Question

(a) Sketch the vector field \(F(x,y) = xi - yj\), several approximated flow lines and flow line equations.

(b) Solve the differential equations to find the flow line that passes through the point \((1,1)\).

Short answer

(a) The vector field \(F(x,y) = xi - yj\) is sketched, several approximated flow lines on vector filed \(F(x,y) = xi - yj\) is drawn and equation of the flow line is \(y = \frac{C}{x}\).

(b) The flow line that passes through the point \((1,1)\) is \(y = \frac{1}{x},x > 0\).

Step-by-step solution

Given data

The functions are \(F(x,y) = xi - yj\) and \(F(x,y) = xi - yj,x = x(t),y = y(t),{\rm{ point }}(1,1).\)

Concept used

Consider a two-dimensional vector is \(F = \langle x,y\rangle \).

The expression for length of the two dimensional vector as follows:

\(|F(x,y)| = \sqrt {{x^2} + {y^2}} \) ……. (1)

Consider the velocity field\(F(x,y)\)as follows:

\(\begin{aligned}F(x,y) &= xi - yj\\ &= \langle x, - y\rangle \end{aligned}\)

The required integration formula with respect to\(x\)as follows:\(\int {\frac{d}{{dx}}} \left( {\frac{1}{x}} \right) = \ln (x)\)

The required integration formula with respect to\(y\)as follows:\(\int {\frac{d}{{dy}}} \left( {\frac{1}{y}} \right) = \ln (y)\)

The required integration formula with respect to \(t\) as follows: \(\int d t = t\)

Find the length of \(F(x,y)\)

(a) Find the length of \(F(x,y)\) use equation (1) as follows:

\(\begin{aligned}{l}|F(x,y)| & = \sqrt {{{(x)}^2} + {{( - y)}^2}} \\|F(x,y)| & = \sqrt {{x^2} + {y^2}} \end{aligned}\)

Consider a certain interval of \(x\) as \(( - 2,2)\) and \(y\) as \(( - 2,2)\) to plot \(F(x,y)\).

The estimated values of \(|F(x,y)|\) and \(F(x,y)\) for different values of \(x\) and \(y\) are shown in Table 1.

Table 1

From table 1, In first quadrant all vectors contains positive \(x\)-components and negative \(y\)-components. In second quadrant all vectors contains negative \(x\)-components and negative \(y\)-components. In third quadrant all vectors contains negative \(x\)-components and positive \(y\)-components. In fourth quadrant all vectors contains positive \(x\)-components and \(y\)-components.

The length of vector is reduces as distance of vector reach the origin.

Sketch the vector field \(F(x,y) = xi - yj\)

Sketch the vector field \(F(x,y) = xi - yj\) as shown in figure 1.

Figure 1

Thus, the vector field \(F(x,y) = xi - yj\) is sketched.

Draw the solid lines in all four quadrants

In Figure 1, draw the solid lines in all four quadrants along with direction of arrowed line in figure 2 as follows:

Figure 2

Thus, the several approximated flow lines on vector filed \(F(x,y) = xi - yj\) is drawn.

Draw the graph of \(y =  \pm \frac{n}{x}\)

Draw the graph of \(y = \pm \frac{n}{x}\) in figure 3 as follows:

Figure 3

Compare figure 2 and figure 3 as follows:

The approximated flow lines are appears as hyperbolas with shape similar to the graph of \(y = \pm \frac{n}{x}\).

If consider \( \pm n\) has the constant value \(C\) then, the flow line equation is \(y = \frac{C}{x}\).

Thus, equation of the flow line is \(y = \frac{C}{x}\).

Differentiate \(x\)  and \(y\)

(b) If \(x = x(t)\) and \(y = y(t)\) are the parametric equations of a flow line, then velocity vector of the flow line at the point \((x,y)\) is \({x^\prime }(t)i + {y^\prime }(t)j\).

Since, the velocity vector coincide with vectors in the vector field, the velocity vector is same as vector field as follows:

\({x^\prime }(t)i + {y^\prime }(t)j = xi - yj\) ……. (2)

Compare the components of \(i\) in equation (2) as follows:

\(\begin{aligned}{}{x^\prime }(t) &= x\\\frac{{dx}}{{dt}} &= x\end{aligned}\) ……. (3)

Compare the components of \(j\) in equation (2) as follows:

\(\begin{aligned}{}{y^\prime }(t)& = - y\\\frac{{dy}}{{dt}} &= - y\end{aligned}\) ……. (4)

Re-arrange the equation (3) as follows:

\(\frac{{dx}}{x} = dt\)

Take integration on both sides as follows:

\(x = \pm {e^{t + {C_1}}}{\rm{ }}\) ……. (5)

Consider \( \pm {e^{{C_1}}}\) has a constant and represented by \(A\) as follows:

Substitute \(A\) for \( \pm {e^{{C_1}}}\) in equation (5) as follows:

\(x = A{e^t}\)

Re-arrange the equation (4) as follows:

\(\frac{{dy}}{y} = - dt\)

Take integration on both sides as follows:

\(y = \pm {e^{ - t + {C_2}}}\) ……. (6)

Consider \( \pm {e^{{C_2}}}\) has a constant and represented by \(B\)as follows:

Substitute \(B\) for \( \pm {e^{{C_2}}}\) in equation (6) as follows:

\(y = B{e^{ - t}}\)

Find the value of \(xy\)

Find the value of \(xy\)as follows:

\(\begin{aligned}{}xy &= A{e^t}B{e^{ - t}}\\ &= AB{e^t}{e^{ - t}}\\& = AB{e^{t - t}}\\ &= AB{e^0}\end{aligned}\)

Re-arrange the equation as follows:

\(xy = AB\) ……. (7)

Since, \(A\) and \(B\) has constant values, the product of \(A\) and \(B\) is also a constant.

Modify equation (7) as follows:

\(xy = {\rm{ cons }}\tan t\) ……. (8)

Find the value of constant

Find the value of "constant" if flow line passes through \((1,1)\) as follows:

Substitute \(1\) for \(x\) and \(1\) for \(y\) in equation (8) as follows:

\(\begin{aligned}{}(1)\;(1) &= {\rm{cons }}\tan t\\1 &= {\rm{ cons }}\tan t\end{aligned}\)

Substitute \(1\) for constant in equation (8) as follows:

\(\begin{aligned}{}xy &= 1\\y &= \frac{1}{x}\end{aligned}\)

The point \((1,1)\) means \(x\) and \(y\) coordinates are located at first quadrant, which means that \(x > 0\).

Therefore, equation of the flow line present at \(x > 0\) is \(y = \frac{1}{x}\).

Thus, the flow line that passes through the point \((1,1)\) is \(y = \frac{1}{x},x > 0\).