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

From UBC Wiki

We apply the Chain rule to calculate

∂G∂t=∂F∂z∂(At)∂t=∂F∂zA∂G∂γ=∂F∂x∂(γ+s)∂γ+∂F∂y∂(γ−s)∂γ=∂F∂x+∂F∂y∂G∂s=∂F∂x∂(γ+s)∂s+∂F∂y∂(γ−s)∂s=∂F∂x−∂F∂y

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

∂2G∂γ2=∂∂x∂F∂x∂(γ+s)∂γ+∂∂y∂F∂x∂(γ−s)∂γ+∂∂x∂F∂y∂(γ+s)∂γ+∂∂y∂F∂y∂(γ−s)∂γ=∂2F∂x2+∂2F∂y∂x+∂2F∂x∂y+∂F2∂y2∂2G∂s2=∂∂x∂F∂x∂(γ+s)∂s+∂∂y∂F∂x∂(γ−s)∂s−∂∂x∂F∂y∂(γ+s)∂s−∂∂y∂F∂x∂(γ−s)∂s=∂2F∂x2−∂2F∂y∂x−∂2F∂x∂y+∂F2∂y2

Now we set together

∂2G∂γ2+∂G2∂s2=∂2F∂x2+∂2F∂y∂x+∂2F∂x∂y+∂F2∂y2+∂2F∂x2−∂2F∂y∂x−∂2F∂x∂y+∂F2∂y2=2∂2F∂x2+2∂2F∂y2=2∂F∂z

where the last step is already given by the problem.

With ∂2G∂γ2+∂G2∂s2=2∂F∂z and ∂G∂t=A∂F∂z


we conclude that A=2 is the value that ∂2G∂γ2+∂G2∂s2=∂G∂t