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

MATH200 December 2012

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Question 02

Assume that the function $\displaystyle F(x,y,z)$ satisfies the equation $\displaystyle {\frac {\partial F}{\partial z}}={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y^{2}}}$ and the mixed partial derivatives $\displaystyle {\frac {\partial ^{2}F}{\partial x\partial y}}$ and $\displaystyle {\frac {\partial ^{2}F}{\partial y\partial x}}$ are equal. Let $\displaystyle A$ be some constant and let $\displaystyle G(\gamma ,s,t)=F(\gamma +s,\gamma -s,At)$. Find the value of $\displaystyle A$ such that $\displaystyle {\frac {\partial G}{\partial t}}={\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial ^{2}G}{\partial s^{2}}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Remember the Chain Rule for function depending on several variables:$\displaystyle {\frac {d}{dt}}F(g(\gamma ,s,t),h(\gamma ,s,t),j(\gamma ,s,t))={\frac {\partial F}{\partial x}}{\frac {\partial g}{\partial t}}+{\frac {\partial F}{\partial y}}{\frac {\partial h}{\partial t}}+{\frac {\partial F}{\partial z}}{\frac {\partial j}{\partial t}}$

Hint 2

Use the Chain Rule to calculate $\displaystyle \ \ {\frac {\partial G}{\partial t}}={\frac {\partial F}{\partial z}}{\frac {\partial (At)}{\partial t}}$, then $\displaystyle {\frac {\partial G}{\partial \gamma }}={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}$ further analogously $\displaystyle {\frac {\partial G}{\partial s}}$ and then the second derivatives applying the chain rule twice.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We apply the Chain rule to calculate $\displaystyle {\begin{aligned}{\frac {\partial G}{\partial t}}&={\frac {\partial F}{\partial z}}{\frac {\partial (At)}{\partial t}}={\frac {\partial F}{\partial z}}A\\{\frac {\partial G}{\partial \gamma }}&={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}={\frac {\partial F}{\partial x}}+{\frac {\partial F}{\partial y}}\\{\frac {\partial G}{\partial s}}&={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial s}}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial s}}={\frac {\partial F}{\partial x}}-{\frac {\partial F}{\partial y}}\end{aligned}}$ For the second derivatives, we apply the Chain rule again to the already calculated derivatives $\displaystyle {\begin{aligned}{\frac {\partial ^{2}G}{\partial \gamma ^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial \gamma }}+{\frac {\partial }{\partial x}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y\partial x}}+{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}\\{\frac {\partial ^{2}G}{\partial s^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial s}}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial s}}-{\frac {\partial }{\partial x}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma +s)}{\partial s}}-{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial s}}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\partial ^{2}F}{\partial x^{2}}}-{\frac {\partial ^{2}F}{\partial y\partial x}}-{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}\end{aligned}}$ Now we set together $\displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y\partial x}}+{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}F}{\partial x^{2}}}-{\frac {\partial ^{2}F}{\partial y\partial x}}-{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}=2{\frac {\partial ^{2}F}{\partial x^{2}}}+2{\frac {\partial ^{2}F}{\partial y^{2}}}=2{\frac {\partial F}{\partial z}}$ where the last step is already given by the problem. With $\displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}=2{\frac {\partial F}{\partial z}}$ and $\displaystyle {\frac {\partial G}{\partial t}}=A{\frac {\partial F}{\partial z}}$ we conclude that $\displaystyle A=2$ is the value that $\displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}={\frac {\partial G}{\partial t}}$

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