial ^{2}G}{\partial s^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial s}}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial s}}-{\frac {\partial }{\partial x}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma +s)}{\partial s}}-{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial s}}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\partial ^{2}F}{\partial x^{2}}}-{\frac {\partial ^{2}F}{\partial y\partial x}}-{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}\end{aligned}}}

Now we set together

${\displaystyle \displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y\partial x}}+{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}F}{\partial x^{2}}}-{\frac {\partial ^{2}F}{\partial y\partial x}}-{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}=2{\frac {\partial ^{2}F}{\partial x^{2}}}+2{\frac {\partial ^{2}F}{\partial y^{2}}}=2{\frac {\partial F}{\partial z}}}$

where the last step is already given by the problem.

With ${\displaystyle \displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}=2{\frac {\partial F}{\partial z}}}$ and ${\displaystyle \displaystyle {\frac {\partial G}{\partial t}}=A{\frac {\partial F}{\partial z}}}$

we conclude that ${\displaystyle \displaystyle A=2}$ is the value that ${\displaystyle \displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}={\frac {\partial G}{\partial t}}}$