Q66P - Mathematical Methods in Physical Sciences

Question

Find  and  as functions of   for the example above, and verify for this case that $\mathit{v}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{a}$ are correctly given by the method of the example.

Step-by-Step Solution

Answer

For this case, that v and a are correctly given by the method of the example, and the value of x and y is given below:

$x=\frac{2t^2-1}{t^2+1},y=\frac{3t}{t^2+1}$

Step 1: Given Information

The function is $z=\frac{i+2t}{t-i}$ .

Step 2: Definition of the Complex number

A complex number can be dictated as:

 $\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}\mathbf{ }$

Where z is the complex number, x and y are real numbers, and i is known as iota, whose value is $\sqrt{\mathbf{-}\mathbf{1}}$ .

 

The modulus of a complex number can be calculated as:

 $\mathbf{\left|}\mathit{z}\mathbf{\right|}\mathbf{=}\mathbf{\surd }\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{)}$

 

Step 3: Find the value of x and y

The complex number is $z=x+iy$ .

$$\begin{gathered} z=x+yi \\ =\frac{i+2t}{t-i}\times \frac{t+i}{t+i} \\ =\frac{2t^2+3ti-1}{t^2+1} \\ =\frac{2t^2-1}{t^2+1}+1\frac{3t}{t^2+1} \end{gathered}$$

The value of x and y is given below:

 $$\begin{gathered} x=\frac{2t^2-1}{t^2+1} \\ y=1\frac{3t}{t^2+1} \end{gathered}$$

Step 4: Find the value of the velocity

The formula for velocity is given below:

 $$\begin{gathered} v\left(t\right)=\left|\frac{dz}{dt}\right| \\ v\left(t\right)=\left|\frac{dx}{dt}+i\frac{ dy}{dt}\right| \end{gathered}$$

 

Find $\frac{dx}{dt}$ as:

 $$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}\left[\frac{2t^2-1}{t^2+1}\right] \\ =\frac{4t(t^2+1)-2t(2t^2-1)}{(t^2+1)} \\ =\frac{6t}{(t^2+1)} \end{gathered}$$

Find $\frac{dy}{dt}$ as:

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\left[\frac{3t}{t^2+1}\right] \\ =\frac{3\left(t^2+1\right)-3t\left(2t\right)}{{(t^2+1)}^2} \\ =\frac{3-3t^2}{{(t^2+1)}^2} \end{gathered}$$ 

 

Substitute the above values in the formula, and the velocity is given below.

$$\begin{gathered} v\left(t\right)=\sqrt{{\left(\frac{6t}{{\left(t^2+1\right)}^2}\right)}^2+{\left(\frac{3-3t^2}{t^2+1}\right)}^2} \\ =\frac{1}{{\left(t^2+1\right)}^2}\times \sqrt{36t^2+\left(9-18t+9t^2\right)} \\ = \frac{1}{{\left(t^2+1\right)}^2}\times \sqrt{{\left(3+3t^2\right)}^2} \\ = \frac{3}{{\left(t^2+1\right)}^2} \end{gathered}$$

Step 5: Find the value of the acceleration

The formula for velocity is given below:

$$\begin{gathered} a\left(t\right)=\left|\frac{d^{2}z}{d{t}^2}\right| \\ =\left|\frac{d^{2}x}{d{t}^2}+i\frac{d^{2}y}{d{t}^2}\right| \end{gathered}$$ 

 

 

Find $\frac{d^{2}x}{d{t}^2}$  as:

 $$\begin{gathered} \frac{d^{2}x}{d{t}^2}=\frac{d}{dt}\left[\frac{6t}{{\left(t^2+1\right)}^2}\right] \\ =\frac{6t{\left(t^2+1\right)}^2-\left(6t\right)\left(4t\right)\left(t^2+1\right)}{{\left(t^2+1\right)}^4} \\ =\frac{6t{\left(t^2+1\right)}^2-\left(6t\right)\left(4t\right)}{{\left(t^2+1\right)}^3} \\ =\frac{6-18t^2}{{\left(t^2+1\right)}^3} \end{gathered}$$

 

 

Find  $\frac{d^{2}y}{d{t}^2}$ as:

  $$\begin{gathered} \frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left[\frac{\left(3-3t^2\right)}{{\left(t^2+1\right)}^2}\right] \\ =\frac{\left(-6t\right){\left(t^2+1\right)}^2-\left(3-3t^2\right)\left(4t\right)\left(t^2+1\right)}{{\left(t^2+1\right)}^4} \\ =\frac{\left(-6t\right){\left(t^2+1\right)}^2-\left(3-3t^2\right)\left(4t\right)}{{\left(t^2+1\right)}^3} \\ =\frac{6t^3-18t}{{\left(t^2+1\right)}^3} \end{gathered}$$

 

Substitute the above values in the formula, and the acceleration is given below:

 $$\begin{gathered} a=\sqrt{{\left(\frac{6t^3-18t^2}{{\left(t^2+1\right)}^3}\right)}^2+{\left(\frac{6t^3-18t}{{\left(t^2+1\right)}^3}\right)}^2} \\ =\frac{1}{{\left(t^2+1\right)}^3}\sqrt{{\left(6t^3-18t^2\right)}^2+{\left(6t^3-18t\right)}^2} \\ =\frac{6}{{\left(t^2+1\right)}^3}\sqrt{{\left(6t^3-18t^2\right)}^2} \\ =\frac{6}{{\left(t^2+1\right)}^{3/2}} \end{gathered}$$

Thus, for this case, that v and a are correctly given by the method of the example, and the value of x and y is given below:

 $x=\frac{2t^2-1}{t^2+1},y=\frac{3t}{t^2+1}$