 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/19/a/solution

Apply the multivariable chain rule to an arbitrary differentiable scalar function of

x, y, z

. With the subscripts indicating the coordinates held fixed, the first operator is

(\partial_{W})_{θ, ϕ} = tan θ cos ϕ (\partial_{x})_{y, z} + tan θ sin ϕ (\partial_{y})_{x, z} + (\partial_{z})_{x, y} .

Holding

W, ϕ

fixed when differentiating with respect to

θ

gives

(\partial_{θ})_{W, ϕ} = W sec^{2} θ cos ϕ (\partial_{x})_{y, z} + W sec^{2} θ sin ϕ (\partial_{y})_{x, z} .

Finally, holding

W, θ

fixed gives

(\partial_{ϕ})_{W, θ} = - W tan θ sin ϕ (\partial_{x})_{y, z} + W tan θ cos ϕ (\partial_{y})_{x, z} .

For example, the coefficient of

\partial_{x}

\partial_{θ}

is precisely

(\partial x / \partial θ)_{W, ϕ}

; all coefficients follow by differentiating the given coordinate functions.

 New to topics? Read the docs here!