Question

Verify equations 4.2.

Short answer

The equations $dx=\frac{1}{2c}\left(1-\cos \theta \right)d\theta$and$x=\frac{1}{2c}\left(\theta -\sin \theta \right)+c\text{'}$are verified.

Step-by-step solution

Given Information

The given equations are$cy={\sin }^2\frac{\theta }{2}$ and $dx=\sqrt{\frac{cy}{1-cy}}dy$.

Definition of Calculus

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

Rewrite the differential equation

Differentiate the equation $cy={\sin }^2\frac{\theta }{2}$.

$$\begin{gathered} cy={\sin }^2\frac{\theta }{2} \\ cy=\frac{1}{2}\left(1-\cos \theta \right) \\ dy=\frac{1}{2c}\sin \theta \\ dy=\frac{1}{c}\sin \frac{\theta }{2}\cos \frac{\theta }{2} \end{gathered}$$

Rewrite the differential equation in term of dx.

$$\begin{gathered} \mathrm{dx}=\sqrt{\frac{\mathrm{cy}}{1-\mathrm{cy}}}\mathrm{dy} \\ \mathrm{dx}=\sqrt{\frac{{\sin }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}{1-{\sin }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}}\frac{1}{\mathrm{c}}\sin \frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}\mathrm{d\theta } \\ \mathrm{dx}=\frac{1}{\mathrm{c}}\frac{\sin {\displaystyle \frac{\mathrm{\theta }}{2}}}{\cos {\displaystyle \frac{\mathrm{\theta }}{2}}}\sin \frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}\mathrm{d\theta } \\ \mathrm{dx}=\frac{1}{\mathrm{c}}{\sin }^2\frac{\mathrm{\theta }}{2}\mathrm{d\theta } \end{gathered}$$

Solve further,

$\mathrm{dx}=\frac{1}{2\mathrm{c}}\left(1-\mathrm{cos\theta }\right)\mathrm{d\theta }.....\left(1\right)$

Integrate the dx

Integrate the equation (1) with respect to $\mathrm{\theta }$.

$$\begin{gathered} \mathrm{x}=\int \frac{1}{2\mathrm{c}}\left(1-\mathrm{cos\theta }\right)\mathrm{d\theta } \\ \mathrm{x}=\frac{1}{2\mathrm{c}}\left(\mathrm{\theta }-\mathrm{sin\theta }\right)+\mathrm{c}\text{'} \end{gathered}$$

Therefore, the equations 4.2, $\mathrm{dx}=\frac{1}{2\mathrm{c}}\left(1-\mathrm{cos\theta }\right)\mathrm{d\theta }$ and $\mathrm{x}=\frac{1}{2\mathrm{c}}\left(\mathrm{\theta }-\mathrm{sin\theta }\right)+\mathrm{c}\text{'}$, are verified.