Question

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r cos θ, r sin θ) in other words g is f in terms of polar coordinates. Suppose that it is known that fx(1, 1) = −2 and fy(1, 1) = 3 Use the chain rule in order to compute the partial derivative gr( √ 2, π/4).

Answer 1

`g_(r)(√(2),(\pi)/(4))=(√(2))/(2)\\`

Explanation:

First, notice that:

`g(√(2),(\pi)/(4))=f(√(2)cos((\pi)/(4)),√(2)sin((\pi)/(4)))\\`

`g(√(2),(\pi)/(4))=f(√(2)((1)/(√(2))),√(2)((1)/(√(2))))\\`

`g(√(2),(\pi)/(4))=f(1,1)\\`

We proceed to use the chain rule to find
`g_(r)(√(2),(\pi)/(4))` using the fact that
`X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)` to find their derivatives:

`g_(r)(r,\theta)=f_(r)(rcos(\theta),rsin(\theta))=f_(x)( rcos(\theta),rsin(\theta))(\delta x)/(\delta r)(r,\theta)+f_(y)(rcos(\theta),rsin(\theta))(\delta y)/(\delta r)(r,\theta)\\`

Because we know
`X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)` then:

`(\delta x)/(\delta r)=cos(\theta)\ and\ (\delta y)/(\delta r)=sin(\theta)`

We substitute in what we had:

`g_(r)(r,\theta)=f_(x)( rcos(\theta),rsin(\theta))cos(\theta)+f_(y)(rcos(\theta),rsin(\theta))sin(\theta)`

Now we put in the values
`r=√(2)\ and\ \theta=(\pi)/(4)` in the formula:

`g_(r)(√(2),(\pi)/(4))=f_(r)(1,1)=f_(x)(1,1)cos((\pi)/(4))+f_(y)(1,1)sin((\pi)/(4))`

Because of what we supposed:

`g_(r)(√(2),(\pi)/(4))=f_(r)(1,1)=-2cos((\pi)/(4))+3sin((\pi)/(4))`

And we operate to discover that:

`g_(r)(√(2),(\pi)/(4))=-2(√(2))/(2)+3(√(2))/(2)`

`g_(r)(√(2),(\pi)/(4))=(√(2))/(2)`

and this will be our answer