Chapter 4: Q16MP (page 239)

If $z = \frac{1}{x} f (\frac{y}{x})$ , prove that $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$ .

Short Answer

Expert verified

It is proved that the equation $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$ $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$ .

Step by step solution

Given Information

The given information is $z = \frac{1}{x} f (\frac{y}{x})$ .

Find the partial differentiation ∂z∂xand ∂z∂y .

Calculate the partial differentiation with respect to $x$ as:

$\frac{\partial z}{\partial x} = \frac{1}{x} f' (\frac{y}{x}) (\frac{- y}{x^{2}}) - \frac{1}{x^{2}} f (\frac{y}{x}) \frac{\partial z}{\partial x} = \frac{- y}{x^{3}} f' (\frac{y}{x}) - \frac{1}{x^{2}} f (\frac{y}{x})$

Now calculate the partial differentiation with respect to $y$ as:

$\frac{\partial z}{\partial y} = \frac{1}{x} f' (\frac{y}{x}) (\frac{1}{x}) + f (\frac{y}{x}) (0) \frac{\partial z}{\partial y} = \frac{1}{x^{2}} f' (\frac{y}{x})$

Perform the required multiplication.

Multiply $\frac{\partial z}{\partial x}$ by $x$ as:

$x \frac{\partial z}{\partial x} = x (\frac{- y}{x^{3}} f' (\frac{y}{x}) - \frac{1}{x^{2}} f (\frac{y}{x})) x \frac{\partial z}{\partial x} = \frac{- y}{x^{2}} f' (\frac{y}{x}) - \frac{1}{x} f (\frac{y}{x})$

Now multiply $\frac{\partial z}{\partial y}$ by as:

$y \frac{\partial z}{\partial y} = \frac{y}{x^{2}} f' (\frac{y}{x})$

Now adding $x \frac{\partial z}{\partial x} = \frac{- y}{x^{2}} f' (\frac{y}{x}) - \frac{1}{x} f (\frac{y}{x})$ and $y \frac{\partial z}{\partial y} = \frac{y}{x^{2}} f' (\frac{y}{x})$ as:

$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = \frac{- y}{x^{2}} f' (\frac{y}{x}) - \frac{1}{x} f (\frac{y}{x}) + \frac{y}{x^{2}} f' (\frac{y}{x}) + \frac{1}{x} f (\frac{y}{x}) x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = \frac{- y}{x^{2}} f' (\frac{y}{x}) + \frac{y}{x^{2}} f' (\frac{y}{x}) - \frac{1}{x} f (\frac{y}{x}) + \frac{1}{x} f (\frac{y}{x}) x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$

Hence, proved that the equation . $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$

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If $z = \frac{1}{x} f (\frac{y}{x})$ , prove that $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$ .

Short Answer

Step by step solution

Given Information

Find the partial differentiation ∂z∂xand ∂z∂y .

Perform the required multiplication.

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If z=1xf(yx), prove that x∂z∂x+y∂z∂y+z=0.

Short Answer

Step by step solution

Given Information

Find the partial differentiation ∂z∂xand ∂z∂y .

Perform the required multiplication.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electromagnetism

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Study anywhere. Anytime. Across all devices.

If $z = \frac{1}{x} f (\frac{y}{x})$ , prove that $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} + z = 0$ .