Chapter 4: Q6MP (page 238)

If $z = r^{2} - x^{2}$ , find ${(\frac{\partial z}{\partial r})}_{θ}, {(\frac{\partial z}{\partial θ})}_{r}, \frac{\partial^{2} z}{\partial r \partial θ}, {(\frac{\partial z}{\partial x})}_{y}$

Short Answer

Expert verified

The value of ${(\frac{\partial z}{\partial r})}_{θ} = 2 r \sin^{2} θ, {(\frac{\partial z}{\partial θ})}_{r} = r^{2} \sin 2 θ, \frac{\partial^{2} z}{\partial r . \partial θ} = 2 r \sin 2 θ and {(\frac{\partial z}{\partial x})}_{y} = 0$

{(\frac{\partial z}{\partial r})}_{θ} = 2 r \sin^{2} θ, {(\frac{\partial z}{\partial θ})}_{r} = r^{2} \sin 2 θ, \frac{\partial^{2} z}{\partial r . \partial θ} = 2 r \sin 2 θ and {(\frac{\partial z}{\partial x})}_{y} = 0

{(\frac{\partial z}{\partial r})}_{θ} = 2 r \sin^{2} θ, {(\frac{\partial z}{\partial θ})}_{r} = r^{2} \sin 2 θ, \frac{\partial^{2} z}{\partial r . \partial θ} = 2 r \sin 2 θ and {(\frac{\partial z}{\partial x})}_{y} = 0

Step by step solution

Given Information

Given that, $z = r^{2} - x^{2}$ .

Formula Used

The value of $r^{2} = x^{2} + y^{2}$ ...(1)

x = r c o s θ y = r s i n θ

Solving the given equation

The given equation

$z = r^{2} - x^{2}$ ...(3)

From equation 1 and 2,

$z = x^{2} + y^{2} - x^{2} z = y^{2}$

Substitute the value of y in the above equation.

$z = {(r \sin θ)}^{2}$

Differentiate with respect to r,

$\frac{\partial z}{\partial r} = 2 r \sin^{2} θ {(\frac{\partial z}{\partial r})}_{θ} = 2 r \sin^{2} θ$

We know that $z = r^{2} \sin^{2} θ$

Differentiate with respect to $θ$

$\frac{\partial z}{\partial θ} = 2 r^{2} \sin θ . \cos θ \frac{\partial z}{\partial θ} = r^{2} \sin 2 θ {(\frac{\partial z}{\partial θ})}_{r} = r^{2} \sin 2 θ$

Solving further

$\frac{\partial^{2} z}{\partial r . \partial θ} = \frac{\partial}{\partial r} (\frac{\partial z}{\partial θ}) \frac{\partial^{2} z}{\partial r . \partial θ} = \frac{\partial}{\partial r} (r^{2} \sin 2 θ) \frac{\partial^{2} z}{\partial r . \partial θ} = 2 r \sin 2 θ$

We know that $z = y^{2}$

$\frac{\partial z}{\partial x} = 0 {(\frac{\partial z}{\partial x})}_{y} = 0$

Hence ${(\frac{\partial z}{\partial r})}_{θ} = 2 r \sin^{2} θ, {(\frac{\partial z}{\partial θ})}_{r} = r^{2} \sin 2 θ, \frac{\partial^{2} z}{\partial r . \partial θ} = 2 r \sin 2 θ and {(\frac{\partial z}{\partial x})}_{y} = 0$ .

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If $z = r^{2} - x^{2}$ , find ${(\frac{\partial z}{\partial r})}_{θ}, {(\frac{\partial z}{\partial θ})}_{r}, \frac{\partial^{2} z}{\partial r \partial θ}, {(\frac{\partial z}{\partial x})}_{y}$

Short Answer

Step by step solution

Given Information

Formula Used

Solving the given equation

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If z=r2-x2, find(∂z∂r)θ,(∂z∂θ)r,∂2z∂r∂θ,(∂z∂x)y

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If $z = r^{2} - x^{2}$ , find ${(\frac{\partial z}{\partial r})}_{θ}, {(\frac{\partial z}{\partial θ})}_{r}, \frac{\partial^{2} z}{\partial r \partial θ}, {(\frac{\partial z}{\partial x})}_{y}$