Chapter 10: Q2P (page 534)

From (10.1) find $\frac{\partial θ}{\partial x} = (\frac{1}{r}) c o s θ c o s ϕ$ and show that $\frac{\partial x}{\partial θ} \neq \frac{\partial θ}{\partial z}$ . Note carefully that $\frac{\partial x}{\partial θ}$ means that $r$ and $ϕ$ are constant, but $\frac{\partial x}{\partial θ}$ means that and are constant. (See Chapter 4, Example 7.6 for further discussion.)

Short Answer

Expert verified

Thus,the required value is $\frac{\partial x}{\partial θ} \neq \frac{\partial θ}{\partial x}$

Step by step solution

Step 1:Determine the function.

$\begin{array}{l} x = r \sin θ \cos ϕ \\ y = r \sin θ \sin ϕ \\ z = r \cos θ \end{array}$

Also,

$r^{2} = x^{2} + y^{2} + z^{2}$ and $θ = \cos^{- 1} (\frac{z}{r})$

The objective is to determine $\frac{\partial θ}{\partial x}$ and to show that $\frac{\partial x}{\partial θ} \neq \frac{\partial θ}{\partial x}$ differentiate $x = r \sin θ \cos ϕ$ with respect to $θ$ .

$\frac{\partial x}{\partial θ} = r \cos θ \cos ϕ$ take r and $ϕ$ as constants.

Thus, the value of $\frac{\partial x}{\partial θ}$ is $\frac{\partial x}{\partial θ} = r \cos θ \cos ϕ ... (1)$

Determine the differentiation.

Differentiate $θ = \cos^{- 1} (\frac{z}{r})$ with respect to.

$\begin{matrix} \frac{\partial θ}{\partial x} = \frac{\partial θ}{\partial r} \frac{\partial r}{\partial x} \\ = \frac{z}{r^{2} \sqrt{1 - \frac{z^{2}}{r^{2}}}} \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ = \frac{z}{r \sqrt{r^{2} - z^{2}}} \frac{x}{r} \Rightarrow (r = \sqrt{x^{2} + y^{2} + z^{2}}) \\ = \frac{z x}{r^{2} \sqrt{r^{2} - z^{2}}} \end{matrix}$

Substitute x and z in the equation,

$\begin{matrix} \frac{\partial θ}{\partial x} = \frac{z x}{r^{2} \sqrt{r^{2} - z^{2}}} \\ \frac{\partial θ}{\partial x} = \frac{(r \cos θ) (r \sin θ \cos ϕ)}{r^{2} \sqrt{r^{2} - {(r \cos θ)}^{2}}} \\ \frac{\partial θ}{\partial x} = \frac{r^{2} \cos θ \sin θ \cos ϕ}{r^{2} \sqrt{r^{2} - r^{2} \cos^{2} θ}} \end{matrix}$

So, we get,

$\begin{matrix} \frac{\partial θ}{\partial x} = \frac{\cos θ \sin θ \cos ϕ}{\sqrt{r^{2} (1 - \cos^{2} θ)}} \\ \frac{\partial θ}{\partial x} = \frac{\cos θ \sin θ \cos ϕ}{\sqrt{r^{2} \sin^{2} θ}} \Rightarrow (\cos^{2} θ + \sin^{2} θ = 1) \\ = \frac{\cos θ \sin θ \cos ϕ}{r \sin θ} \\ = \frac{\cos θ \cos ϕ}{r} \end{matrix}$

Thus, the value of $\frac{\partial θ}{\partial x}$ is $\frac{\partial θ}{\partial x} = \frac{1}{r} \cos θ \cos ϕ ... (2)$

From equation (1) and (2) observe that,

$\frac{\partial x}{\partial θ} \neq \frac{\partial θ}{\partial x}$

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Short Answer

Step by step solution

Step 1:Determine the function.

Determine the differentiation.

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From (10.1) find ∂θ∂x=(1r)cosθcosϕand show that∂x∂θ≠∂θ∂z. Note carefully that ∂x∂θmeans thatr and ϕare constant, but ∂x∂θmeans that and are constant. (See Chapter 4, Example 7.6 for further discussion.)

Short Answer

Step by step solution

Step 1:Determine the function.

Determine the differentiation.

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