Q. 31 - Calculus

Question

sketch the parametric curve by eliminating the parameter 

$x=\cosh t,y=\sinh t,t\in \mathrm{ℝ}$

Step-by-Step Solution

Answer

The equation after elimination of the parameter is$x^2-y^2=1\text{ or }{y}^2=x^2-1$

Step 1: Given Information

The parametric equations are$x=\cosh t,y=\sinh t,t\in \mathrm{ℝ}$

Step 2: Calculation

Consider the parametric equations $x=\cosh t,y=\sinh t,t\in \mathrm{ℝ}$.

The objective is to sketch the parametric curve by eliminating the parameter.

Take the equation $x=\cosh t$.

Square the equation on both sides.

$x^2={\cosh }^{2}t$

Take the equation $y=\sinh t$.

Square the equation on both sides. then

$y^2={\sinh }^{2}t$

Now subtract the equations $y^2={\sinh }^{2}t$ from$x^2={\cosh }^{2}t$.

Thus.

$$\begin{gathered} x^2-y^2={\cosh }^{2}t-{\sinh }^{2}t \\ {x}^2-y^2=1 \left[Since {\cosh }^{2}t-{\sinh }^{2}t=1\right] \\ {x}^2-1=y^2 \\ {y}^2=x^2-1 \end{gathered}$$

In order to draw the graph of the equation assume $x=1,2,3,4$.

Substitute $x=1$in the equation$y^2=x^2-1$.

Then,

$$\begin{gathered} y^2=1^2-1 \\ y=0 \\ (x,y)=(1,0) \end{gathered}$$

Step 3: Further simplification

Substitute $x=2$in the equation $y^2=x^2-1$

Then,

$$\begin{gathered} y^2=(2{)}^2-1 \\ {y}^2=4-1 \\ {y}^2=3 \\ y=\pm \sqrt{3} \\ (x,y)=(2,-\sqrt{3}\left)\right(2,\sqrt{3}) \end{gathered}$$

Substitute $x=3$in the equation $y^2=x^2-1$

Then,

$$\begin{gathered} y^2=(3{)}^2-1 \\ {y}^2=9-1 \\ y=\sqrt{8}=\pm 2\sqrt{2} \\ \left(x_{,}y\right)=(3,-2\sqrt{2}\left)\right(3.2\sqrt{2}) \end{gathered}$$

Substitute $x=4$ in the equation$y^2=x^2-1$.

Then

$$\begin{gathered} y^2=(4{)}^2-1 \\ {y}^2=16-1 \\ y=\pm \sqrt{15} \left[\text{ since }{y}^2=15\Rightarrow y=\pm \sqrt{15}\right] \\ \left(x_{,}y\right)=(4,-\sqrt{15}\left)\right(4,\sqrt{15}) \end{gathered}$$

The graphical representation by using the points $(1,0)(2,-\sqrt{3})(2,\sqrt{3})(3,-2\sqrt{2})(3,2\sqrt{2})(4,-\sqrt{15})(4,\sqrt{15})$is as follows,

Therefore, the equation after the elimination of the parameter is $x^2-y^2=1\text{ or }{y}^2=x^2-1$