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tatyana61 [14]
3 years ago
14

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 co

s 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).
Mathematics
1 answer:
Olenka [21]3 years ago
3 0

Answer:

g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\

Step-by-step explanation:

First, notice that:

g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\

g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\

g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\

We proceed to use the chain rule to find g_{r}(\sqrt{2},\frac{\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))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\

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

\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\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=\sqrt{2}\ and\ \theta=\frac{\pi}{4} in the formula:

g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})

Because of what we supposed:

g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})

And we operate to discover that:

g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}

g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}

and this will be our answer

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