- #1

Math100

- 756

- 205

- Homework Statement
- If ## F(x, y')=\sqrt{x^2+y'^2} ##, find ## \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial y'}, \frac{dF}{dx} ## and ## \frac{d}{dx}(\frac{\partial F}{\partial y'}) ##. Also show that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial}{\partial y'}(\frac{dF}{dx}) ##.

- Relevant Equations
- None.

Note that ## \frac{\partial F}{\partial x}=\frac{2x}{2\sqrt{x^2+y'^2}}=\frac{x}{\sqrt{x^2+y'^2}}, \frac{\partial F}{\partial y}=0, \frac{\partial F}{\partial y'}=\frac{2y'}{2\sqrt{x^2+y'^2}}=\frac{y'}{\sqrt{x^2+y'^2}} ##.

Now we have ## \frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}y'+\frac{\partial F}{\partial y'}y"=\frac{x+y'y"}{\sqrt{x^2+y'^2}} ##.

Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{d}{dx}(\frac{y'}{\sqrt{x^2+y'^2}})=\frac{\sqrt{x^2+y'^2}\cdot \frac{d}{dx}(y')-y'\cdot \frac{d}{dx}(\sqrt{x^2+y'^2})}{x^2+y'^2}=\frac{y"\cdot \sqrt{x^2+y'^2}-y'(\frac{x+y'y"}{\sqrt{x^2+y'^2}})}{\sqrt{x^2+y'^2}}=\frac{y"(x^2+y'^2)-y'(x+y'y")}{(x^2+y'^2)^{\frac{3}{2}}} ##.

Also ## \frac{\partial}{\partial y'}(\frac{dF}{dx})=\frac{\partial}{\partial y'}(\frac{x+y'y"}{\sqrt{x^2+y'^2}})=\frac{\sqrt{x^2+y'^2}\cdot \frac{\partial}{\partial y'}(x+y'y")-(x+y'y")\cdot \frac{\partial}{\partial y'}(\sqrt{x^2+y'^2})}{x^2+y'^2}=\frac{\sqrt{x^2+y'^2}\cdot y"-(x+y'y")\cdot (\frac{y'}{\sqrt{x^2+y'^2}})}{x^2+y'^2}=\frac{\sqrt{x^2+y'^2}(\sqrt{x^2+y'^2}\cdot y")-y'(x+y'y")}{\sqrt{x^2+y'^2}}\cdot \frac{1}{x^2+y'^2}=\frac{y"(x^2+y'^2)-y'(x+y'y")}{(x^2+y'^2)^{\frac{3}{2}}} ##.

Therefore, ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial}{\partial y'}(\frac{dF}{dx}) ##.

Now we have ## \frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}y'+\frac{\partial F}{\partial y'}y"=\frac{x+y'y"}{\sqrt{x^2+y'^2}} ##.

Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{d}{dx}(\frac{y'}{\sqrt{x^2+y'^2}})=\frac{\sqrt{x^2+y'^2}\cdot \frac{d}{dx}(y')-y'\cdot \frac{d}{dx}(\sqrt{x^2+y'^2})}{x^2+y'^2}=\frac{y"\cdot \sqrt{x^2+y'^2}-y'(\frac{x+y'y"}{\sqrt{x^2+y'^2}})}{\sqrt{x^2+y'^2}}=\frac{y"(x^2+y'^2)-y'(x+y'y")}{(x^2+y'^2)^{\frac{3}{2}}} ##.

Also ## \frac{\partial}{\partial y'}(\frac{dF}{dx})=\frac{\partial}{\partial y'}(\frac{x+y'y"}{\sqrt{x^2+y'^2}})=\frac{\sqrt{x^2+y'^2}\cdot \frac{\partial}{\partial y'}(x+y'y")-(x+y'y")\cdot \frac{\partial}{\partial y'}(\sqrt{x^2+y'^2})}{x^2+y'^2}=\frac{\sqrt{x^2+y'^2}\cdot y"-(x+y'y")\cdot (\frac{y'}{\sqrt{x^2+y'^2}})}{x^2+y'^2}=\frac{\sqrt{x^2+y'^2}(\sqrt{x^2+y'^2}\cdot y")-y'(x+y'y")}{\sqrt{x^2+y'^2}}\cdot \frac{1}{x^2+y'^2}=\frac{y"(x^2+y'^2)-y'(x+y'y")}{(x^2+y'^2)^{\frac{3}{2}}} ##.

Therefore, ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial}{\partial y'}(\frac{dF}{dx}) ##.