Line Integral (4-4-2020)

Calculus Level pending

Consider a vector field $\vec{F} = (F_x, F_y, F_z) = (y, z, x)$ . Determine the line integral of the vector field over a straight-line path from $(x_1, y_1, z_1) = (1,1,1)$ to $(x_2, y_2, z_2) = (2,3,4)$ .

$\large{\int_C \vec{F} \cdot \vec{d \ell}}$

The answer is 11.5.

1 solution

Karan Chatrath
Apr 5, 2020

Unlike the previous problem, this one does not lead to an exact differential:

The equation of the line joining the given points is:

$\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{3} = t$

Here, $t$ is a parameter. This implies that the line can be expressed parametrically as follows:

$x = t+1 \implies dx = dt$ $y = 2t+1 \implies dy = 2dt$ $z = 3t+1 \implies dz = 3dt$

Consider the given vector field:

$\vec{F} = y \ \hat{i} + z \ \hat{j} + x \ \hat{k}$ $d\vec{l} = dx \ \hat{i} + dy \ \hat{j} + dz \ \hat{k}$ $\implies \vec{F} \cdot d\vec{l} = y \ dx + z \ dy + x \ dz$

Re-writing and simplifying the above dot product in terms of $t$ and realising that $t$ varies from $0$ to $1$ as one moves from the start to endpoint leads to the required integral which is:

$\boxed{\int_{0}^{1} \left(11t + 6\right)dt = \frac{23}{2}}$

Hello! I am struggling with your quantum well problem. I obtain the minimum energy value as 13.1542857. Is this a correct value?Moreover, the solution to the ODE I obtain is:

For negative x: $\psi(x) = A \cos\left(x\sqrt{E-1}\right) + B \sin\left(x\sqrt{E-1}\right)$

For positive x: $\psi(x) = C \cos\left(x\sqrt{E-5}\right) + D \sin\left(x\sqrt{E-5}\right)$

Applying boundary conditions gives:

$A \cos\left(\sqrt{E-1}\right) - B \sin\left(\sqrt{E-1}\right)=0$ $C \cos\left(\sqrt{E-5}\right) + D \sin\left(\sqrt{E-5}\right)=0$

At $x=0$ the wave function must be continuous and differentiable which leads to two more equations. Thus, 4 linear equations for $A$ , $B$ , $C$ and $D$ and the value of energy for which a system has non-trivial solutions has to be evaluated.

Additional two equations based on continuously differentiable conditions are: $A = C$ $B\sqrt{E-1}=D\sqrt{E-5}$

Hence my result for energy. What am I doing wrong?

Thanks in advance!

Karan Chatrath - 1 year, 2 months ago

$E \approx 5$ seems to work as well. Don't want to exhaust my last attempt, yet, though... Just want to be sure if my thinking is correct.

Karan Chatrath - 1 year, 2 months ago

Hello. Sorry for the delayed reply. I didn't get a notification, and just happened to stumble upon this note. I swept the value of $E$ , and for each $E$ , I started at $x = -1$ and integrated to find the value of the wave function at $x = 1$ . The lowest energy I found which satisfied the $\Psi(1) = 0$ requirement was $E \approx 5.078$

Steven Chase - 1 year, 2 months ago

And I get $13.15$ for the second lowest allowed energy

Steven Chase - 1 year, 2 months ago

Line Integral (4-4-2020)

The answer is 11.5.

1 solution

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