## Related questions with answers

The flow lines (or streamlines) of a vector field are the paths followed by a particle whose velocity field is the given vector field. Thus the vectors in a vector field are tangent to the flow lines. If parametric equations of a flow line are x = x(t), y = y(t) explain why these functions satisfy the differential equations dx/dt=x and dy/dt=-y. Then solve the differential equations to find an equation of the flow line that passes through the point (1, 1).

Solutions

Verified(b) Let $x=x(t)$ and $y=y(t)$ the parametric equations of a flow line. By definition of the flow lines, the velocity vector of the flow line at the point $(x, y)$ is is equal to the given vector field.

As the velocity vector is given by $\displaystyle \mathbf{v}=\langle d x(t)/ d t, d y(t) / d t \rangle$, then we have $\mathbf{v}=\mathbf{F}(x,y)$ or

$\begin{equation} \langle d x(t)/ d t, d y(t) / d t \rangle=\mathbf{F}(x,y)=\langle x, -y\rangle \end{equation}$

Therefore, the stream lines satisfy the differential equations

$\begin{equation} d x / d t=x, \qquad d y / d t=-y \end{equation}$

and

$\begin{equation} d y / d t=-y \end{equation}$

To solve (2) we write it as

$d x / x=d t$

Hence, integrating we obtain

$\ln |x|=t+A$

or

$\begin{equation} x=\pm e^{t+A}=e^{A} e^{t}=C_{1}e^t \end{equation}$

where $A$ and $C_{1}$ are constants.

We can solve (3) in a similar way to obtain

$\begin{equation} y=C_{2} e^{-t} \end{equation}$

where $C_{2}$ is a constant.

Taking the product between (4) and (5) we have

$x y=C_{1} e^{t} C_{2} e^{-t}=C_{1}C_{2}=K$

with $K$ a constant.

Therefore, the equations for the flow lines are

$\begin{equation} y=\frac{K}{x} \end{equation}$

If the flow line passes through (1,1) we obtain from (6)

$(1)=\frac{K}{1}$

which gives $K=1$. Then, the flow line through this point has equation

$y=\frac{1}{x},\quad x>0$

At any point $\dfrac{dx}{dt}$ is the $x$ component of the velocity.

At any point $\dfrac{dy}{dt}$ is the $y$ component of the velocity.

Here it is given that Velocity at the point $(x, y$) is $\textbf{F}(x, y) = x\textbf{ i }-y\textbf{ j }$

Note that, the $x$ component of the velocity is $x$ and

the $y$-component of the velocity is $-y$

Therefore, we can write

$\dfrac{dx}{dt} = x \rightarrow \textbf{(1)}$

$\dfrac{dy}{dt} = -y \rightarrow \textbf{(2)}$

Let $x=x(t)$ and $y=y(t)$ be the parametric equations of a flow line. The definition of the flow lines gives us that the velocity vector of the flow line at $(x,y)$ is equal to the given vector field.

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Thomas' Calculus

14th Edition•ISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir#### Calculus: Early Transcendentals

8th Edition•ISBN: 9781285741550 (6 more)James Stewart#### Calculus: Early Transcendentals

9th Edition•ISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson#### Calculus

9th Edition•ISBN: 9781337624183Daniel K. Clegg, James Stewart, Saleem Watson## More related questions

1/4

1/7