Chapter 4: Q5MP (page 238)

Given $z = y^{2}$ - $2 x^{2}$ , Find ${(\frac{\partial z}{\partial x})}_{r}, {(\frac{\partial z}{\partial θ})}_{x}, \frac{\partial^{2} z}{\partial x \partial θ}$ .

Short Answer

Expert verified

The value of ${(\frac{\partial z}{\partial x})}_{r} = - 6 x, {(\frac{\partial z}{\partial θ})}_{x} = 2 x^{2} tanθ . \sec^{2} θ \frac{\partial^{2} z}{\partial x \partial θ} = 4 x tanθ . \sec^{2} θ$ .

Step by step solution

Given Information

Given that $z = y^{2} - 2 x^{2}$ and the polar coordinates are $x = r \cos θ, y = r \sin θ$

Formula Used

The value of $r^{2} = x^{2} + y^{2}$ .

Finding the values of (∂z∂x)r,(∂z∂θ)x,∂2z∂x∂θ

Using the formula,

$\frac{\partial z}{\partial x} = - 6 x {(\frac{\partial z}{\partial x})}_{r} = - 6 x$

We know that $\tan θ = \frac{y}{x} \Rightarrow y = x \tan θ$

Then

$z = x^{2} \tan^{2} θ - 2 x^{2} \frac{\partial z}{\partial θ} = 2 x^{2} \tan^{2} θ - 2 x^{2} {(\frac{\partial z}{\partial θ})}_{x} = 2 x^{2} \tan θ . s e c^{2} θ$

Solving further

$\frac{\partial^{2} z}{\partial x \partial θ} = \frac{\partial}{\partial x} (\frac{\partial z}{\partial θ}) \frac{\partial^{2} z}{\partial x \partial θ} = \frac{\partial}{\partial x} (2 x^{2} \tan θ . s e c^{2} θ) \frac{\partial^{2} z}{\partial x \partial θ} = 4 x \tan θ . s e c^{2} θ$

Hence ${(\frac{\partial z}{\partial x})}_{r} = - 6 x, {(\frac{\partial z}{\partial θ})}_{x} = 2 x^{2} tanθ . \sec^{2} θ \frac{\partial^{2} z}{\partial x \partial θ} = 4 x tanθ . \sec^{2} θ$ .

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Given $z = y^{2}$ - $2 x^{2}$ , Find ${(\frac{\partial z}{\partial x})}_{r}, {(\frac{\partial z}{\partial θ})}_{x}, \frac{\partial^{2} z}{\partial x \partial θ}$ .

Short Answer

Step by step solution

Given Information

Formula Used

Finding the values of (∂z∂x)r,(∂z∂θ)x,∂2z∂x∂θ

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Step by step solution

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Formula Used

Finding the values of (∂z∂x)r,(∂z∂θ)x,∂2z∂x∂θ

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Given $z = y^{2}$ - $2 x^{2}$ , Find ${(\frac{\partial z}{\partial x})}_{r}, {(\frac{\partial z}{\partial θ})}_{x}, \frac{\partial^{2} z}{\partial x \partial θ}$ .