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Problem 105

Show that if a function \(\mathrm{F}: \mathrm{R}^{2} \rightarrow \mathrm{R}\) satisfies $\partial \mathrm{F} / \partial \mathrm{x}=\partial \mathrm{F} / \partial \mathrm{y}$, then \(F(x, y)=f(x+y)\) where \(f\) is some differentiable function of one variable, \(\mathrm{s}=\mathrm{x}+\mathrm{y}\)

Expert verified

To show that a function \( F: R^2 \rightarrow R \) satisfying \( \frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} \) is of the form \( F(x, y) = f(x+y) \), we first introduce a new variable \( s = x + y \). Then, we calculate the partial derivative of \( F \) with respect to \( s \), and using the chain rule, we obtain \( \frac{\partial F}{\partial s} = \frac{\partial F}{\partial x}(1+1) = 2\frac{\partial F}{\partial x} \). Integrating the partial derivative with respect to \( s \) yields \( F(s) = \int 2\frac{\partial F}{\partial x} (dx + dy) \). By separating the integral and integrating both parts, we get \( F(s) = f(x) + g(y) \), where \( f(x) \) and \( g(y) \) are the antiderivatives of each integral. Given that \( s = x + y \), we can express \( F \) as a function of \( s \), \( F(x, y) = f(s) - g(y) \), which simplifies to \( F(x, y) = f(x+y) \). This confirms our claim.

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Chapter 4

1 1 ] dill [ 2 Let \(\mathrm{u}(\mathrm{x}, \mathrm{y})\) and $\mathrm{v}(\mathrm{x}, \mathrm{y})\( be defined as functions of \)\mathrm{x}\( and \)\mathrm{y}$ by the equations \(u \cos v-x=0\) (1) \(u \sin v-y=0\) (2) Find \((\partial \mathrm{u} / \partial \mathrm{x}) \quad\) and \(\quad(\partial \mathrm{v} / \partial \mathrm{x})\)

Chapter 4

Use differentials to compute a) \((\partial z / \partial \mathrm{x}),(\partial z / \partial y)\) where \(z=\left[\left(x^{2}-1\right) / y\right]\) b) $(\partial r / \partial x),(\partial r / \partial y),(\partial x / \partial r)\( where \)r=\sqrt{\left(x^{2}+y^{2}\right)}$ c) $(\partial \mathrm{z} / \partial \mathrm{x}),(\partial \mathrm{z} / \partial \mathrm{y})\( where \)\mathrm{z}=\arctan (\mathrm{y} / \mathrm{x})$

Chapter 4

Show that the line normal to the surface given by $\mathrm{F}(\mathrm{x}, \mathrm{y}, \mathrm{z})=0\( at a point \)\left(\mathrm{x}_{0}, \mathrm{y}_{0}, \mathrm{z}_{0}\right)$ has direction ratios $(\partial \mathrm{F} / \partial \mathrm{x})|[(\mathrm{x}) 0,(\mathrm{y}) 0,(z) 0]:(\partial \mathrm{F} / \partial \mathrm{y})|[(\mathrm{x}) 0,(\mathrm{y}) 0,(\mathrm{z}) 0]:(\partial \mathrm{F} / \partial \mathrm{z}) \mid[(\mathrm{x}) 0,(\mathrm{y}) 0,(\mathrm{z}) 0]$ What are these ratios when \(z=f(x, y)\) is a solution to \(\mathrm{F}(\mathrm{x}, \mathrm{y}, \mathrm{z})=0 ?\)

Chapter 4

Let \(u=r^{3}\) be a scalar field where \(r\) is the distance \(O P\) from the origin 0 to a variable point \(\mathrm{P}\), in \(\mathrm{R}^{3}\). Find the gradient of \(u\) at \(P\) without resorting to rectangular coordinates.

Chapter 4

Find the line orthogonal to the graph of $\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{xy}\( at the point \)\left(\mathrm{x}_{0}, \mathrm{y}_{0}, \mathrm{z}_{0}\right)=(-2,3,-6)$.

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