Q 17. - Calculus

Question

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(a) Give parametric equations for $L$

(b) Write an equation for $L$ in symmetric form.

Step-by-Step Solution

Answer

Part (a) $x\left(t\right)=2+7 t, y\left(t\right)=3-5 t, z\left(t\right)=2 t$

Part (b) $\frac{x-2}{7}=\frac{y-3}{-5}=\frac{z}{2}$

Part (a) Step 1: Given information

The line $\mathcal{L}$ determined by the equation $r\left(t\right)=(2+7t,3-5t,2t),-\infty \le t\le \infty$

Part (a) Step 2: Calculation

The goal is to write the equation $\mathcal{L}$ as a set of parametric equations.

In the form of vector parametrization, the equation of line $\mathcal{L}$ is

$r\left(t\right)=(2+7 t, 3-5 t, 2 t)$

The vector function $r\left(t\right)$ in three -dimensional plane represents $r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right)$

Then, $r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right)=(2+7 t, 3-5 t, 2 t)$

Thus, the parametric equations of $\mathcal{L}$ are $x\left(t\right)=2+7 t, y\left(t\right)=3-5 t, z\left(t\right)=2 t$

Therefore, the answer is $x\left(t\right)=2+7 t, y\left(t\right)=3-5 t, z\left(t\right)=2 t$

Part (b) Step 1: Calculation

The goal is to write the symmetric form of the equation $\mathcal{L}$

Remove the parameter $t$ from the parametric equations of the line $\mathcal{L}$ to write the symmetric form.

The parametric equations are $x\left(t\right)=2+7 t, y\left(t\right)=3-5 t, z\left(t\right)=2 t$

Take $x\left(t\right)=2+7 t$

$x=2+7 t$

Add to both sides of the equation$-2$

$$\begin{gathered} x-2=2+7t-2 \\ x-2=\backslash \mathrm{not}2+7t-22 \\ x-2=7t \end{gathered}$$

On both sides of the equation, multiply by $7$

$$\begin{gathered} \frac{x-2}{7}=\frac{7t}{7} \\ \frac{x-2}{7}=t \dots \dots \left(1\right) \end{gathered}$$

Take $y\left(t\right)=3-5 t$

$y=3-5 t$

Add to both sides of the equation$-3$

$$\begin{gathered} y-3=3-5t-3 \\ y-3=\backslash \mathrm{not}\beta -5t-\backslash \mathrm{not}\beta \\ y-3=-5t \end{gathered}$$

Part (b) Step 2: Calculation

Multiply by$-5$ on both sides of the equation.

$$\begin{gathered} \frac{y-3}{-5}=\frac{-5t}{-5} \\ \frac{y-3}{-5}=t\dots ..\left(2\right) \end{gathered}$$

Take $z\left(t\right)=2 t$

$z=2 t$

Multiply by $2$on both sides of the equation.

$$\begin{gathered} \frac{z}{2}=\frac{2t}{2} \\ \frac{z}{2}=t \end{gathered}$$

Equating the equations $\left(1\right),\left(2\right)\left(3\right)$ that are $\frac{x-2}{7}=t,\frac{y-3}{-5}=t,\frac{z}{2}=t$ The following is an example of how they could be written.

Thus,

$\frac{x-2}{7}=\frac{y-3}{-5}=\frac{z}{2}=t$

Thus, the symmetric equations are $\frac{x-2}{7}=\frac{y-3}{-5}=\frac{z}{2}$ Therefore, the required answer is $\frac{x-2}{7}=\frac{y-3}{-5}=\frac{z}{2}$